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#### Question 1:

The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:

 No. of calls (x): 0 1 2 3 4 5 6 No. of intervals (f): 15 24 29 49 54 43 39

Compute the mean number of calls per intervals.

Let the assume mean be.

We know that mean,

Here, we have .

Putting the values in the formula, we get

Hence, the mean number of calls per interval is 3.54.

#### Question 2:

Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below: Find the mean number of heads per toss

 No. of heads per toss (x): 0 1 2 3 4 5 No. of tosses (f): 38 144 342 287 164 25

Let the assume mean be.

We know that mean

Now, we have

Putting the values above in formula, we have

Hence, the mean number of heads per toss is 2.47.

#### Question 3:

The following table given the number of branches and number of plants in the garden of a school.

 No. of branches (x): 2 3 4 5 6 No. of plants (f): 49 43 57 38 13

Let the assume mean be.

We know that mean,

Now, we have

Putting the values in the above formula, we get

Hence, the mean number of branches per plant is approximately 3.62.

#### Question 4:

The following table gives the number of children of 150 families in a village

 No. of children (x): 0 1 2 3 4 5 No. of families (f): 10 21 55 42 15 7

Find the average number of children per family

Let the assume mean be.

We know that mean,

Now, we have

Putting the values above in formula, we get

Hence, the average number of children per family is 2.35.

#### Question 5:

The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:

 Marks (x): 15 20 22 24 25 30 33 38 45 Frequency (f): 5 8 11 20 23 18 13 3 1

Find the average number of marks.

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in the above formula, we get

Hence, the average number of marks is 26.08.

#### Question 6:

The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:

 No. of students absent (x): 0 1 2 3 4 5 6 7 No. of days (f): 1 4 10 50 34 15 4 2

Find the mean number of students absent per day.

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in the above formula,

Hence, the mean number of students absent per day is approximately 3.53.

#### Question 7:

In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:

 No. of misprints per page (x): 0 1 2 3 4 5 No. of pages (f): 154 95 36 9 5 1

Find the average number of misprints per page.

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in above formula, we have

Hence, the mean number of students absent per day is 0.73.

#### Question 8:

The following distribution gives the number of accidents met by 160 workers in a factory during a month.

 No. of accidents (x): 0 1 2 3 4 No. of workers (f): 70 52 34 3 1

Find the average number of accidents per worker.

Let the assume mean be.

We know that mean,

Now, we have .

Putting the values in the above formula, we get

Hence, the average number of accidents per worker is 0.83.

#### Question 9:

Find the mean from the following frequency distribution of marks at a test in statistics:

 Marks (x): 5 10 15 20 25 30 35 40 45 50 No. of students (f): 15 50 80 76 72 45 39 9 8 6

Let the assumed mean beand .

We know that mean,

Now, we have.

Putting the values in the above formula, we get

Hence, the mean marks is 22.075.

#### Question 1:

The following tables gives the distribution of total household expenditure (in rupees) of manual workers in a city.

 Expenditure (in rupees) (xi) Frequency (fi) Expenditure (in rupees) (xi) Frequency (fi) 100-150 150-200 200-250 250-300 24 40 33 28 300-350 350-400 400-450 450-500 30 22 16 7

Find the average expenditure (in rupees) per household

Given:

Let the assumed mean be A = 275 and h = 50.

We know that mean,

Now, we have.

Putting the values in the above formula, we get

Hence, the average expenditureper household is 266.25.

#### Question 2:

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of a plants in 20 houses in a locality. Find the mean number of plants per house.

 Number of plants: 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Number of houses: 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

We may prepare the table as shown:

We know that mean,

Hence, mean = 8.1

Direct method is easier than other methods. Therefore, we used direct method.

#### Question 3:

Consider the following distribution of daily wages of 50 workers of a factory.

 Daily wages (in Rs.) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Let the assumed mean be A = 120 and h = 20.

We know that mean,

Now, we have .

Putting the values in the above formula, we have

Hence, the mean daily wage of the workers is Rs 145.20.

#### Question 4:

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarised as follows. Find the mean heat beats per minute for these women, choosing a suitable method.

 Number of heart beats per minute: 65-58 68-71 71-74 74-77 77-80 80-83 83-86 Number of women: 2 4 3 8 7 4 2

Let the assumed mean be A = 75.5 and h = 3.

 No. of Heart Beats per Min: Mid Value $\left({x}_{i}\right)$ No. of Women $\left({f}_{i}\right)$: ${d}_{i}={x}_{i}-A$ ${u}_{i}=\frac{1}{h}×{d}_{i}$ ${f}_{i}{u}_{i}$ 65-68 66.5 2 $-$9 $-$3 $-$6 68-71 69.5 4 $-$6 $-$2 $-$8 71-74 72.5 3 $-$3 $-$1 $-$3 74-77 75.5 = A 8 0 0 0 77-80 78.5 7 3 1 7 80-83 81.5 4 6 2 8 83-86 84.5 2 9 3 6 $\sum {f}_{i}=30$ $\sum {f}_{i}{u}_{i}=4$

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean heart beats per minute for women is 75.9.

#### Question 5:

Find the mean of each of the following frequency distributions :

 Class interval: 3 - 5 5 - 7 7 - 9 9 - 11 11 - 13 Frequency: 5 10 10 7 8
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 Class Frequency(${f}_{i}$) Class Mark(${x}_{i}$) ${f}_{i}{x}_{i}$ 3 - 5 5 4 20 5 - 7 10 6 60 7 - 9 10 8 80 9 - 11 7 10 70 11 - 13 8 12 96 $\sum {f}_{i}=40$ $\sum {f}_{i}{x}_{i}$ = 326

Mean, $\overline{)x}=\frac{\sum {f}_{i}{x}_{i}}{\sum {f}_{i}}=\frac{326}{40}=8.15$
Thus, mean = 8.15

#### Question 6:

Find the mean of each of the following frequency distributions :

 Class interval: 50-70 70-90 90-110 110-130 130-150 150-170 Frequency: 18 12 13 27 8 22

Let the assumed mean be A = 100 and h = 20.

We know that mean

Now we have

Putting the values in the above formula, we get

Hence, the mean is 112.20.

#### Question 7:

Find the mean of each of the following frequency distributions :

 Class interval: 0−8 8−16 16−24 24−32 32−40 Frequency: 6 7 10 8 9

Let the assumed mean be A = 20 and h = 8.

We know that mean,

Now, we have.

Putting the values in the above formula, we get

Hence, the mean is 21.4.

#### Question 8:

Find the mean of each of the following frequency distributions :

 Class interval: 0−6 6−12 12−18 18−24 24−30 Frequency: 7 5 10 12 6

Let the assumed mean be A = 15 and h = 6.

We know that mean,

Now, we have.

Putting the values in the above formula, we get

Hence, the mean is 15.75.

#### Question 9:

Find the mean of each of the following frequency distributions :

 Class interval: 0−10 10−20 20−30 30−40 40−50 Frequency: 9 12 15 10 14

Let the assumed mean be A = 25 and h = 10.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 26.333.

#### Question 10:

Find the mean of each of the following frequency distributions :

 Class interval: 0−8 8−16 16−24 24−32 32−40 Frequency: 5 9 10 8 8

Let the assumed mean be A = 20 and h = 8.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 21.

#### Question 11:

Find the mean of each of the following frequency distributions :

 Class interval: 0−8 8−16 16−24 24−32 32−40 Frequency: 5 6 4 3 2

Let the assumed mean be A = 20 and h = 8.

We know that mean,

Now, we have.

Putting the values in the above formula, we get

Hence, the mean is 16.4.

#### Question 12:

Find the mean of each of the following frequency distributions :

 Class interval: 10−30 30−50 50−70 70−90 90−110 110−130 Frequency: 5 8 12 20 3 2

Let the assumed mean be A = 60 and h = 20.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 65.6.

#### Question 13:

Find the mean of each of the following frequency distributions :

 Class interval: 25−35 35−45 45−55 55−65 65−75 Frequency: 6 10 8 12 4

Let the assumed mean be A = 50 and h = 10.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 49.5.

#### Question 14:

Find the mean of each of the following frequency distributions :

 Classes: 25−29 30−34 35−39 40−44 45−49 50−54 55−59 Frequency: 14 22 16 6 5 3 4

The given series is an inclusive series. Firstly, make it an exclusive series.

 Class Interval Mid-Value(xi) Frequency(fi) ${u}_{i}=\frac{{d}_{i}}{h}=\frac{{d}_{i}}{5}$ ${f}_{i}{u}_{i}$ 24.5$-$29.5 27 14 −15 −3 −42 29.5$-$34.5 32 22 −10 −2 −44 34.5$-$39.5 37 16 −5 −1 −16 39.5$-$44.5 A = 42 6 0 0 0 44.5$-$49.5 47 5 5 1 5 49.5$-$54.5 52 3 10 2 6 54.5$-$59.5 57 4 15 3 12 $\sum _{}^{}{f}_{i}=70$ $\sum _{}^{}{f}_{i}{u}_{i}=-79$

Let the assumed mean be A = 42 and h = 5.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 36.357.

#### Question 15:

For the following distribution, calculate mean using all suitable methods:

 Size of item: 1−4 4−9 9−16 16−27 Frequency: 6 12 26 20

Direct Method:

We may prepare the table as shown:

We know that mean,

Hence, the mean is 13.25.

Short-Cut Method:

We may prepare the table as shown:

 Size of Item Mid value(xi) Frequency(fi) ${f}_{i}{d}_{i}$ 1–4 2.5 −10 6 −60 4–9 6.5 −6 12 −72 9–16 12.5 = A 0 26 0 16–27 21.5 9 20 180 N = $\sum _{}{f}_{i}=64$ $\sum _{}{f}_{i}{d}_{i}=48$

Let the assumed mean be A = 12.5.

We know that mean, $\overline{)X}=A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$=12.5+\frac{48}{64}\phantom{\rule{0ex}{0ex}}=12.5+0.75\phantom{\rule{0ex}{0ex}}=13.25$

Hence, the mean is 13.25.

Step-deviation method cannot be used to evaluate the mean of the distribution as the width of the class intervals are not equal. Here, h is not fixed.

#### Question 16:

The weekly observations on cost of living index in a certain city for the year 2004 − 2005 are given below. Compute the weekly cost of living index.

 Cost of living Index Number of Students Cost of living Index Number of Students 1400−1500 1500−1600 1600−1700 5 10 20 1700−1800 1800−1900 1900−2000 9 6 2

Let the assumed mean be A = 1650 and h = 100.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 1663.46.

#### Question 17:

The following table shows the marks scored by 140 students in an examination of a certain paper.

 Marks: 0−10 10−20 20−30 30−40 40−50 Number of students: 20 24 40 36 20

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

We may prepare the table as shown:

(i) Direct method
We know that mean,
$=\frac{3620}{140}\phantom{\rule{0ex}{0ex}}=25.857$

Hence, the mean is 25.857.

(ii) Short-cut method

Let the assumed mean A = 25.

We know that mean,

Hence, the mean is 25.857.

(iii) Step deviation method

Let the assumed mean A = 25 and h = 10.

We know that mean,

Hence, the mean is 25.857.

#### Question 18:

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2.

 Class: 0−20 20−40 40−60 60−80 80−100 100−120 Frequency: 5 f1 10 f2 7 9

It is given that mean = 62.8 and .

Let the assumed mean A = 50 and h = 20.

.....(1)

We know that mean,

Now, we have , , .

Putting the values in the above formula, we have

.....(2)

Putting the value of in (2), we get

Putting the value of in (1), we get

Hence, the missing frequency f1 = 8 and f2 = 12.

#### Question 19:

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find the missing frequency.

 Class: 11−13 13−15 15−17 17−19 19−21 21−23 23−25 Frequency: 7 6 9 13 − 5 4

Given: Mean = 18

Suppose the missing frequency is x.

Let the assumed mean A = 18 and h = 2.

We know that mean,

Now, we have,,.

Putting the values in the above formula, we have

$18=18+2\left(\frac{x-20}{x+44}\right)\phantom{\rule{0ex}{0ex}}⇒2\left(\frac{x-20}{x+44}\right)=0\phantom{\rule{0ex}{0ex}}⇒x-20=0\phantom{\rule{0ex}{0ex}}⇒x=20$

Thus, the missing frequency is 20.

#### Question 20:

If the mean of the following distribution is 27, find the value of p.

 Class: 0−10 10−20 20−30 30−40 40−50 Frequency: 8 p 12 13 10

Given: Mean = 27

Let the assumed mean A = 25 and h = 10.

We know that mean,

Now, we have,,

Putting the values in the above formula, we have

Thus, the value of p is 7.

#### Question 21:

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Number of mangoes: 50−52 53−55 56−58 59−61 62−64 Number of boxes: 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

The given series is an inclusive series. Firstly, make it an exclusive series.

 Class Interval Mid-Value(xi) Frequency(fi) ${u}_{i}=\frac{{d}_{i}}{h}=\frac{{d}_{i}}{3}$ ${f}_{i}{u}_{i}$ 49.5$-$52.5 51 15 −6 −2 −30 52.5$-$55.5 54 110 −3 −1 −110 55.5$-$58.5 57 135 0 0 0 58.5$-$61.5 60 115 3 1 115 61.5$-$64.5 63 25 6 2 50 $\sum _{}^{}{f}_{i}=400$ $\sum _{}^{}{f}_{i}{u}_{i}=25$

Let the assumed mean be A = 57 and h = 3.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is approximately 57.19.

#### Question 22:

The table below shows the daily expenditure on food of 25 household in a locality

 Daily expenditure (in Rs): 100−150 150−200 200−250 250−300 300−350 Number of households: 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Let the assumed mean a = 225 and h = 50.

 Daily expenditure (in Rs) fi xi di = xi − 225 fiui 100−150 4 125 − 100 − 2 − 8 150−200 5 175 − 50 − 1 − 5 200−250 12 225 0 0 0 250−300 2 275 50 1 2 300−350 2 325 100 2 4 Total $\sum _{}{f}_{i}=$25 $\sum _{}{f}_{i}{u}_{i}=$− 7

Now,

Therefore, mean daily expenditure on food is Rs 211.

#### Question 23:

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

 Concentration of SO2 (in ppm) Frequency 0.00−0.04 0.04−0.08 0.08−0.12 0.12−0.16 0.16−0.20 0.20−0.24 4 9 9 2 4 2

Find the mean of concentration of SO2 in the air.

Let the assumed mean A = 0.1 and h = 0.04.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

$=0.10+0.04\left[\frac{1}{30}×\left(-1\right)\right]\phantom{\rule{0ex}{0ex}}=0.10-\frac{0.04}{30}\phantom{\rule{0ex}{0ex}}=0.10-0.001\phantom{\rule{0ex}{0ex}}=0.099$

Hence, the mean concentration of SO2 in the air is 0.099 ppm.

#### Question 24:

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
(i)

 Number of days: 0 − 6 6 − 12 12 − 18 18 − 24 24 − 30 30 − 36 36 − 42 Number of students: 10 11 7 4 4 3 1

(ii)
 Number of days: 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40 Number of students: 11 10 7 4 4 3 1

(i) Let the assume mean = 17.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean number of days a student was absent is 12.475.

(ii)

 Number of Days Mid-Value (xi) Number of Students (fi) fixi 0−6 3 11 33 6−10 8 10 80 10−14 12 7 84 14−20 17 4 68 20−28 24 4 96 28−38 33 3 99 38−40 39 1 39 $\sum _{}{f}_{i}=40$ $\sum _{}{f}_{i}{x}_{i}=499$

We know

Mean$=\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}$
∴ Mean number of days a student was absent$=\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}=\frac{499}{40}=12.475$

Hence, the mean number of days a student was absent is 12.475.

#### Question 25:

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (in %): 45−55 55−65 65−75 75−85 85−95 Number of cities: 3 10 11 8 3

Let the assumed mean A = 70 and h = 10.

We know that mean,

Now, we have

Putting the values in the above formula, we have

Hence, the mean literacy rate is approximately 69.43%.

#### Question 26:

The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .

Age below (in years) :    30      40      50      60    70    80

Number of persons:       100     220   350  750   950   1000

Let the assumed mean A = 55 and h = 10.

 Marks Mid-Value Frequency ${\mathbit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{55}}{\mathbf{10}}$ ${\mathbit{f}}_{\mathbit{i}}{\mathbit{u}}_{\mathbit{i}}$ 20–30 25 100 −3 −300 30–40 35 120 −2 −240 40–50 45 130 −1 −130 50–60 55 400 0 0 60–70 65 200 1 200 70–80 75 50 2 100 N = 1000 $\sum _{}{f}_{i}{u}_{i}=-370$

We know that mean,
Now, we have

#### Question 27:

If the mean of the following frequency distribution is 18 , find the missing frequency .

Class interval :          11-13     13-15    15-17    17-19    19-21    21-23     23-25

Frequency:                  3              6            9          13        f              5            4

It is given that mean of the data is 18.
Let the assumed mean A = 18 and h = 2.

 Marks Mid-Value(xi) Frequency ${\mathbit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{18}}{\mathbf{2}}$ ${\mathbit{f}}_{\mathbit{i}}{\mathbit{u}}_{\mathbit{i}}$ 11–13 12 3 −3 −9 13–15 14 6 −2 −12 15–17 16 9 −1 −9 17–19 18 13 0 0 19–21 20 f 1 f 21–23 22 5 2 10 23–25 24 4 3 12 N = 40 + f $\sum _{}{f}_{i}{u}_{i}=-8+f$

We know that mean,
Now, we have
$18=18+2\left[\frac{1}{\left(40+f\right)}×\left(-8+f\right)\right]\phantom{\rule{0ex}{0ex}}⇒0=\frac{-16+2f}{40+f}\phantom{\rule{0ex}{0ex}}⇒-16+2f=0\phantom{\rule{0ex}{0ex}}⇒f=8$

#### Question 28:

Find the missing frequencies in the following distribution , if the sum of the frequencies is 120 and the mean is 50 .
Class:             0-20      20-40       40-60         60-80        80-100

Frequency:     17           f             32              f2             19

It is given that mean of the data is 50.
Let the assumed mean A = 50 and h = 10.

 Marks Mid-Value(xi) Frequency ${\mathbit{u}}_{\mathbit{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{50}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathbit{i}}{\mathbit{u}}_{\mathbit{i}}$ 0–20 10 17 −2 −34 20–40 30 f1 −1 −f1 40–60 50 32 0 0 60–80 70 f2 1 f2 80–100 90 19 2 38 N = 68 + f1 + f2 $\sum _{}{f}_{i}{u}_{i}=4-{f}_{1}+{f}_{2}$

We know that mean,
Now, we have N = 68 + f1 + f2, $\sum _{}{f}_{i}{u}_{i}=4-{f}_{1}+{f}_{2}$.

Now, N = 68 + f1 + f2 = 120

#### Question 29:

The daily income of a sample of 50 employees are tabulated as follows:
Income  (in â‚ą):        1-200          201-400          401-600        601-800

No.of employees :     14                15                    14                7

Find the mean daily income of employees.

Let the assumed mean A = 500.5 and h = 200.

 Marks Mid-Value(xi) Frequency ${\mathbit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{500}\mathbf{.}\mathbf{5}}{\mathbf{200}}$ ${\mathbit{f}}_{\mathbit{i}}{\mathbit{u}}_{\mathbit{i}}$ 1–200 100.5 14 −2 −28 201–400 300.5 15 −1 −15 401–600 500.5 14 0 0 601–800 700.5 7 1 7 N = 50 $\sum _{}{f}_{i}{u}_{i}=-36$

We know that mean,
Now, we have
$\overline{X}=500.5+200\left[\frac{1}{50}×\left(-36\right)\right]\phantom{\rule{0ex}{0ex}}=500.5-144\phantom{\rule{0ex}{0ex}}=356.5$

#### Question 30:

The marks obtained by 110 students in an examination are given below:

 Marks: 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 Frequency: 14 16 28 23 18 8 3

Find the mean marks of the students.

Let the assumed mean, A be 47.5.

Here, h = 5

 Marks Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{\mathbf{i}}\mathbf{=}{\mathbit{x}}_{\mathbf{i}}\mathbf{-}\mathbf{47}\mathbf{.}\mathbf{5}$ ${\mathbit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{47}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 30–35 32.5 14 −15 −3 −42 35–40 37.5 16 −10 −2 −32 40–45 42.5 28 −5 −1 −28 45–50 47.5 23 0 0 0 50–55 52.5 18 5 1 18 55–60 57.5 8 10 2 16 60–65 62.5 3 15 3 9 $\sum _{}{f}_{i}=110$ $\sum _{}{f}_{i}{u}_{i}=-59$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean marks of the students

$=47.5+5×\left(\frac{-59}{110}\right)\phantom{\rule{0ex}{0ex}}=47.5-2.68\phantom{\rule{0ex}{0ex}}=44.82$

Hence, the mean marks of the students is 44.82.

#### Question 1:

Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

First of all arranging the data in ascending order of magnitude, we have

Here,, which is an odd number

Therefore, median is the value of

#### Question 2:

The following is the distribution of height of students of a certain class in a certain city:

 Height (in cms): 160−162 163−165 166−168 169−171 172−174 No. of students: 15 118 142 127 18

Find the median height.

First we prepare the following cummulative table to compute the median.

Now,

Thus, the cumulative frequency just greater than 210 is 275 and the corresponding class is .

Therefore, is the median class.

We know that,

Hence, the median height is approximately 167.1 cm.

#### Question 3:

Following is the distribution of I.Q. of 100 students. Find the median I.Q.

 I.Q.: 55−64 65−74 75−84 85−94 95−104 105−114 115−124 125−134 135-144 No. of students: 1 2 9 22 33 22 8 2 1

Here, the frequency table is given in inclusive form. Transforming the given table into exclusive form and prepare the cumulative frequency table.

 IQ Frequency(fi) Cummulative Frequency(c.f.) 54.5−64.5 1 1 64.5−74.5 2 3 74.5−84.5 9 12 84.5−94.5 22 34 94.5−104.5 33 67 104.5−114.5 22 89 114.5−124.5 8 97 124.5−134.5 2 99 134.5−144.5 1 100 N = 100

Here,

So,

Thus, the cumulative frequency just greater than 50 is 67 and the corresponding class is 94.5−104.5.

Therefore, 94.5−104.5 is the median class.

Here, l = 94.5, f = 33, F = 34 and h = 9

We know that,

$=94.5+\left(\frac{50-34}{33}\right)×10\phantom{\rule{0ex}{0ex}}=94.5+\frac{160}{33}\phantom{\rule{0ex}{0ex}}=94.5+4.85\phantom{\rule{0ex}{0ex}}=99.35$

Hence, the median is 99.35.

#### Question 4:

Calculate the median salary of the following data giving salaries of 280 persons:

 Salary (in thousands) 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 No. of Persons 49 133 63 15 6 7 4 2 1

 Salary (In thousand Rs) Frequency CF 5 – 10 49 49 10 – 15 133 182 15 – 20 63 245 20 – 25 15 260 25 – 30 6 266 30 – 35 7 273 35 – 40 4 277 40 – 45 2 279 45 – 50 1 280

$\frac{N}{2}=\frac{280}{2}=140$
The cumulative frequency which is greater than and nearest to 140 is 182.
Median class = 10 – 15

We also have,

l (lower limit of median class) = 10

h (class size) = 5

n (number of observations) = 280

cf = (cumulative frequency of the class preceding the median class) = 49

f (frequency of median class) = 133

Median for grouped data is given by the formula :

Median =

where is frequency of median class and $cf$ is cumulative frequency of previous class.

Median = 10 +  = Rs 13.42.
Disclaimer: The obtained answer does not match the answer in the book.

#### Question 5:

Calculate the median from the following data:

 Marks below: 10 20 30 40 50 60 70 80 No. of students: 15 35 60 84 96 127 198 250

First we prepare the following cummulative table to compute the median.

Here,

So,

Thus, the cumulative frequency just greater than 125 is 127 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

Hence, the median is 59.35.

#### Question 6:

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

 Age in years: 0−10 10−20 20−30 30−40 40−50 No. of persons: 5 25 ? 18 7

Let the frequency of the class 20−30 be.It is given that median is 35 which lies in the class 20−30. So 20−30 is the median class.

Now, lower limit of median class

Height of the class

Frequency of median class

Cumulative frequency of preceding median class

Total frequency

Formula to be used to calculate median,

Where,

- Lower limit of median class

- Height of the class

- Frequency of median class

- Cumulative frequency of preceding median class

- Total frequency

Put the values in the above,

Hence, the required frequency is 25.

#### Question 7:

The following table gives the frequency distribution of married women by age at marriage:

 Age (in years) Frequency Age (in years) Frequency 15−19 20−24 25−29 30−34 35−39 53 140 98 32 12 40−44 45−49 50−54 55−59 60 and above 9 5 3 3 2

Calculate the median and interpret the results.

Here, the frequency table is given in inclusive form. So, we first transform it into exclusive form by subtracting and adding h/2 to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and upper limit of the previous class.

We have, N = 357
So, N/2 = 178.5

Thus, the cumulative frequency just greater than 178.5 is 193 and the corresponding class is 19.5−24.5.

Therefore, 19.5−24.5 is the median class.

Here, l = 19.5, f = 140, F = 193 and h = 5

We know that

$=19.5+\left(\frac{178.5-53}{140}\right)×5\phantom{\rule{0ex}{0ex}}=19.5+\frac{125.5}{140}×5\phantom{\rule{0ex}{0ex}}=19.5+\frac{125.5}{28}\phantom{\rule{0ex}{0ex}}=\frac{546+125.5}{28}\phantom{\rule{0ex}{0ex}}=23.98$

Hence, the median age 23.98 years.

Thus, nearly half the women were married between the age of 19.5 years and 24.5 years.

#### Question 8:

The following table gives the distribution of the life time of 400 neon lamps:

 Lite time: (in hours) Number of lamps 1500−2000 2000−2500 2500−3000 3000−3500 3500−4000 4000−4500 4500−5000 14 56 60 86 74 62 48

Find the median life.

We prepare the cumulative frequency table, as given below.

We have,

So,

Now, the cumulative frequency just greater than 200 is 216 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

Hence, the median life of the lamps is approximately 3406.98 hours.

#### Question 9:

The distribution below gives the weight of 30 students in a class. Find the median weight of students:

 Weight (in kg): 40−45 45−50 50−55 55−60 60−65 65−70 70−75 No. of students: 2 3 8 6 6 3 2

We prepare the cumulative frequency table, as given below.

We have,

So,

Now, the cumulative frequency just greater than 15 is 19 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

Hence, the median weight of students is 56.67 kg.

#### Question 10:

Find the missing frequencies and the median for the following distribution if the mean is 1.46.

 No. of accidents: 0 1 2 3 4 5 Total Frequency: 46 ? ? 25 10 5 200

(1) Let the missing frequencies be x and y.

Given:

.....(1)

We know that mean,

.....(2)

Solving (1) and (2), we get

Therefore,

Hence, the missing frequencies are 38 and 76.

(2) Calculation of median.

Now, we have.

So,

Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.

Hence, the median is 1.

#### Question 11:

An incomplete distribution is given below :

 Variable: 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency: 12 30 − 65 − 25 18

You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.

(i) Let the missing frequencies be x and y.

First we prepare the following cummulative table.

Here,

So,

It is given that the median is 46.

Therefore, is the median class.

We know that

Also,

Putting the value of x, we get

Hence, the missing frequencies are 34 and 46.

(ii)

We may prepare the table as shown.

We know that mean,

Now, we have.

Putting the values in the above formula, we have

Hence, the mean is 45.87.

#### Question 12:

If the median of the following frequency distribution is 28.5 find the missing frequencies:

 Class interval: 0−10 10−20 20−30 30−40 40−50 50−60 Total Frequency: 5 f1 20 15 f2 5 60

Given: Median = 28.5

We prepare the cumulative frequency table, as given below.

Now, we have

.....(1)

Also,

Since the median = 28.5 so the median class is 20−30.

Here,

We know that

Putting the value of in (1), we get

Hence, the missing frequencies are 7 and 8.

#### Question 13:

The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observation in the data:

 Class interval Frequency Class interval Frequency 0−100 100−200 200−300 300−400 400−500 2 5 f1 12 17 500−600 600−700 700−800 800−900 900−1000 20 f2 9 7 4

Given: Median = 525

We prepare the cumulative frequency table, as given below.

Now, we have

.....(1)

So,

Since median = 525 so the median class is .

Here,

We know that

Putting the value of in (1), we get

Hence, the missing frequencies are 9 and 15.

#### Question 14:

If the median of the following data is 32.5, find the missing frequencies.

 Class interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Total Frequency: f1 5 9 12 f2 3 2 40

Given: Median = 32.5

We prepare the cumulative frequency table, as given below.

Now, we have

.....(1)

Also,

Since median = 32.5 so the median class is .

Here,

We know that

Putting the value of in (1), we get

Hence, the missing frequencies are 3 and 6.

#### Question 15:

Compute the median for each of the following data:

(i)

 Marks No. of students Less than 10 Less than 30 Less than 50 Less than 70 Less than 90 Less than 110 Less than 130 Less than 150 0 10 25 43 65 87 96 100

(ii)
 Marks No. of students More than 150 More than 140 More than 130 More than 120 More than 110 More than 100 More than 90 More than 80 0 12 27 60 105 124 141 150

(i)

We prepare the cumulative frequency table, as given below.

Now, we have

So,

Now, the cumulative frequency just greater than 50 is 65 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

Hence, the median is 76.36.

Note: The first class in the table can be omitted also.

(ii)

We prepare the cumulative frequency table, as given below.

Now, we have

So.

Thus, the cumulative frequency just greater than 75 is 105 and the corresponding class is .

Therefore, is the median class.

(Because class interval given in descending order)

We know that

= 120 − 3.333

= 116.67   (approx)

Hence, the median is 116.67.

#### Question 16:

A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

 Height in cm Number of Girls Less than 140 Less than 145 Less than 150 Less than 155 Less than 160 Less than 165 4 11 29 40 46 51

Find the median height.

We prepare the cumulative frequency, table as given below.

Now, we have

So,

Now, the cumulative frequency just greater than 25.5 is 40 and the corresponding class is .

Therefore, is the median class.

We know that

= 150 − 1.59
= 148.41

Hence, the median height is 148.41 cm.

#### Question 17:

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

 Age in years Number of policy holders Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60 2 6 24 45 78 89 92 98 100

We prepare the cumulative frequency, table as given below.

Now, we have

So,

Now, the cumulative frequency just greater than 50 is 78 and the corresponding class is .

Therefore, is the median class.

Here,

We know that

Hence, the median age is 35.76 years.

#### Question 18:

The length of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

 Length (in mm): 118−126 127−135 136−144 145−153 154−162 163−171 172−180 No. of leaves 3 5 9 12 5 4 2

Find the mean length of leaf.

Calculation for mean.

 Length (in mm) Mid-Values(xi) Number of Leaves(fi ) fi xi 117.5–126.5 122 3 366 126.5–135.5 131 5 655 135.5–144.5 140 9 1260 144.5–153.5 149 12 1788 153.5–162.5 158 5 790 162.5–171.5 167 4 668 171.5–180.5 176 2 353 N = 40 $\sum _{}{f}_{i}{x}_{i}=$5880

Mean length of the leaf = $\frac{1}{N}\sum _{}^{}{f}_{i}{x}_{i}=\frac{1}{40}×5880=147$

Calculation for median.

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

 Length (in mm) Number of Leaves(fi ) Cumulative Frequency (c.f.) 117.5–126.5 3 3 126.5–135.5 5 8 135.5–144.5 9 17 144.5–153.5 12 29 153.5–162.5 5 34 162.5–171.5 4 38 171.5–180.5 2 40 N = 40

Now, we have

So,

Now, the cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5–153.5.

Therefore, 144.5–153.5 is the median class.

Here,

We know that

$=144.5+\left(\frac{20-17}{12}\right)×9\phantom{\rule{0ex}{0ex}}=144.5+\frac{27}{12}\phantom{\rule{0ex}{0ex}}=144.5+2.25\phantom{\rule{0ex}{0ex}}=146.75$

Hence, the median length of leaf is 146.75 mm.

Disclaimer: If the question asks for the mean length of the leaf, then the answer is 147 mm whereas if the question asks fro the median length of the leaf, then the answer is 146.75 mm, which is same as the answer given in the book.

#### Question 19:

An incomplete distribution is given as follows:

 Variable: 0−10 10−20 20−30 30−40 40−50 50 − 60 60 − 70 Frequency: 10 20 ? 40 ? 25 15

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Let the frequency of the class 20−30 be and that of class 40−50 be. The total frequency is 170. So,

So,           .....(1)

It is given that median is 35 which lies in the class 30-40. So 30-40 is the median class.

Now, lower limit of median class

Height of the class

Frequency of median class

Cumulative frequency of preceding median class

Total frequency

Formula to be used to calculate median,

Where,

- Lower limit of median class

- Height of the class

- Frequency of median class

- Cumulative frequency of preceding median class

- Total frequency

Put the values in the above,

Using equation (1), we have

Therefore,

#### Question 20:

The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.

Class interval:       0-6      6-12         12-18           18-24         24-30

Frequency:              4          x               5                 y                  1

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

 Class interval Frequency(fi ) Cumulative Frequency(c.f.) 0–6 4 4 6–12 x 4 + x 12–18 5 9 + x 18–24 y 9 + x + y 24–30 1 10 + x + y 10 + x + y = 20

Median = 14.4
It lies in the interval 12–18, so the median class is 12–18.
Now, we have

We know that

$14.4=12+\frac{6×\left(10-4-x\right)}{5}\phantom{\rule{0ex}{0ex}}⇒12=36-6x\phantom{\rule{0ex}{0ex}}⇒6x=24\phantom{\rule{0ex}{0ex}}⇒x=4$
Now,
10 + x + y = 20
$⇒x+y=10\phantom{\rule{0ex}{0ex}}⇒y=10-4=6$

#### Question 21:

The median of the following data is 50. Find the values of p and q  , if the sum of all  the frequencies is 90 .

Marks:        20-30        30-40       40-50        50-60        60-70         70-80       80-90

Frequency:    p               15            25            20               q                 8              10

The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have

 Class interval Frequency(fi ) Cumulative Frequency (c.f.) 20–30 p p 30–40 15 p + 15 40–50 25 p + 40 50–60 20 p + 60 60–70 q p + q + 60 70–80 8 p + q + 68 80–90 10 p + q + 78 78 + p + q = 90

Median = 50
It lies in the interval 50–60, so the median class is 50–60.
Now, we have

We know that

#### Question 1:

Find the mode of the following data:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

The frequency table for the given data

 Value x 3 4 5 6 7 8 9 Frequency f 4 2 5 2 2 1 2

We observe that the value 5 has the maximum frequency.

Hence, the mode of data is 5.

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

The frequency table for the given data

 Value x 3 4 5 6 7 8 9 Frequency f 5 2 4 2 2 1 2

We observe that the value 3 has the maximum frequency.

Hence, the mode of data is 3.

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

The frequency table for the given data

 Value x 8 15 18 19 20 24 25 26 Frequency f 1 4 1 1 1 2 1 1

We observe that the value 15 has the maximum frequency.

Hence, the mode of data is 15.

#### Question 2:

The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:

 Shirt Size: 37 38 39 40 41 42 43 44 Number of persons: 15 25 39 41 36 17 15 12

Find the modal shirt size worn by the group.

 Shirt Size 37 38 39 40 41 42 43 44 Number of persons 15 25 39 41 36 17 15 12

Here, shirt size 40 has the maximum number of persons.

Hence, the mode shirt size is 40.

#### Question 3:

Find the mode of the following distribution.

(i)

 Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency: 5 8 7 12 28 20 10 10

(ii)
 Class-interval: 10−15 15−20 20−25 25−30 30−35 35−40 Frequency: 30 45 75 35 25 15

(iii)
 Class-interval: 25−30 30−35 35−40 40−45 45−50 50−55 Frequency: 25 34 50 42 38 14

(iv)
 Class: 0−10 10−20 20−30 30−40 40−50 50−60 60–70 Frequency: 8 10 10 16 12 6 7

(i) Here, maximum frequency is 28 so the modal class is 40−50.

Therefore,

(ii) Here, maximum frequency is 75 so the modal class is 20−25.

Therefore,

(iii) Here, maximum frequency is 50 so the modal class is 35−40.

Therefore,

(iv)
Here, maximum frequency is 16 so the modal class is 30−40.

Therefore,

Lower limit of the modal class, l = 30

Width of the modal class, h = 10

Frequency of the modal class, f = 16

Frequency of the class preceding the modal class,  f1 = 10

Frequency of the class following the modal class,  f2 = 12

∴ Mode of the distribution

$=l+\left(\frac{f-{f}_{1}}{2f-{f}_{1}-{f}_{2}}\right)×h\phantom{\rule{0ex}{0ex}}=30+\left(\frac{16-10}{2×16-10-12}\right)×10\phantom{\rule{0ex}{0ex}}=30+\frac{6}{10}×10\phantom{\rule{0ex}{0ex}}=36$

Hence, the mode of the distribution is 36.

#### Question 4:

Compare the modal ages of two groups of students appearing for an entrance test:

 Age (in years): 16−18 18−20 20−22 22−24 24−26 Group A: 50 78 46 28 23 Group B: 54 89 40 25 17

 Age (in years) Group ‘A’ Group ‘B’ 16–18 50 54 18–20 78 89 20–22 46 40 22–24 28 25 24–25 23 17

For group “A”

The maximum frequency is 78 so the modal class is 18–20.

Therefore,

For group “B”

The maximum frequency 89 so modal class 18–20.

Therefore,

Thus, the modal age of group A is 18.93 years whereas the modal age of group B is 18.83 years.

#### Question 5:

The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.

 Marks: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 80−90 90−100 Frequency: 3 5 16 12 13 20 5 4 1 1

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 80−90 90−100 Frequency 3 5 16 12 13 20 5 4 1 1

Here, the maximum frequency is 20 so the modal class is 50−60.

Therefore,

Now,

Thus, the mode of the marks obtained by the students in science is 53.17.

#### Question 6:

The following is the distribution of height of students of a certain class in a certain city:

 Height (in cms): 160−162 163−165 166−168 169−171 172−174 Number of students: 15 118 142 127 18

Find the average height of maximum number of students.

The given data is an inclusive series. So, firstly convert it into an exclusive series as given below.

 Height (in cm) 159.5−162.5 162.5−165.5 165.5−168.5 168.5−171.5 171.5−174.5 Number of students 15 118 142 127 18

Here, the maximum frequency is 142 so the modal class is 165.5−168.5.

Therefore,

Now,

= 165.5 + 1.85
= 167.35

Thus, the average height of maximum number of students is 167.35 cm.

#### Question 7:

The following table show the ages of the patients admitted in a hospital during a year:

 Age (in years): 5−15 15−25 25−35 35−45 45−55 55−65 Number of patients: 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

 Age (in years) 5−15 15−25 25−35 35−45 45−55 55−65 Number of patients 6 11 21 23 14 5

Here, the maximum frequency is 23 so the modal class is 35−45.

Therefore,

Thus, the mode of the ages of the patients is 36.8 years.

Calculation for mean.

 Age (in years) Mid-Values(x) Number of patients(f) fx 5−15 10 6 60 15−25 20 11 220 25−35 30 21 630 35−45 40 23 920 45−55 50 14 700 55−65 60 5 300

Mean = $\frac{\sum _{}fx}{\sum _{}f}=\frac{2830}{80}=35.37$

Thus, the mean age of the patients is 35.37 years.

The mean age of the patients is less than the modal age of the patients.

#### Question 8:

The following data is gives the information on the observed lifetimes (in hours) of 225 electrical components:

 Lifetimes (in hours) 0−20 20−40 40−60 60−80 80−100 100−120 No. of components 10 35 52 61 38 29

Determine the modal lifetimes of the components.

 Lifetimes (in hours) 0−20 20−40 40−60 60−80 80−100 100−120 No. of components 10 35 52 61 38 29

Here, the maximum frequency of electrical components is 61 so the modal class is 60−80.

Therefore,

Thus, the modal lifetimes of the components is 65.625 hours.

#### Question 9:

The following table gives the daily income of 50 workers of  a factory :

Daily income (in â‚ą)          100-120        120-140       140-160        160-180       180-200

Number of workers :             12              14                  8                    6                  10

Find the mean, mode and median  of the above data .

Let the assumed mean a be 150.

The table for the given data can be drawn as

 Class Interval Number of Workers(fi) Classmark(xi) di = xi − 150 fidi Cumulative Frequency(c.f.) 100−120 12 110 −40 −480 12 120−140 14 130 −20 −280 26 140−160 8 150 0 0 34 160−180 6 170 20 120 40 180−200 10 190 40 400 50 Total 50 −240

Mean is given by .

Thus, the mean of the given data is 145.2.

It can be seen in the data table that the maximum frequency is 14. The class corresponding to this frequency is 120−140.

∴ Modal class = 120 − 140

Lower limit of modal class (l) = 120

Class size (h) = 140 − 120 = 20

Frequency of modal class (f1) = 14

Frequency of class preceding the modal class (f1) = 12

Frequency of class succeeding the modal class (f1) = 8

Mode is given by

Thus, the mode of the given data is 125.

Here, number of observations (n) = 50

This observation lies in class interval 120−140.

Therefore, the median class is 120−140.

Lower limit of median class (l) = 120

Cumulative frequency of class preceding the median class(c.f.) = 12

Frequency of median class(f) = 14

Median of the data is given by

Thus, the median of the given data is 138.57.

#### Question 10:

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

 Number of students per Teacher Number of States/U.T Number of students per Teacher Number of State/ U.T 15−20 20−25 25−30 30−35 3 8 9 10 35−40 40−45 45−50 50−55 3 0 0 2

 Number of students per teacher Number of states/U.T. (fi) xi f i xi 15−20 3 17.5 52.5 20−25 8 22.5 180.0 25−30 9 27.5 247.5 30−35 10 32.5 325.0 35−40 3 37.5 112.5 40−45 0 42.5 0 45−50 0 47.5 0 50−55 2 52.5 105.0

Here, the maximum frequency is 10 so the modal class is 30−35.

Therefore,

Thus, the mode of the data is 30.6.

Now,
Mean of the data = $\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}=\frac{1022.5}{35}=29.2$
Thus, the mean of the data is 29.2.

The mode of the number of students per teacher in the states is more than the mean of the number of students per teacher in the states.

#### Question 11:

Find the mean, median and mode of the following data:

 Classes: 0−50 50−100 100−150 150−200 200−250 250−300 300−350 Frequency: 2 3 5 6 5 3 1

 Classes Frequency (fi) xi fi xi C.f. 0−50 2 25 50 2 50−100 3 75 225 5 100−150 5 125 625 10 150−200 6 175 1050 16 200−250 5 225 1125 21 250−300 3 275 825 24 300−350 1 325 325 25 $\sum _{}{f}_{i}{x}_{i}=4225$

Here, the maximum frequency is 6 so the modal class 150−200.

Therefore,

Thus, the mean of the data is 169.

Thus, the median of the data is 170.83.

= 175

Thus, the mode of the data is 175.

#### Question 12:

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data:

 Number of cars: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency: 7 14 13 12 20 11 15 8

The given data is shown below.

 Number of cars Frequency 0−10 7 10−20 14 20−30 13 30−40 12 40−50 20 50−60 11 60−70 15 70−80 8

Here, the maximum frequency is 20 so the modal class is 40−50.

Therefore,

$\therefore \mathrm{Mode}=l+\frac{f-{f}_{1}}{2f-{f}_{1}-{f}_{2}}×h\phantom{\rule{0ex}{0ex}}=40+\frac{20-12}{2×20-12-11}×10\phantom{\rule{0ex}{0ex}}=40+\frac{8}{17}×10\phantom{\rule{0ex}{0ex}}=40+4.7\phantom{\rule{0ex}{0ex}}=44.7$

Thus, the mode of the data is 44.7.

#### Question 13:

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption: (in units) 65−85 85−105 105−125 125−145 145−165 165−185 185−205 No. of consumers: 4 5 13 20 14 8 4

The given data is shown below.

 Monthly Consumption (in units) No. of consumers (f i) xi fi xi C.f. 65−85 4 75 300 4 85−105 5 95 475 9 105−125 13 115 1495 22 125−145 20 135 2700 42 145−165 14 155 2170 56 165−185 8 175 1400 64 185−205 4 195 780 68

Here, the maximum frequency is 20 so the modal class is 125−145.

Therefore,

Thus, the mode of the monthly consumption of electricity is 135.76 units.

Mean = = 137.05

Thus, the mean of the monthly consumption of electricity is 137.05 units.

Here,

Total number of consumers, N = 68 (even)

Then, $\frac{N}{2}=34$

∴ Median

= 137

Thus, the median of the monthly consumption of electricity is 137 units.

#### Question 14:

100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

 Number of letters: 1−4 4−7 7−10 10−13 13−16 16−19 Number of surnames: 6 30 40 16 4 4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, fund the modal size of the surnames.

Consider the following table.

 Number of letters No. of surname xi fi xi C.f. 1−4 6 2.5 15 6 4−7 30 5.5 165 36 7−10 40 8.5 340 76 10−13 16 11.5 184 92 13−16 4 14.5 58 96 16−19 4 17.5 70 100

Here, the maximum frequency is 40 so the modal class is 7−10.

Therefore,

Thus, the modal sizes of the surnames is 7.88.

Thus, the mean number of letters in the surnames is 8.32.

Median

Thus, the median number of letters in the surnames is 8.05.

#### Question 15:

Find the mean, median and mode of the following data:

 Classes: 0−20 20−40 40−60 60−80 80−100 100−120 120−140 Frequency: 6 8 10 12 6 5 3

Consider the following data.

 Class Frequency (fi) xi fi xi C.f. 0−20 6 10 60 6 20−40 8 30 240 14 40−60 10 50 500 24 60−80 12 70 840 36 80−100 6 90 540 42 100−120 5 110 550 47 120−140 3 130 390 50

Here, the maximum frequency is 12 so the modal class is 60−80.

Therefore,

Thus, the median of the data is 61.66.

Thus, the mean of the data is 62.4.

Mode

Thus, the mode of the data is 65.

#### Question 16:

The following data gives the distribution of total monthly household expenditure of 200 families of a villages. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

 Expenditure (in Rs.) Frequency Expenditure (in Rs.) Frequency 1000−1500 1500−2000 2000−2500 2500−3000 24 40 33 28 3000−3500 3500−4000 4000−4500 4500−5000 30 22 16 7

 Expenditure Frequency (f i) xi f ixi 1000−1500 24 1250 30000 1500−2000 40 1750 70000 2000−2500 33 2250 74250 2500−3000 28 2750 77000 3000−3500 30 3250 97500 3500−4000 22 3750 82500 4000−4500 16 4250 68000 4500−5000 7 4750 33250

Here, the maximum frequency is 40 so the modal class is 1500−2000.

Therefore,

= 1500 + 347.83
= Rs 1847.83

Thus, the modal monthly expenditure of the families is Rs 1847.83.

Now,

Mean monthly expenditure of the families

Thus, the mean monthly expenditure of the families is Rs 2662.50.

#### Question 17:

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

 Runs scored Number of batsman Runs scored Number of Batsman 3000−4000 4000−5000 5000−6000 6000−7000 4 18 9 7 7000−8000 8000−9000 9000−10000 10000−11000 6 3 1 1

Find the mode of data.

The given data is shown below.

 Runs scored Number of batsmen 3000−4000 4 4000−5000 18 5000−6000 9 6000−7000 7 7000−8000 6 8000−9000 3 9000−10000 1 10000−11000 1

Here, the maximum frequency is 18 so the modal class is 4000-5000.

Therefore,

$\therefore \mathrm{Mode}=l+\frac{f-{f}_{1}}{2f-{f}_{1}-{f}_{2}}×h\phantom{\rule{0ex}{0ex}}=4000+\frac{18-4}{2×18-4-9}×1000\phantom{\rule{0ex}{0ex}}=4000+\frac{14}{23}×1000\phantom{\rule{0ex}{0ex}}=4000+608.7\phantom{\rule{0ex}{0ex}}=4608.7$

Thus, the mode of the data (or runs scored) is 4608.7.

#### Question 18:

The frequency distribution table of agriculture holdings in a village is given below:

Area of land (in hectares):       1-3      3-5     5-7     7-9     9-11     11-13

Number of families:                 20       45      80      55        40         12

Find the modal agricultural holdings of the village .

The maximum class frequency is 80. The class corresponding to this frequency is 5–7.
So, the modal class is 5–7.
l (the lower limit of modal class) = 5
f1 (frequency of the modal class) = 80
f0 (frequency of the class preceding the modal class) = 45
f2 (frequency of the class succeeding the modal class) = 55
h (class size) = 2
$\mathrm{Mode}=l+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h\phantom{\rule{0ex}{0ex}}=5+\left(\frac{80-45}{2×80-45-55}\right)×2\phantom{\rule{0ex}{0ex}}=5+\frac{35}{60}×2\phantom{\rule{0ex}{0ex}}=5+\frac{35}{30}\phantom{\rule{0ex}{0ex}}=6.2$

Hence, the modal agricultural holdings of the village is 6.2 hectares.

#### Question 19:

The monthly income of 100 families are given as below :
Income in ( in  â‚ą)                                     Number of families

0-5000                                                           8
5000-10000                                                        26
10000-15000                                                      41
15000-20000                                                      16
20000-25000                                                      3
25000-30000                                                      3
30000-35000                                                      2
35000-40000                                                      1
Calculate the modal income.

The maximum class frequency is 41. The class corresponding to this frequency is 10000–15000.
So, the modal class is 10000–15000.
l (the lower limit of modal class) = 10000
f1 (frequency of the modal class) = 41
f0 (frequency of the class preceding the modal class) = 26
f2 (frequency of the class succeeding the modal class) = 16
h (class size) = 5000
$\mathrm{Mode}=l+\left(\frac{{f}_{1}-{f}_{0}}{2{f}_{1}-{f}_{0}-{f}_{2}}\right)×h\phantom{\rule{0ex}{0ex}}=10000+\left(\frac{41-26}{2×41-26-16}\right)×5000\phantom{\rule{0ex}{0ex}}=10000+\frac{15}{40}×5000\phantom{\rule{0ex}{0ex}}=10000+1875\phantom{\rule{0ex}{0ex}}=11875$
Thus, the modal income is â‚ą11875.

#### Question 1:

Calculate the mean for the following distribution :

 x: 5 6 7 8 9 f: 4 8 14 11 3

Given:

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

Hence, mean

#### Question 2:

Find the mean of the following data:

 x: 19 21 23 25 27 29 31 f: 13 15 16 18 16 15 13

Given:

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

Hence, mean

#### Question 3:

If the mean of the following data is 20.6. Find the value of p.

 x: 10 15 p 25 35 f: 3 10 25 7 5

Given:

Also, mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, p

#### Question 4:

If the mean of the following data is 15, find p.

 x: 5 10 15 20 25 f: 6 p 6 10 5

Given:

Also, mean = 15

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method

Hence, p

#### Question 5:

Find the value of p for the following distribution whose mean is 16.6.

 x: 8 12 15 p 20 25 30 f: 12 16 20 24 16 8 4

Given:

Mean = 16.6

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method

Hence, p

#### Question 6:

Find the missing value of p for the following distribution whose mean is 12.58

 x: 5 8 10 12 p 20 25 f: 2 5 8 22 7 4 2

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method

Hence, p

#### Question 7:

Find the missing frequency (p) for the following distribution whose mean is 7.68.

 x: 3 5 7 9 11 13 f: 6 8 15 p 8 4

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

$303+9p=314.88+7.68p\phantom{\rule{0ex}{0ex}}9p-7.68p=314.88-303$

Hence, p

#### Question 8:

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

 Age (in years): 15 16 17 18 19 20 No. of students: 3 8 10 10 5 4

Given:

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

Hence, the mean age of the students

#### Question 9:

Candidate of four schools appear in a mathematics test. The data were as follows:

 Schools No. of Candidates Average Score I II III IV 60 48 Not available 40 75 80 55 50

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.

Given:

Mean score of the candidates = 66

Let the number of candidates that appeared from school III be x.

First of all prepare the frequency table in such a way that its first column consists of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,
$66=\frac{10340+55x}{260}$

By using cross multiplication method,

Hence, the number of candidates that appeared from school III is 124.

#### Question 10:

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

 No. of heads per toss No. of tosses 0 1 2 3 4 5 38 144 342 287 164 25 Total 1000

Given:

First of all prepare the frequency table in such a way that its first column consist of the numnber of heads per tosses and the second column the corresponding number of tosses .

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denote by and in the third column to obtain.

We know that mean,
$=\frac{2470}{1000}\phantom{\rule{0ex}{0ex}}=2.47$

Hence, the mean number of heads per toss is 2.47.

#### Question 11:

The arithmetic mean of the following data is 14. Find the value of k.

 xi: 5 10 15 20 25 fi: 7 k 8 4 5

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, k = 6

#### Question 12:

The arithmetic mean of the following data is 25, find the value of k.

 xi: 5 15 25 35 45 fi: 3 k 3 6 2

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, k = 4

#### Question 13:

If the mean of the following data is 18.75. Find the value of p.

 xi: 10 15 p 25 30 fi: 5 10 7 8 2

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,

By using cross multiplication method,

Hence, p = 20

#### Question 14:

Find the value of p, if the mean of the following distribution is 20.

 x: 15 17 19 20+p 23 f: 2 3 4 5p 6

Given:

Mean

First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

We know that mean,
$20=\frac{295+5p\left(20+p\right)}{15+5p}$

By using cross multiplication method,

$5{p}^{2}+100p+295=300+100p\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=300-295=5\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=1\phantom{\rule{0ex}{0ex}}⇒p=1$

Hence, p

#### Question 15:

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

 x: 10 30 50 70 90 f: 17 f1 32 f2 19 Total 120

Given:

Mean

First of all prepare the frequency table in such a way that its first column consists of the values of the variate and the second column the corresponding frequencies.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.

Then, sum of all entries in the column second and denoted by and in the third column to obtain.

Now,

..... (1)

We know that mean,

By using cross multiplication method,

..... (2)

Putting the value of from equation (1) in (2), we get

Therefore,

Putting the value of in equation (1), we get

Hence,

#### Question 1:

Draw an ogive by less than method for the following data:

 No. of rooms: 1 2 3 4 5 6 7 8 9 10 No. of houses: 4 9 22 28 24 12 8 6 5 2

Firstly, we prepare the cumulative frequency table by less than method.

 No. of rooms No. of houses (f) Cumulative frequency 1 4 4 2 9 13 3 22 35 4 28 63 5 24 87 6 12 99 7 8 107 8 6 113 9 5 118 10 2 120

Now, plot the less than ogive using the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118) and (10, 120).

#### Question 2:

The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

 Marks No. of student Marks No. of students 600−640 640−680 680−720 720−760 16 45 156 284 760−800 800−840 840−880 172 59 18

Prepare a cumulative frequency table by less than method and draw an ogive.

 Marks No. of students Marks less then Cumulative frequency Suitable Points 600-640 16 640 16 (640,16) 640-680 45 680 61 (680,61) 680-720 156 720 217 ( 720, 217) 720-760 284 470 501 (760, 501) 760-800 172 800 673 ( 800, 673) 800-840 59 840 732 ( 840,732) 840-880 18 880 750 ( 880,750)

Now, we draw the less than ogive with suitable points.

#### Question 3:

Draw an ogive to represent the following frequency distribution:

 Class-interval: 0−4 5−9 10−14 15−19 20−24 Frequency: 2 6 10 5 3

Firstly, prepare the cumulative frequency table.

 Class Interval No. of students Less than Cumulative frequency Suitable points 0-4 2 4 2 (4, 2) 5-9 6 9 8 (9, 8) 10-14 10 14 18 (14, 18) 15-19 5 19 23 (19, 23) 20-24 3 24 26 (24, 26)

Now, plot the less than ogive using the suitable points.

#### Question 4:

The monthly profits (in Rs.) of 100 shops are distributed as follows:

 Profits per shop: 0−50 50−100 100−150 150−200 200−250 250−300 No. of shops: 12 18 27 20 17 6

Draw the frequency polygon for it.

Firstly, we make a cumulative frequency table.

 Profit per shop No. of shop More than profit Cumulative frequency Suitable points 0-50 12 0 100 (0, 100) 50-100 18 50 88 (50, 88) 100-150 27 100 70 (100, 70) 150-200 20 150 43 (150, 43) 200-250 17 200 23 (200, 23) 250-300 6 250 6 (250, 6)

Now, plot the frequency polygon (or more than ogive) using suitable points.

#### Question 5:

The following distribution gives the daily income of 50 workers of a factory:

 Daily income in (Rs): 100−120 120−140 140−160 160−180 180−200 Number of workers: 12 14 8 6 10

Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.

 Daily income (in Rs.) No. of workers (f) Daily income (less than) Cumulative frequency Suitable points 100−120 12 120 12 (120, 12) 120−140 14 140 26 (140, 26) 140−160 8 160 34 (160, 34) 160−180 6 180 40 (180, 40) 180−200 10 200 50 (200, 50)

Now, plot the less than ogive with suitable points.

#### Question 6:

The following table gives production yield per hectare of wheat of 100 farms of a village:

 Production yield in kg per hectare: 50−55 55−60 60−65 65−70 70−75 75−80 Number of farms: 2 8 12 24 38 16

Draw 'less than' ogive and 'more than' ogive.

Prepare a table for less than type.

 Production yield No. of farms Production yield (less than) Cumulative frequency Suitable points 50−55 2 55 2 (55, 2) 55−60 8 60 10 (60, 10) 60−65 12 65 22 (65, 22) 65−70 24 70 46 (70, 46) 70−75 38 75 84 (75, 84) 75−80 16 80 100 (80, 100)

Now, plot the less than ogive using suitable points.

Again, prepare a table for more than type.

 Production yield No. of farms Production yield (more than) Cumulative frequency Suitable points 50-55 2 50 100 (50, 100) 55-60 8 55 98 (55, 98) 60-65 12 60 90 (60, 90) 65-70 24 65 78 (65, 78) 70-75 38 70 54 (70, 54) 75-80 16 75 16 (75, 16)

Now, plot the more than ogive with suitable points.

#### Question 7:

During the medical check up of 35 students of a class, their weights were recorded as follows:

 Weight (in kg) Number of students Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52 0 3 5 9 14 28 32 35

Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.

Prepare the table for cumulative frequency for less than type.

 Weight (in kg) No. of students Weight (less than) Cumulative frequency Suitable points 36–38 0 38 0 (38, 0) 38–40 3 40 3 (40, 3) 40–42 2 42 5 (42, 5) 42–44 4 44 9 (44, 9) 44–46 5 46 14 (46, 14) 46–48 14 48 28 (48, 28) 48–50 4 50 32 (50, 32) 50–52 3 52 35 (52, 35)

Now, draw the less than ogive using suitable points.

Here, =

From (0, 17.5) draw a line parallel to horizontal axis, which intersects the graph at the point M having x-coordinate as 46.5.

Therefore, the median is 46.5 kg.

Calculation for median using formula.

Now,

So, 46–48 is the median class.

Here,

Hence, both the methods give same result.

#### Question 8:

The annual rainfall record of a city for 66 days is given in the following table :
Rainfall (in cm ):        0-10      10-20    20-30      30-40      40-50     50-60

Number of days :        22            10        8              15          5              6

Calculate the median rainfall using ogives of more than type and less than type.

Prepare a table for less than type.

 Rainfall (in cm) Number of Days Rainfall (Less Than) Cumulative Frequency Suitable Points 0−10 22 10 22 (10, 22) 10−20 10 20 32 (20, 32) 20−30 8 30 40 (30, 40) 30−40 15 40 55 (40, 55) 40−50 5 50 60 (50, 60) 50−60 6 60 66 (60, 66)

Now, plot the less than ogive using suitable points.

Here, N = 66.
$\therefore \frac{N}{2}=33$
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting the curve at Q. From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinate of R is 21.25.

Thus, the median rainfall is 21.25 cm.

Let us now prepare a table for more than type.

 Rainfall (in cm) Number of Days Rainfall (More Than) Cumulative Frequency Suitable Points 0−10 22 0 66 (0, 66) 10−20 10 10 44 (10, 44) 20−30 8 20 34 (20, 34) 30−40 15 30 26 (30, 26) 40−50 5 40 11 (40, 11) 50−60 6 50 6 (50, 6)

Now, plot the more than ogive with suitable points.

Here, N = 66.
$\therefore \frac{N}{2}=33$
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the y-axis. Let the point be P. From this point draw a line parallel to the x-axis cutting the curve at Q. From this point Q, draw a line parallel to the y-axis and meeting the x-axis at the point R. The x-coordinate of R is 21.25.

Thus, the median rainfall is 21.25 cm.

#### Question 9:

Change the following distribution to a 'More than type' distribution. Hence, draw the more than type' ogive for this distribution.

 Class – interval: 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 Frequency: 10 8 12 24 6 25 15

The cumulative frequency table of 'More than type' is given below.

 Class Interval Cumulative Frequency More than 20 100 More than 30 90 More than 40 82 More than 50 70 More than 60 46 More than 70 40 More than 80 15

Mark the lower class class limits along the x-axis and the cumulative frequencies along y-axis.

Plot the points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40) and (80, 15). Join the points to obtain the 'More than type' ogive.

#### Question 10:

The following distribution gives daily income of 50 workers of a factory:

 Daily income(in â‚ą): 200 – 220 220 – 240 240 – 260 260 – 280 280 – 300 Number of workers: 12 14 8 6 10

Convert the distribution above to a 'less than type' cumulative frequency distribution and draw its ogive.

The cumulative frequency table of 'Less than type' is given below.

 Daily income(in â‚ą) Cumulative Frequency Less than 220 12 Less than 240 26 Less than 260 34 Less than 280 40 Less than 300 50

Mark the upper class class limits along the x-axis and the cumulative frequencies along y-axis.

Plot the points (220, 12), (240, 26), (260, 34), (280, 40) and (300, 50). Join the points to obtain the 'Less than type' ogive.

#### Question 11:

The following table gives the height of trees:

 Height No. of trees Less than 7 Less than 14 Less than 21 Less than 28 Less than 35 Less than 42 Less than 49 Less than 56 26 57 92 134 216 287 341 360

Draw 'less than' ogive and 'more than' ogive.

Consider the following table.

 Height (less than) Height-class No. of trees Cumulative frequency Suitable points 7 0-7 26 26 (7, 26) 14 7-14 31 57 (14, 57) 21 14-21 35 92 (21, 92) 28 21-28 42 134 (28, 134) 35 28-35 82 216 (35, 216) 42 35-42 71 287 (42, 287) 49 42-49 54 341 (49, 341) 56 49-56 19 360 (56, 360)

Now, draw the less than ogive using suitable points.

Now, prepare the cumulative frequency table for more than series.

 Height No. of trees Height (more than) Cumulative frequency Suitable points 0-7 26 0 360 (0, 360) 7-14 31 7 334 (7, 334) 14-21 35 14 303 (14, 303) 21-28 42 21 268 (21, 268) 28-35 82 28 226 (28, 226) 35-42 71 35 144 (35, 144) 42-49 54 42 73 (42, 73) 49-56 19 49 54 (49, 54)

Now, draw the more than ogive using suitable points.

#### Question 12:

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

 Profit (in lakhs in Rs) Number of shops (frequency) More than or equal to 5 More than or equal to 10 More than or equal to 15 More than or equal to 20 More than or equal to 25 More than or equal to 30 More than or equal to 35 30 28 16 14 10 7 3

Draw both ogives for the above data and hence obtain the median.

Firstly, we prepare the cumulative frequency table for less than type.

 Profit (In lakh in Rs.) No. of shops (frequency) Profit (less than) Cumulative frequency Suitable points 5-10 2 10 2 (10, 2) 10-15 12 15 14 (15, 14) 15-20 2 20 16 (20, 16) 20-25 4 25 20 (25, 20) 25-30 3 30 23 (30, 23) 30-35 4 35 27 (35, 27) 35-40 3 40 30 (40, 30)

Again, prepare the cumulative frequency table for more than type.

 Profit (In lakh in Rs.) No. of shops (frequency) Profit (more than) Cumulative frequency Suitable points 5-10 2 5 30 (5, 30) 10-15 12 10 28 (10, 28) 15-20 2 15 16 (15, 16) 20-25 4 20 14 (20, 14) 25-30 3 25 10 (25, 10) 30-35 4 30 7 (30, 7) 35-40 3 35 3 (35, 3)

Now, “more than ogive” and “less than ogive” can be drawn as follows:

The x-coordinate of the point of intersection of the “more-than ogive” and “less-than ogive” gives the median of the given distribution..

So, the corresponding median is Rs 17.5 lakh.

#### Question 1:

Which of the following is not a measure of central tendency?

(a) Mean
(b) Median
(c) Mode
(d) Standard deviation

Standard deviation is not a measure of central tendency.

Hence, correct option is (d).

#### Question 2:

The algebraic sum of the deviations of a frequency distribution from its mean is

(a) always positive
(b) always negative
(c) 0
(d) a non-zero number

The algebraic sum of the deviations of a frequency distribution from its mean is zero.

Hence, the correct option is (c).

#### Question 3:

The arithmetic mean of 1, 2, 3, ... , n is

(a) $\frac{n+1}{2}$
(b) $\frac{n-1}{2}$
(c) $\frac{n}{2}$
(d) $\frac{n}{2}+1$

Arithmetic mean of 1, 2, 3, ... , n

$=\frac{1+2+3+...+n}{n}\phantom{\rule{0ex}{0ex}}=\frac{\frac{n\left(n+1\right)}{2}}{n}\phantom{\rule{0ex}{0ex}}=\frac{n+1}{2}$

Hence, the correct option is (a).

#### Question 4:

For a frequency distribution, mean, median and mode are connected by the relation

(a) Mode = 3 Mean − 2 Median
(b) Mode = 2 Median − 3 Mean
(c) Mode = 3 Median − 2 Mean
(d) Mode = 3 Median + 2 Mean

The relation between mean, median and mode is

Mode = 3 Median − 2 Mean

Hence, the correct option is (c).

#### Question 5:

Which of the following cannot be determined graphically?

(a) Mean
(b) Median
(c) Mode
(d) None of these

‘Mean’ cannot be determined by graphically.

Hence, the correct option is (a).

#### Question 6:

The median of a given frequency distribution is found graphically with the help of

(a) Histogram
(b) Frequency curve
(c) Frequency polygon
(d) Ogive

The median of a given frequency distribution is found graphically with the help of ‘Ogive’.

Hence, the correct option is (d).

#### Question 7:

The mode of a frequency distribution can be determined graphically from

(a) Histogram
(b) Frequency polygon
(c) Ogive
(d) Frequency curve

The mode of frequency distribution can be determined graphically from “Histogram”.

Hence, the correct option is (a).

#### Question 8:

Mode is

(a) least frequency value
(b) middle most value
(c) most frequent value
(d) None of these

Mode is “Most frequent value”.

Hence, the correct option is (c).

#### Question 9:

The mean of n observation is $\overline{)\mathrm{X}}$. If the first item is increased by 1, second by 2 and so on, then the new mean is

(a) $\overline{)\mathrm{X}}+n$
(b) $\overline{)\mathrm{X}}+\frac{n}{2}$
(c) $\overline{)\mathrm{X}}+\frac{n+1}{2}$
(d) None of these

Let be the n observations.

Mean$=\overline{)X}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{n}$

$⇒{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)X}$

If the first item is increased by 1, second by 2 and so on.

Then, the new observations are .

New mean =
$=\frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}+\left(1+2+3+...+n\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)X}+\frac{n\left(n+1\right)}{2}}{n}\phantom{\rule{0ex}{0ex}}=\overline{)X}+\frac{n+1}{2}$

Hence, the correct answer is (c).

#### Question 10:

One of the methods of determining mode is

(a) Mode = 2 Median − 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median − 2 Mean
(d) Mode = 3 Median + 2 Mean

We have,

Mode = 3 Median − 2 Mean

Hence, the correct option is (c).

#### Question 11:

If the mean of the following distribution is 2.6, then the value of y is

 Variable (x): 1 2 3 4 5 Frequency 4 5 y 1 2

(a) 3
(b) 8
(c) 13
(d) 24

Consider the following table.

 Variable (x) f fx 1 4 4 2 5 10 3 y 3y 4 1 4 5 2 10

Now,

y = 8

Hence, the correct option is (b).

#### Question 12:

The relationship between mean, median and mode for a moderately skewed distribution is

(a) Mode = 2 Median − 3 Mean
(b) Mode = Median − 2 Mean
(c) Mode = 2 Median − Mean
(d) Mode = 3 Median −2 Mean

Mode = 3Median − 2Mean

Hence, the correct option is (d).

#### Question 13:

The mean of a discrete frequency distribution xi / fi, i = 1, 2, ......, n is given by

(a)
(b)
(c)
(d) $\frac{\sum _{i=1}^{n}{f}_{i}{x}_{i}}{\sum _{1=1}^{n}i}$

The mean of discrete frequency distribution is

Hence, the correct option is (a).

#### Question 14:

If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, the x =

(a) 1
(b) 2
(c) 6
(d) 4

The given observations are x, x + 3, x + 6, x + 9, and x + 12.

Now,

Hence, the correct option is (d).

#### Question 15:

If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =

(a) 27
(b) 25
(c) 28
(d) 30

The given observations are 24, 25, 26, x + 2, x + 3, 30, 31, 34.

Median = 27.5

Here, n = 8

Hence, the correct option is (b).

#### Question 16:

If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x =

(a) 15
(b) 16
(c) 17
(d) 18

The given observations arranged in ascending order are

Hence, the correct option is (c).

#### Question 17:

The median of first 10 prime numbers is

(a) 11
(b) 12
(c) 13
(d) 14

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

n = 10 (even)

Hence, the correct option is (b).

#### Question 18:

If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =

(a) 44
(b) 45
(c) 46
(d) 48

 Value 34 43 48 60 64 x Frequency 1 2 2 1 1 1

It is given that the mode of the given date is 43. So, it is the value with the maximum frequency.

Now, this is possible only when x = 43. In this case, the frequency of the observation 43 would be 3.

Hence,

x + 3 = 46

Hence, the correct option is (c).

#### Question 19:

If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =

(a) 15
(b) 16
(c) 17
(d) 19

 Value 14 15 16 17 19 x Frequency 1 2 2 2 1 1

It is given that the mode of the data is 15. So, it is the observation with the maximum frequency.

This is possible only when x = 15. In this case, the frequency of 15 would be 3.

Hence, the correct answer is (a).

#### Question 20:

The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m − 1 and median q. Then, p + q =

(a) 4
(b) 5
(c) 6
(d) 7

Consider the numbers 3, 2, 2, 4, 3, 3, p.

Mean = $\frac{3+2+3+4+3+3+p}{7}$

$⇒7×\left(4-1\right)=17+p\phantom{\rule{0ex}{0ex}}⇒21=17+p\phantom{\rule{0ex}{0ex}}⇒p=4$

Arranging the numbers 3, 2, 2, 4, 3, 3, 4 in ascending order, we have

2, 2, 3, 3, 3, 4, 4

q = 3

So,

Hence, the correct option is (d).

#### Question 21:

If the mean of frequency distribution is 8.1 and Σfixi = 132 + 5k, Σfi = 20, then k =

(a) 3
(b) 4
(c) 5
(d) 6

Given:

Σfixi = 132 + 5k, Σfi = 20 and mean = 8.1.

Then,

Hence, the correct option is (d).

#### Question 22:

If the mean of 6, 7, x, 8, y, 14 is 9, then

(a) x + y = 21
(b) x + y = 19
(c) xy = 19
(d) xy = 21

The given observations are 6, 7, x, 8, y, 14.

Mean = 9          (Given)

$⇒\frac{6+7+x+8+y+14}{6}=9\phantom{\rule{0ex}{0ex}}⇒35+x+y=54\phantom{\rule{0ex}{0ex}}⇒x+y=54-35=19$

Hence, the correct option is (b).

#### Question 23:

The mean of n observation is $\overline{)x}$. If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is

(a) $\overline{)x}+\left(2n+1\right)$
(b) $\overline{)x}+\frac{n+1}{2}$
(c) $\overline{)x}+\left(n+1\right)$
(d) $\overline{)x}-\frac{n+1}{2}$

Let be the n observations.

Mean$=\overline{)x}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{n}$

$⇒{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)x}$

If the first item is increased by 1, the second by 2, the third by 3 and so on.

Then, the new observations are .

New mean =
$=\frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}+\left(1+2+3+...+n\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)x}+\frac{n\left(n+1\right)}{2}}{n}\phantom{\rule{0ex}{0ex}}=\overline{)x}+\frac{n+1}{2}$

Hence, the correct answer is (c).

#### Question 24:

If the mean of first n natural numbers is $\frac{5n}{9}$, then n =

(a) 5
(b) 4
(c) 9
(d) 10

Given:
Mean of first n natural number =

$⇒\frac{1+2+3+...+n}{n}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{n\left(n+1\right)}{2}}{n}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{n+1}{2}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}⇒9n+9=10n\phantom{\rule{0ex}{0ex}}⇒n=9$

Hence, the correct option is (c).

#### Question 25:

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is

(a) 25
(b) 18
(c) 20
(d) 22

Given: Mean = 24 and Mode = 12

We know that

Mode = 3Median − 2Mean

⇒ 12 = 3Median − 2 × 24

⇒ 3Median = 12 + 48 = 60

⇒ Median = 20

Hence, the correct option is (c).

#### Question 26:

The mean of first n odd natural number is

(a) $\frac{n+1}{2}$
(b) $\frac{n}{2}$
(c) n
(d) n2

The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).

∴ Mean of first n odd natural numbers

Hence, the correct answer is (c).

#### Question 27:

The mean of first n odd natural numbers is $\frac{{n}^{2}}{81}$, then n =
(a) 9
(b) 81
(c) 27
(d) 18

The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).

∴ Mean of first n odd natural numbers

Now,
Mean of first n odd natural numbers = $\frac{{n}^{2}}{81}$                    (Given)
$\therefore n=\frac{{n}^{2}}{81}\phantom{\rule{0ex}{0ex}}⇒n=81$

Hence, the correct option is (b).

#### Question 28:

If the difference of mode and median of a data is 24, then the difference of median and mean is

(a) 12
(b) 24
(c) 8
(d) 36

Given: Mode − Median = 24

We know that

Mode = 3Median − 2Mean

Now,

Mode − Median = 2(Median − Mean)

⇒ 24 = 2(Median − Mean)

⇒ Median − Mean = 12

Hence, the correct option is (a).

#### Question 29:

If the arithmetic mean, 7, 8, x, 11, 14 is x, then x =

(a) 9
(b) 9.5
(c) 10
(d) 10.5

The given observations are 7, 8, x, 11, 14.

Mean = x       (Given)

Now,

$\mathrm{Mean}=\frac{7+8+x+11+14}{5}\phantom{\rule{0ex}{0ex}}⇒x=\frac{40+x}{5}\phantom{\rule{0ex}{0ex}}⇒5x=40+x\phantom{\rule{0ex}{0ex}}⇒4x=40\phantom{\rule{0ex}{0ex}}⇒x=10$

Hence, the correct option is (c).

#### Question 30:

If mode of a series exceeds its mean by 12, then mode exceeds the median by

(a) 4
(b) 8
(c) 6
(d) 10

Given: Mode − Mean = 12

We know that

Mode = 3Median − 2Mean

∴ Mode − Mean = 3(Median − Mean)

⇒ 12 = 3(Median − Mean)

⇒ Median − Mean = 4       .....(1)

Again,

Mode = 3Median − 2Mean

⇒ 2Mode = 6Median − 4Mean

⇒ Mode − Mean + Mode = 6Median − 5Mean

⇒ 12 + (Mode − Median) = 5(Median − Mean)

⇒12 + (Mode − Median) = 20                              [Using (1)]

⇒ Mode − Median = 20 − 12 = 8

Hence, the correct option is (b).

#### Question 31:

If the mean of first n natural number is 15, then n =

(a) 15
(b) 30
(c) 14
(d) 29

Given:

Mean of first n natural numbers = 15

$⇒\frac{1+2+3+...+n}{n}=15\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{n\left(n+1\right)}{2}}{n}=15\phantom{\rule{0ex}{0ex}}⇒\frac{n+1}{2}=15\phantom{\rule{0ex}{0ex}}⇒n+1=30\phantom{\rule{0ex}{0ex}}⇒n=29$

Hence, the correct option is (d).

#### Question 32:

If the mean of observation , then the mean of x1 + a, x2 + a, ....., xn + a is

(a) $a\overline{)x}$
(b) $\overline{)x}-a$
(c) $\overline{)x}+a$
(d) $\frac{\overline{)x}}{a}$

The mean of .

$\therefore \frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}}{n}=\overline{)x}\phantom{\rule{0ex}{0ex}}⇒{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)x}$

Mean of x1 + a, x2 + a, ... , xn + a

$=\frac{\left({x}_{1}+a\right)+\left({x}_{2}+a\right)+\left({x}_{3}+a\right)+...+\left({x}_{n}+a\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{\left({x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}\right)+\left(a+a+a+...+a\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)x}+na}{n}\phantom{\rule{0ex}{0ex}}=\overline{)x}+a$

Hence, the correct option is (c).

#### Question 33:

Mean of a certain number of observation is $\overline{)x}$. If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is

(a) $\frac{\overline{)x}}{m}+n$
(b) $\frac{\overline{)x}}{n}+m$
(c) $\overline{)x}+\frac{n}{m}$
(d) $\overline{)x}+\frac{m}{n}$

Let be k observations.

Mean of the observations = $\overline{)x}$

If each observation is divided by m and increased by n, then the new observations are

∴ Mean of new observations

Hence, the correct option is (a).

#### Question 34:

If =

(a) 23
(b) 24
(c) 27
(d) 25

Given:

Now, = $\frac{{x}_{i}-25}{10}$

Therefore, h = 10 and A = 25

We know that

Hence, the correct option is (c).

#### Question 35:

If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by

(a) 2
(b) 1.5
(c) 1
(d) 0.5

The given data set is

If 35 is removed, then the new data set is

30, 34, 36, 37, 38, 39, 40

n = 7 (odd)

Therefore,

Increase in median

Hence, the correct option is (d).

#### Question 36:

While computing mean of grouped data , we assume that the frequencies are
(a) evenly distributed over all the classes.
(b) centred at the class marks of the classes .
(c) centred at the upper limit of the classes.
(d) centred at the lower limit of the classes.

We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option (b).

#### Question 37:

In the formula    , for finding the mean of grouped frequency distribution ${u}_{i}$ =
(a)      (b)        (c)      (d)

It is given that .
This is the formula to find mean of any data by step deviation method.

Here,

Hence, the correct answer is option (c).

#### Question 38:

For the following distribution :

Class:               0-5         5-10        10-15        15-20        20-25

Frequency:        10            15            12            20              9

the sum of the lower limits  of the median and modal class is
(a)  15      (b)  25       (c)   30      (d)   35

 Class Frequency Cumulative Frequency 0–5 10 10 5–10 15 25 10–15 12 37 15–20 20 57 20–25 9 66
Here, N = 66.
$\therefore \frac{N}{2}=33$, which lies in the interval 10–15.
So, the lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15–20.
Therefore, the lower limit of modal class is 15.
So, the required sum is 10 + 15 = 25.
Hence, the correct answer is option (b).

#### Question 39:

For the following distribution:
Below:                       10         20       30      40      50     60

Number of students:      3          12        27      57      75      80

the modal class is

(a) 10-20        (b)  20-30       (c)  30-40        (d)    50-60

 Below Class Interval Cumulative Frequency Frequency 10 0–10 3 3 20 10–20 12 9 30 20–30 27 15 40 30–40 57 30 50 40–50 75 18 60 50–60 80 5

Here, N = 80.
$\therefore \frac{N}{2}=40$, which lies in the interval 30–40.
Therefore, the modal class is 30–40.
Hence, the correct answer is option (c).

#### Question 40:

Consider the following frequency distributions:
Class:              65-85          85-105        105-125    125-145     145-165    165-185   - 185-205

Frequency:      4                  5                    13              20              14              7                  4

The difference of the upper limit of the median class and the lower limit of the modal class is
(a)  0     (b)    19       (c)   20        (d) 38

 Class Frequency Cumulative frequency 65–85 4 4 85–105 5 9 105–125 13 22 125–145 20 42 145–165 14 56 165–185 7 63 185–205 4 67

Here, N = 67.
$\therefore \frac{N}{2}=33.5$, which lies in the interval 125–145.
Therefore, the lower limit of the median class is 125.
The highest frequency is 20, which lies in the interval 125–145.
Therefore, the upper limit of modal class is 145.
So, the required difference is 145 − 125 = 20.
Hence, the correct answer is option (c).

#### Question 41:

In the formula , for finding the mean of grouped data ${{d}_{i}}^{{\text{'}}^{s}}$  are derivations from $a$ of
(a)  lower limits of classes        (b)  upper limits ofclasses
(c)  mid-points of classes          (d) frequency of the class marks

The given formula represents the formula to find the mean by assumed mean method.
Here, d${x}_{i}-a$ where xi is the ith observation and a is assumed mean.
So, di's are the deviation from a of mid-points of the classes.
Hence, the correct answer is option (c).

#### Question 42:

The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its
(a)  mean       (b)  median    (c) mode     (d)   all the three above

The less than ogive and more than ogive when drawn on the same graph intersect at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median.
Thus, the abscissa of the point of intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.
Hence, the correct answer is option (b).

#### Question 43:

Consider the following frequency distribution :
Class:               0-5               6-11               12-17               18-23                24-29

Frequency:       13                  10                 15                      8                     11

The upper limit of the median class is
(a) 17         (b)   17.5       (c)   18           (d)    18.5

The given classes in the table are non-continuous. So, we first make the classes continuous by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit in each class.

 Class Frequency Cumulative Frequency 0.5–5.5 13 13 5.5–11.5 10 23 11.5–17.5 15 38 17.5–23.5 8 46 23.5–29.5 11 57

Now, from the table we see that N = 57.
So, $\frac{N}{2}=\frac{57}{2}=28.5$
28.5 lies in the class 11.5–17.5.
The upper limit of the interval 11.5–17.5 is 17.5.
Hence, the correct answer is option (b).

#### Question 1:

If the mean of x, y, z is y, then mean of x and z is ________.

It is given that the mean of x, y, z is y.

Mean of x and z $=\frac{x+z}{2}=\frac{2y}{2}=y$         [Using (1)]

If the mean of x, y, z is y, then mean of x and z is ____y____.

#### Question 2:

The mode of the data 2, x, 3, 4, 5, 2, 4, 6 where x > 2, is ________.

Disclaimer: The question has been modified to match the answer given in the book as given below. The answer provided for the question is 4.

The mode of the data 2, x, 3, 4, 5, 2, 4, 6, 4 where x > 2, is ________.

Solution:

Mode is the value which occur most frequently in a set of observations.

Here, x is a variable which can take any value greater than 2. It can be 3 or 4 or 5 or .... .

However, in this set of observations, the value 4 occurs most frequently i.e. 3 times.

So, the mode of the given data is 4.

The mode of the data 2, x, 3, 4, 5, 2, 4, 6, 4 where x > 2, is ____4____.

Solution for question given in the book.

Mode is the value which occur most frequently in a set of observations.

Here, x is a variable which can take any value greater than 2. It can be 3 or 4 or 5 or .... .

However, in this set of observations, the value 2 and 4 occurs most frequently i.e. 2 times.

So, the mode of the given data is 2 and 4.

When x = 3, then the data set is 2, 3, 3, 4, 5, 2, 4, 6. So, the mode is 2, 3 and 4.

When x = 4, then the data set is 2, 4, 3, 4, 5, 2, 4, 6. So, the mode is 4.

When x = 5, then the data set is 2, 5, 3, 4, 5, 2, 4, 6. So, the mode is 2, 4 and 5.

When x = 6, then the data set is 2, 6, 3, 4, 5, 2, 4, 6. So, the mode is 2, 4 and 6.

For x = 7, 8, 9,...

The mode is 2 and 4.

#### Question 3:

The mean of first 673 natural numbers is _________.

The first 673 natural numbers are 1, 2, 3, ..., 673.

These are in AP with a = 1, l = 673 and n = 673.

∴ Sum of first 673 natural numbers

Now,

Mean of first 673 natural numbers

The mean of first 673 natural numbers is ___337___.

#### Question 4:

The mean of the observations 425, 430, 435, ..., 495 is ___________.

The observations 425, 430, 435, ..., 495 are in AP.

Here, a = 425 and d = 5.

Let the number of observation be n.

Sum of the given observations

∴ Mean of the given observations

The mean of the observations 425, 430, 435, ..., 495 is _____460______.

#### Question 5:

If the mean and median of a unimodal data are 34.5 and 32.5 respectively, then mode of the data is __________.

Mean of the data = 34.5

Median of the data = 32.5

Now,

Mode = 3 × Median − 2 × Mean
$=3×32.5-2×34.5\phantom{\rule{0ex}{0ex}}=97.5-69\phantom{\rule{0ex}{0ex}}=28.5$

If the mean and median of a unimodal data are 34.5 and 32.5 respectively, then mode of the data is ___28.5___.

#### Question 6:

The mean of first n odd natural numbers is _________.

The first n odd natural numbers are 1, 3, 5, ..., 2n − 1.

This forms an AP with a = 1 and l = 2n − 1.

Sum of first n odd natural numbers

∴ Mean of first n odd natural numbers

The mean of first n odd natural numbers is ____n____.

#### Question 7:

The mean of the observations 1, 3, 5, 7, 9, ..., 99 is _________.

The observations 1, 3, 5, 7, 9, ..., 99 are first 50 odd natural numbers.

We know that the mean of first n odd natural numbers is n.

Here, n = 50

So, the mean of the observations 1, 3, 5, 7, 9, ..., 99 is 50.

The mean of the observations 1, 3, 5, 7, 9, ..., 99 is ____50____.

#### Question 8:

If the mean of first n natural numbers is 20, then n = __________.

The first n natural numbers are 1, 2, 3, ..., n.

This forms an AP with a = 1 and d = 1.

Sum of first n natural numbers = 1 + 2 + 3 + ... + n =

∴ Mean of first n natural numbers =

Now,

If the mean of first n natural numbers is 20, then n = _____39_____.

#### Question 9:

If a mode exceeds a mean by 12, then the mode exceeds median by ___________.

Given: Mode − Mean = 12        .....(1)

We know that,

Mode = 3Median − 2Mean

∴ Mode − Mean = 3(Median − Mean)

⇒ 3(Median − Mean) = 12

⇒ Median − Mean = 4           .....(2)

Subtracting (2) from (1), we get

(Mode − Mean) − (Median − Mean) = 12 − 4

⇒ Mode − Median = 8

So, the mode exceeds median by 8.

If a mode exceeds a mean by 12, then the mode exceeds median by ____8____.

#### Question 10:

If the median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m, then the median of the observations x3, x4, x5, x6 (where x1 < x2 < x3 < x4 < x< x< x< x8) is _________.

If the number of observations i.e. n is even, then

Median = Mean of $\left(\frac{n}{2}\right)\mathrm{th}$ observation and $\left(\frac{n}{2}+1\right)\mathrm{th}$ observation

Now,

The median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m. Here, n = 8.

Consider the observations x3, x4, x5, x6. Here, n = 4.

∴ Median of the observations  x3, x4, x5, x6 =     [Using (1)]

If the median of the observations x1, x2, x3, x4, x5, x6, x7, x8 is m, then the median of the observations x3, x4, x5, x6 (where x1 < x2 < x3 < x4 < x< x< x< x8) is ____m____.

#### Question 11:

If mode – median = 2, then median – mean = ___________.

We know that,

Mode = 3Median − 2Mean

⇒ Mode − Median = 2(Median − Mean)

∴ 2(Median − Mean) = 2                   (Mode − Median = 2)

⇒ Median − Mean = 1

If mode – median = 2, then median – mean = ____1____.

#### Question 12:

If the average of a, b,c,d is the average of b and c, then the value of a – b – c + d is ___________.

Average of a, b, c, d = $\frac{a+b+c+d}{4}$

Average of b and c = $\frac{b+c}{2}$

It is given that,

Average of a, b, c, d = Average of b and c

$⇒\frac{a+b+c+d}{4}=\frac{b+c}{2}\phantom{\rule{0ex}{0ex}}⇒a+b+c+d=2b+2c\phantom{\rule{0ex}{0ex}}⇒a+b+c+d-2b-2c=0\phantom{\rule{0ex}{0ex}}⇒a-b-c+d=0$

If the average of a, b, c, d is the average of b and c, then the value of a – b – c + d is _____0_____.

#### Question 13:

Given that a, b, c, d are non-zero integers such that a < b < c < d. If the mean and median of a, b, c, d are equal to zero, then a = ________ and b = ________.

Mean of a, b, c, d $=\frac{a+b+c+d}{4}$

Now,

Median of a, b, c, d = Mean of 2nd observation and 3rd observation $=\frac{b+c}{2}$

From (1) and (2), we get

$a+d=0⇒a=-d$

Also, $b+c=0⇒b=-c$

Given that a, b, c, d are non-zero integers such that a < b < c < d. If the mean and median of a, b, c, d are equal to zero, then a = ___−d___ and b = ___−c___.

#### Question 14:

If a < b < 2a and mean and median of a, b and 2a are 15 and 12 respectively, then a = _________.

The given observations are a, b and 2a.

Mean of the given observations = 15

If the number of observations i.e. n is odd, then

Median = $\left(\frac{n+1}{2}\right)\mathrm{th}$ observation

Here, n = 3.

Median of the given observations = 2nd observation = b = 12         .....(2)

From (1) and (2), we get

$3a+12=45\phantom{\rule{0ex}{0ex}}⇒3a=45-12=33\phantom{\rule{0ex}{0ex}}⇒a=11$

If a < b < 2a and mean and median of a, b and 2a are 15 and 12 respectively, then a = ____11____.

#### Question 15:

If the mean of 26, 19, 15, 24 and x is x, then median of the data is __________.

The given data is 26, 19, 15, 24 and x.

Mean of the given data = x

$\therefore \frac{26+19+15+24+x}{5}=x\phantom{\rule{0ex}{0ex}}⇒84+x=5x\phantom{\rule{0ex}{0ex}}⇒4x=84\phantom{\rule{0ex}{0ex}}⇒x=21$

The given data is 26, 19, 15, 24 and 21.

Arrange the data in increasing order.

15, 19, 21, 24, 26

Here, n = 5

∴ Median of the data = $\left(\frac{5+1}{2}\right)\mathrm{th}$ observation = 3rd observation = 21

If the mean of 26, 19, 15, 24 and x is x, then median of the data is _____21_____.

#### Question 1:

Define mean.

Mean

The mean of a set of observation is equal to sum of observations divided by the number of observations.

(Where = sum of the observation and n = number of observations)

#### Question 2:

What is the algebraic sum of deviation of a frequency distribution about its mean?

The algebraic sum of deviation of a frequency distribution about its mean is zero.

#### Question 3:

Which measure of central tendency is given by the x-coordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?

Median

#### Question 4:

What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?

We know that the abscissa of the point of intersection of two ogives gives the median.

From the given figure, it can be seen that both the ogives intersect at the point (4, 15).

∴ Median of the data = 4

#### Question 5:

Write the empirical relation between mean, mode and median.

The empirical relation between mean, median and mode is

Mode = 3 Median − 2 Mean

#### Question 6:

Which measure of central tendency can be determine graphically?

Median can be determined graphically.

#### Question 7:

Write the modal class for the following frequency distribution:

 Class-interval: 10−15 15−20 20−25 25−30 30−35 35−40 Frequency: 30 35 75 40 30 15

 Class Interval 10−15 15−20 20−25 25−30 30−35 35−40 Frequency 30 35 75 40 30 15

Here, the maximum frequency is 75 and the corresponding class-interval is 20−25.

Therefore, 20−25 is the modal class.

#### Question 8:

A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.

Here, N = 40

So,

Draw a line parallel to x-axis from the point (0, 20), intersecting the graph at point P.

Now, draw PM from P on the x-axis. The x-coordinate of M gives us the median.

∴ Median = 50

#### Question 9:

Write the median class for the following frequency distribution:

 Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency: 5 8 7 12 28 20 10 10

We are given the following table.

 Class Interval Frequency Cumulative Frequency 0−10 5 5 10−20 8 13 20−30 7 20 30−40 12 32 40−50 28 60 50−60 20 80 60−70 10 90 70−80 10 100 N = 100

Here, N = 100

$\therefore \frac{N}{2}=50$

The cumulative frequency just greater than 50 is 60.

So, the median class is 40−50.

#### Question 10:

In the graphical representation of a frequency distribution, if the distance between mode and mean is k times the distance between median and mean, then write the value of k.

We have,

Mode = 3 Median − 2 Mean

⇒ Mode − Mean = 3 Median − 3 Mean

⇒ Mode − Mean = 3(Median − Mean)             .....(1)

It is given that,

Mode − Mean = k (Median − Mean)               .....(2)

From (1) and (2), we get

k = 3

#### Question 11:

Find the class marks of classes 10−25 and 35−55.

Class mark of 10−25

Class mark of 35−55

#### Question 12:

Write the median class of the following distribution:

 Class-interval: 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency: 4 4 8 10 12 8 4

Consider the following distribution table.

 Class Frequency C.F. 0−10 4 4 10−20 4 8 20−30 8 16 30−40 10 26 40−50 12 38 50−60 8 46 60−70 4 50 N = 50

Here,

The cumulative frequency just greater than 25 is 26.

So, the median class is 30−40.

#### Question 13:

If the mean of first n natural numbers is 15, then find n