Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Maths Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1:

If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.   [CBSE 2014]

Mean of given observations =

Hence, the value of x is 7.

#### Question 2:

If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?

​Mean of given observations =

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500

Then, new mean = $\frac{500}{25}=20$

Thus, the new mean will be 20.

#### Question 3:

Compute the mean of the following frequency distribution:

 Class 10 − 30 30 − 50 50 − 70 70 − 90 90 − 110 Frequency 15 18 25 10 2

Here, h = 20

Let the assumed mean, A be 60.

 Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{\mathit{i}}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{60}$ ${\mathbit{u}}_{\mathit{i}}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{60}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 10−30 20 15 −40 −2 −30 30−50 40 18 −20 −1 −18 50−70 60 25 0 0 0 70−90 80 10 20 1 10 90−110 100 2 40 2 4 $\underset{}{\sum {f}_{i}=70}$ $\underset{}{\sum {f}_{i}{u}_{i}=-34}$

We know

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

∴ Mean of the given frequency distribution

Thus, the mean of the given frequency distribution is approximately 50.29.

#### Question 4:

Compute the mean of the following frequency distribution:

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

Here, h = 20

Let the assumed mean, A be 70.

 Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{70}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{70}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−20 10 6 −60 −3 −18 20−40 30 8 −40 −2 −16 40−60 50 10 −20 −1 −10 60−80 70 12 0 0 0 80−100 90 6 20 1 6 100−120 110 5 40 2 10 120−140 130 3 60 3 9 $\underset{}{\sum {f}_{i}=50}$ $\underset{}{\sum {f}_{i}{u}_{i}=-19}$

We know

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

∴ Mean of the given frequency distribution

$=70+20×\left(\frac{-19}{50}\right)\phantom{\rule{0ex}{0ex}}=70-7.6\phantom{\rule{0ex}{0ex}}=62.4$

Thus, the mean of the given frequency distribution is 62.4.

#### Question 5:

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.

 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

Here, h = 6

Let the assumed mean, A be 21.

 Number of days Mid-Values(xi) Number of Students (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{21}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{21}}{\mathbf{6}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−6 3 10 −18 −3 −30 6−12 9 11 −12 −2 −22 12−18 15 7 −6 −1 −7 18−24 21 4 0 0 0 24−30 27 4 6 1 4 30−36 33 3 12 2 6 36−42 39 1 18 3 3 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-46}$

We know

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

∴ Mean number of days a student was absent

$=21+6×\left(\frac{-46}{40}\right)\phantom{\rule{0ex}{0ex}}=21-6.9\phantom{\rule{0ex}{0ex}}=14.1$

Thus, the mean number of days a student was absent is 14.1.

#### Question 6:

The daily expenses of 40 families has been shown by the following frequency distribution:

 Expenses (in ₹) 500 − 700 700 − 900 900 − 1100 1100 − 1300 1300 − 1500 Number of families 6 8 10 9 7

Find the mean daily expenses.

Here, h = 200

Let the assumed mean, A be ₹1000.

 Expenses (in ₹) Mid-Values(xi) Number of Families(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{1000}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{1000}}{\mathbf{200}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 500−700 600 6 −400 −2 −12 700−900 800 8 −200 −1 −8 900−1100 1000 10 0 0 0 1100−1300 1200 9 200 1 9 1300−1500 1400 7 400 2 14 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=3}$

We know

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

∴ Mean daily expenses

$=1000+200×\left(\frac{3}{40}\right)\phantom{\rule{0ex}{0ex}}=1000+15\phantom{\rule{0ex}{0ex}}=₹1015$

Thus, the mean daily expenses is ₹1015.

#### Question 7:

During a medical checkup, the number of heartbeats per minute of 40 patients were recorded and summarised as follows:

 Number of heartbeats per minute 64 − 68 68 − 72 72 − 76 76 − 80 80 − 84 84 – 88 Number of patients 6 8 10 12 3 1

Find the mean heartbeats per minute for these patients.

Here, h = 4

Let the assumed mean, A be 78.

 Number of Heartbeats per Minute Mid-Values(xi) Number of Patients (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{78}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{78}}{\mathbf{4}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 64−68 66 6 −12 −3 −18 68−72 70 8 −8 −2 −16 72−76 74 10 −4 −1 −10 76−80 78 12 0 0 0 80−84 82 3 4 1 3 84−88 86 1 8 2 2 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-39}$

We know

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

∴ Mean heartbeats per minute for these patients

$=78+4×\left(\frac{-39}{40}\right)\phantom{\rule{0ex}{0ex}}=78-3.9\phantom{\rule{0ex}{0ex}}=74.1$

Thus, the mean heartbeats per minute for these patients is 74.1.

#### Question 8:

The table given below shows the age distribution of 1000 persons who visited a marketing centre on a Sunday.

 Age (in years) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 – 60 60 – 70 Number of persons 105 222 220 138 102 113 100

Find the mean age of the persons visiting the marketing centre on that day.

Here, h = 10

Let the assumed mean, A be 35.

 Age (in years) Mid-Values(xi) Number of persons (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{35}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{35}}{\mathbf{10}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−10 5 105 −30 −3 −315 10−20 15 222 −20 −2 −444 20−30 25 220 −10 −1 −220 30−40 35 138 0 0 0 40−50 45 102 10 1 102 50−60 55 113 20 2 226 60−70 65 100 30 3 300 $\underset{}{\sum {f}_{i}=1000}$ $\underset{}{\sum {f}_{i}{u}_{i}=-351}$

We know

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

∴ Mean age of the persons visiting the marketing centre on that day

$=35+10×\left(\frac{-351}{1000}\right)\phantom{\rule{0ex}{0ex}}=35-3.51\phantom{\rule{0ex}{0ex}}=31.49$

Thus, the mean age of the persons visiting the marketing centre on that day 31.49 years.

#### Question 9:

The arithmetic mean of the following frequency distribution is 53. Find the value of x.

 Class 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 Frequency 12 15 32 x 13

 Class Mid-Values(xi) Frequency (fi) fi xi 0−20 10 12 120 20−40 30 15 450 40−60 50 32 1600 60−80 70 x 70x 80−100 90 13 1170 $\underset{}{\sum {f}_{i}=72+x}$ $\underset{}{\sum {f}_{i}{x}_{i}=3340+70x}$

Mean of the given frequency distribution = 53

We know

Mean $=\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}$

$\therefore \frac{3340+70x}{72+x}=53\phantom{\rule{0ex}{0ex}}⇒3340+70x=3816+53x\phantom{\rule{0ex}{0ex}}⇒70x-53x=3816-3340$
$⇒17x=476\phantom{\rule{0ex}{0ex}}⇒x=\frac{476}{17}=28$

Thus, the value of x is 28.

#### Question 10:

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹18, find the missing frequency f.

 Daily pocket allowance (in ₹) 11−13 13−15 15−17 17−19 19−21 21−23 23−25 Number of children 7 6 9 13 f 5 4

The given data is shown as follows:

 Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fixi 11−13 7 12 84 13−15 6 14 84 15−17 9 16 144 17−19 13 18 234 19−21 f 20 20f 21−23 5 22 110 23−25 4 24 96 Total ∑ fi = 44 + f ∑ fixi = 752 + 20f

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒18=\frac{752+20f}{44+f}\phantom{\rule{0ex}{0ex}}⇒18\left(44+f\right)=752+20f\phantom{\rule{0ex}{0ex}}⇒792+18f=752+20f\phantom{\rule{0ex}{0ex}}⇒20f-18f=792-752\phantom{\rule{0ex}{0ex}}⇒2f=40\phantom{\rule{0ex}{0ex}}⇒f=20$

​Hence, the value of f is 20.

#### Question 11:

If the mean of the following frequency distribution is 54, find the value of p.   [CBSE 2006C]

 Class 0−20 20−40 40−60 60−80 80−100 Frequency 7 p 10 9 13

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fixi 0−20 7 10 70 20−40 p 30 30p 40−60 10 50 500 60−80 9 70 630 80−100 13 90 1170 Total ∑ fi = 39 + p ∑ fixi = 2370 + 30p

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒54=\frac{2370+30p}{39+p}\phantom{\rule{0ex}{0ex}}⇒54\left(39+p\right)=2370+30p\phantom{\rule{0ex}{0ex}}⇒2106+54p=2370+30p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106\phantom{\rule{0ex}{0ex}}⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=11$

​Thus, the value of is 11.

#### Question 12:

The mean of the following data is 42, find the missing frequencies and y if the sum of the frequencies is 100.

 Class interval 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 7 10 x 13 y 10 14 9

The given data is shown as follows:

 Class interval Frequency (fi) Class mark (xi) fixi 0−10 7 5 35 10−20 10 15 150 20−30 x 25 25x 30−40 13 35 455 40−50 y 45 45y 50−60 10 55 550 60−70 14 65 910 70−80 9 75 675 Total ∑ fi = 63 + x + y ∑ fixi = 2775 + 25x + 45y

Sum of the frequencies = 100

Now, The mean of given data is given by

If x = 12, then y = 37 − 12 = 25

​Thus, the value of is 12 and y is 25.

#### Question 13:

The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹188.

 Expenditure (in ₹) 140−160 160−180 180−200 200−220 220−240 Number of families 5 25 f1 f2 5

The given data is shown as follows:

 Expenditure (in ₹) Number of families (fi) Class mark (xi) fixi 140−160 5 150 750 160−180 25 170 4250 180−200 f1 190 190f1 200−220 f2 210 210f2 220−240 5 230 1150 Total ∑ fi = 35 + f1 + f2 ∑ fixi = 6150 + 190f1 + 210f2

Sum of the frequencies = 100

Now, The mean of given data is given by

If f1 = 50, then f2 = 65 − 50 = 15

​Thus, the value of f1 is 50 and f2 is 15.

#### Question 14:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

 Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 f1 12 f2 8 5

 Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-20 7 10 70 20-40 f1 30 30 f1 40-60 12 50 600 60-80 18- f1 70 1260-70 f1 80-100 8 90 720 100-120 5 110 550 $\sum {f}_{i}=50$ $\sum \left({f}_{i}×{x}_{i}\right)=3200-40{f}_{1}$

#### Question 15:

If the mean of the following frequency distribution is 188, find the missing frequencies x and y, if the sum of all frequencies is 100.

 Class 0 – 80 80 – 160 160 – 240 240 – 320 320 – 400 Frequency 20 25 x y 10

Here, h = 80

Let the assumed mean, A be 200.

 Class Mid-Values(xi) Frequency(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{200}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{200}}{\mathbf{80}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−80 40 20 −160 −2 −40 80−160 120 25 −80 −1 −25 160−240 200 x 0 0 0 240−320 280 y 80 1 y 320−400 360 10 160 2 20 $\underset{}{\sum {f}_{i}=x+y+55}$ $\underset{}{\sum {f}_{i}{u}_{i}=-45+y}$

$\underset{}{\sum {f}_{i}=x+y+55}=100$         (Given)

We know

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

$\therefore 200+80×\left(\frac{-45+y}{100}\right)=188$          (Given)

$⇒\frac{-45+y}{100}=\frac{188-200}{80}=-0.15\phantom{\rule{0ex}{0ex}}⇒-45+y=-15\phantom{\rule{0ex}{0ex}}⇒y=45-15=30$

Substituting the value of y in (1), we get

x + 30 = 45

x = 45 − 30 = 15

Thus, the missing frequencies x and y are 15 and 30, respectively.

#### Question 16:

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

 Literacy rate (%) 45−55 55−65 65−75 75−85 85−95 Number of cities 4 11 12 9 4

Using Direct method, the given data is shown as follows:

 Literacy rate (%) Number of cities (fi) Class mark (xi) fixi 45−55 4 50 200 55−65 11 60 660 65−75 12 70 840 75−85 9 80 720 85−95 4 90 360 Total ∑ fi = 40 ∑ fixi = 2780

The mean of given data is given by

​Thus, the mean literacy rate is 69.5%.

#### Question 17:

Find the mean, using assumed-mean method:

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of students 12 18 27 20 17 6

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-25\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-10 12 5 -20 -240 10-20 18 15 -10 -180 20-30 27 25=A 0 0 30-40 20 35 10 200 40-50 17 45 20 340 50-60 6 55 30 180 $\sum {f}_{i}=100$ $\sum \left({f}_{i}×{d}_{i}\right)=300$

#### Question 18:

Find the mean, using assumed-mean method:

 Class 100-120 120-140 140-160 160-180 180-200 Frequency 10 20 30 15 5

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-150\right)$ $\left({f}_{i}×{d}_{i}\right)$ 100-120 10 110 -40 -400 120-140 20 130 -20 -400 140-160 30 150=A 0 0 160-180 15 170 20 300 180-200 5 190 40 200 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{d}_{i}\right)=-\text{3}00$

#### Question 19:

Find the mean, using assumed-mean method:

 Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 20 35 52 44 38 31

 Class Frequency $\left({f}_{i}\right)$ Mid Values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-50\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-20 20 10 -40 -800 20-40 35 30 -20 -700 40-60 52 50=A 0 0 60-80 44 70 20 880 80-100 38 90 40 1520 100-120 31 110 60 1860 $\sum {f}_{i}=220$ $\sum \left({f}_{i}×{d}_{i}\right)=2760$

#### Question 20:

Find the mean of the following frequency distribution using step-deviation method.

 Class 0−10 10−20 20−30 30−40 40−50 Frequency 7 10 15 8 10

Let us choose a = 25, h = 10, then di = xi − 25 and ui$\frac{{x}_{i}-25}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 25 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{25}}{\mathbf{10}}$ fiui 0−10 7 5 −20 −2 −14 10−20 10 15 −10 −1 −10 20−30 15 25 0 0 0 30−40 8 35 10 1 8 40−50 10 45 20 2 20 Total ∑ fi = 50 ∑ fiui = 4

The mean of given data is given by

​Thus, the mean is 25.8.

#### Question 21:

Find the mean of the following data, using step-deviation method.

 Class 5−15 15−25 25−35 35−45 45−55 55−65 65−75 Frequency 6 10 16 15 24 8 7

Let us choose a = 40, h = 10, then di = xi − 40 and ui = $\frac{{x}_{i}-40}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 40 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{40}}{\mathbf{10}}$ fiui 5−15 6 10 −30 −3 −18 15−25 10 20 −20 −2 −20 25−35 16 30 −10 −1 −16 35−45 15 40 0 0 0 45−55 24 50 10 1 24 55−65 8 60 20 2 16 65−75 7 70 30 3 21 Total ∑ fi = 86 ∑ fiui = 7

The mean of given data is given by

​Thus, the mean is 40.81.

#### Question 22:

​The weights of tea in 70 packets are shown in the following table:

 Weight (in grams) 200−201 201−202 202−203 203−204 204−205 205−206 Number of packets 13 27 18 10 1 1

Find the mean weight of packets using step-deviation method.                     [CBSE 2013]

​Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = $\frac{{x}_{i}-202.5}{1}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of packets (fi) Class mark (xi) di = xi − 202.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{202}\mathbf{.}\mathbf{5}}{\mathbf{1}}$ fiui 200−201 13 200.5 −2 −2 −26 201−202 27 201.5 −1 −1 −27 202−203 18 202.5 0 0 0 203−204 10 203.5 1 1 10 204−205 1 204.5 2 2 2 205−206 1 205.5 3 3 3 Total ∑ fi = 70 ∑ fiui = −38

The mean of given data is given by

​Hence, the mean is 201.96 g.

#### Question 23:

Find the mean of the following frequency distribution using a suitable method.

 Class 20−30 30−40 40−50 50−60 60−70 Frequency 25 40 42 33 10

Let us choose a = 45, h = 10, then di = xi − 45 and ui = $\frac{{x}_{i}-45}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 45 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ fiui 20−30 25 25 −20 −2 −50 30−40 40 35 −10 −1 −40 40−50 42 45 0 0 0 50−60 33 55 10 1 33 60−70 10 65 20 2 20 Total ∑ fi = 150 ∑ fiui = −37

The mean of given data is given by

​Thus, the mean is 42.534.

#### Question 24:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

 Age (in years) 18-24 24-30 30-36 36-42 42-48 48-54 Number of workers 6 8 12 8 4 2

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-33\right)}{6}$ $\left({f}_{i}×{u}_{i}\right)$ 18-24 6 21 −2 −12 24-30 8 27 −1 −8 30-36 12 33 = A 0 0 36-42 8 39 1 8 42-48 4 45 2 8 48-54 2 51 3 6 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=2$

#### Question 25:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

 Class 500-520 520-540 540-560 560-580 580-600 600-620 Frequency 14 9 5 4 3 5

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-550\right)}{20}$ $\left({f}_{i}×{u}_{i}\right)$ 500-520 14 510 −2 −28 520-540 9 530 −1 −9 540-560 5 550 = A 0 0 560-580 4 570 1 4 580-600 3 590 2 6 600-620 5 610 3 15 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=-12$

#### Question 26:

Find the mean age from the following frequency distribution:

 Age (in years) 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of person 4 14 22 16 6 5 3

Converting the series into exclusive form, we get:

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-42\right)}{5}$ $\left({f}_{i}×{u}_{i}\right)$ 24.5-29.5 4 27 −3 −12 29.5-34.5 14 32 −2 −28 34.5-39.5 22 37 −1 −22 39.5-44.5 16 42 = A 0 0 44.5-49.5 6 47 1 6 49.5-54.5 5 52 2 10 54.5-59.5 3 57 3 9 $\sum {f}_{i}=70$ $\sum \left({f}_{i}×{u}_{i}\right)=-37$

#### Question 27:

The following table shows the age distribution of patients of malaria in a village during a particular month.

 Age (in years) 5-14 15-24 25-34 35-44 45-54 55-64 No. of cases 6 11 21 23 14 5

Find the average age of the patients.

Converting the series into exclusive form, we get:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-29.5\right)}{10}$ $\left({f}_{i}×{u}_{i}\right)$ 4.5-14.5 6 9.5 −2 −12 14.5-24.5 11 19.5 −1 −11 24.5-34.5 21 29.5 = A 0 0 34.5-44.5 23 39.5 1 23 44.5-54.5 14 49.5 2 28 54.5-64.5 5 59.5 3 15 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{u}_{i}\right)=43$

#### Question 28:

Weight of 60 eggs were recorded as given below:

 Weight (in grams) 75−79 80−84 85−89 90−94 95−99 100−104 105−109 Number of eggs 4 9 13 17 12 3 2

Calculate their mean weight to the nearest gram.

Let us choose a = 92, h = 5, then di = xi − 92 and ui = $\frac{{x}_{i}-92}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi − 92 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{92}}{\mathbf{5}}$ fiui 74.5−79.5 4 77 −15 −3 −12 79.5−84.5 9 82 −10 −2 −18 84.5−89.5 13 87 −5 −1 −13 89.5−94.5 17 92 0 0 0 94.5−99.5 12 97 5 1 12 99.5−104.5 3 102 10 2 6 104.5−109.5 2 107 15 3 6 Total ∑ fi = 60 ∑ fiui = −19

The mean of given data is given by

​Thus, the mean weight to the nearest gram is 90 g.

#### Question 29:

​The following table shows the marks scored by 80 students in an examination:

 Marks Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Number of students 3 10 25 49 65 73 78 80

Calculate the mean marks correct to 2 decimal places.

Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = $\frac{{x}_{i}-17.5}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi − 17.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ fiui 0−5 3 3 2.5 −15 −3 −9 5−10 10 7 7.5 −10 −2 −14 10−15 25 15 12.5 −5 −1 −15 15−20 49 24 17.5 0 0 0 20−25 65 16 22.5 5 1 16 25−30 73 8 27.5 10 2 16 30−35 78 5 32.5 15 3 15 35−40 80 2 37.5 20 4 8 Total ∑ fi = 80 ∑ fiui = 17

The mean of given data is given by

​Thus, the mean marks correct to 2 decimal places is 18.56.

#### Question 1:

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.    [CBSE 2014]

 Age (in years) 0−15 15−30 30−45 45−60 60−75 Number of patients 5 20 40 50 25

We prepare the cumulative frequency table, as shown below:

 Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0−15 5 5 15−30 20 25 30−45 40 65 45−60 50 115 60−75 25 140 Total N = ∑ fi = 140

Now, N = 140 $⇒\frac{N}{2}=70$.

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.

Thus, the median class is 45−60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Hence, the median age is 46.5 years.

#### Question 2:

Compute the median from the following data:

 Marks 0-7 7-14 14-21 21-28 28-35 35-42 42-49 Number of students 3 4 7 11 0 16 9

 Class Frequency (f) Cumulative frequency 0-7 3 3 7-14 4 7 14-21 7 14 21-28 11 25 28-35 0 25 35-42 16 41 42-49 9 50 N=∑f=50

#### Question 3:

The following table shows the daily wages of workers in a factory:

 Daily wages (in Rs) 0-100 100-200 200-300 300-400 400-500 Number of workers 40 32 48 22 8

Find the median daily wage income of the workers.

 Class Frequency(f) Cumulative frequency 0-100 40 40 100-200 32 72 200-300 48 120 300-400 22 142 400-500 8 150 N=∑f=150

#### Question 4:

Calculate the median from the following frequency distribution:

 Class 5-10 10-15 15-20 20-25 20-30 30-35 35-40 40-45 Frequency 5 6 15 10 5 4 2 2

 Class Frequency(f) Cumulative frequency 5-10 5 5 10-15 6 11 15-20 15 26 20-25 10 36 25-30 5 41 30-35 4 45 35-40 2 47 40-45 2 49 N=∑f=49

#### Question 5:

Given below is the number of units of electricity consumed in a week in a certain locality:

 Consumption (in units) 65-85 85-105 105-125 125-145 145-165 165-185 195-205 Number of consumers 4 5 13 20 14 7 4

Calculate the median

 Class Frequency(f) 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 N=∑f=67

#### Question 6:

Calculate the median from the following data:

 Height (in cm) 135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 No. of boys 6 10 18 22 20 15 6 3

 Class Frequency(f) Cumulative frequency 135-140 6 6 140=145 10 16 145-150 18 34 150-155 22 56 155-160 20 76 160-165 15 91 165-170 6 97 170-175 3 100 N=∑f=100

#### Question 7:

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

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 25 ? 18 7

 Class Frequency (fi) c.f 0-10 5 5 10-20 25 30 20-30 x x+30 30-40 18 x+48 40-50 7 x+55

#### Question 8:

The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

 Class 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Frequency 12 a 12 15 b 6 6 4

We prepare the cumulative frequency table, as shown below:

 Class Frequency (fi) Cumulative frequency (cf) 0−5 12 12 5−10 a 12 + a 10−15 12 24 + a 15−20 15 39 + a 20−25 b 39 + a + b 25−30 6 45 + a + b 30−35 6 51 + a + b 35−40 4 55 + a + b Total N = ∑fi = 70

Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,

55 a + = 70 ⇒ a + = 15     ...(1)

Median is 16, which lies in 15−20. So, the median class is 15−20.

l = 15, h = 5, N = 70, f = 15 and cf = 24 + a

Now,

$\mathrm{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)×h\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{\frac{70}{2}-\left(24+a\right)}{15}\right)×5\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{35-24-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{11-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16-15=\frac{11-a}{3}\phantom{\rule{0ex}{0ex}}⇒1×3=11-a\phantom{\rule{0ex}{0ex}}⇒a=11-3\phantom{\rule{0ex}{0ex}}⇒a=8$

b = 15 a    [From (1)]
b = 15 − 8
b = 7

Hence, a = 8 and b = 7.

#### Question 9:

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

 Runs scored 2500−3500 3500−4500 4500−5500 5500−6500 6500−7500 7500−8500 Number of batsmen 5 x y 12 6 2

We prepare the cumulative frequency table, as shown below:

 Runs scored Number of batsmen (fi) Cumulative frequency (cf) 2500−3500 5 5 3500−4500 x 5 + x 4500−5500 y 5 + x + y 5500−6500 12 17 + x + y 6500−7500 6 23 + x + y 7500−8500 2 25 + x + y Total N = ∑fi = 60

Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,

25 x + = 60 ⇒ x + = 35     ...(1)

Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

∴ y = 35 − x    [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.

#### Question 10:

If the median of the following frequency distribution is 32.5, find the values of f1 and f2.

 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

 Class Frequency(f) Cumulative frequency 0-10 ${f}_{1}$ ${f}_{1}$ 10-20 5 ${f}_{1}$+5 20-30 9 ${f}_{1}$+14 30-40 12 ${f}_{1}$+26 40-50 ${f}_{2}$ ${f}_{1}+{f}_{2}$+26 50-60 3 ${f}_{1}+{f}_{2}$+29 60-70 2 ${f}_{1}+{f}_{2}$+31 N=∑f=40

#### Question 11:

Calculate the median for the following data:

 Age (in years) 19-25 26-32 33-39 40-46 47-53 54-60 Frequency 35 96 68 102 35 4

First, we will convert the data into exclusive form.

 Class Frequency(f) Cumulative frequency 18.5-25.5 35 35 25.5-32.5 96 131 32.5-39.5 68 199 39.5-46.5 102 301 46.5-53.5 35 336 53.5-60.5 4 340 N=∑f=340

#### Question 12:

Find the median wages for the following frequencies distribution:

 Wages per day (in Rs) 61-70 71-80 81-90 91-100 101-110 111-120 No. of women workers 5 15 20 30 20 8

Converting the given data into exclusive form, we get:

 Class Frequency(f) Cumulative frequency 60.5-70.5 5 5 70.5-80.5 15 20 80.5-90.5 20 40 90.5-100.5 30 70 100.5-110.5 20 90 110.5-120.5 8 98 N=∑f=98

#### Question 13:

Find the median from the following data:

 Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35 35-40 41-45 Frequency 7 10 16 32 24 16 11 5 2

Converting into exclusive form, we get:

 Class Frequency(f) Cumulative frequency 0.5-5.5 7 7 5.5-10.5 10 17 10.5-15.5 16 33 15.5-20.5 32 65 20.5-25.5 24 89 25.5-30.5 16 105 30.5-35.5 11 116 35.5-40.5 5 121 40.5-45.5 2 123 N=∑f=123

#### Question 14:

Find the median from the following data:

 Marks No. of students Below 10 12 Below 20 32 Below 30 57 Below 40 80 Below 50 92 Below 60 116 Below 70 164 Below 80 200

 Class Cumulative frequency Frequency (f) 0-10 12 12 10-20 32 20 20-30 57 25 30-40 80 23 40-50 92 12 50-60 116 24 60-70 164 48 70-80 200 36 N=∑f=200

#### Question 1:

Find the mode of the following frequency distribution:         [CBSE 2014]

 Marks 10−20 20−30 30−40 40−50 50−60 Frequency 12 35 45 25 13

Here the maximum class frequency is 45, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now,

modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f1) of the modal class = 45,

frequency (f0) of class preceding the modal class = 35,

frequency (
f2) of class succeeding the modal class = 25

Now, let us substitute these values in the formula:

Hence, the mode is 33.33.

#### Question 2:

Compute the mode of the following data:              [CBSE 2013]

 Class 0−20 20−40 40−60 60−80 80−100 Frequency 25 16 28 20 5

Here the maximum class frequency is 28, and the class corresponding to this frequency is 40−60. So, the modal class is 40−60.

Now,

Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,

frequency (f1) of the modal class = 28,

frequency (f0) of class preceding the modal class = 16,

frequency (f2
) of class succeeding the modal class = 20.

Now, let us substitute these values in the formula:

Hence, the mode is 52.

#### Question 3:

Heights of students of Class X are given in the foloowing frequency distribution:

 Height (in cm) 150−155 155−160 160−165 165−170 170−175 Number of students 15 8 20 12 5

Find the modal height.
Also, find the mean height. Compare and interpret the two measures of central tendency.    [CBSE 2014]

Here the maximum class frequency is 20, and the class corresponding to this frequency is 160−165. So, the modal class is 160−165.

Now,

Modal class =
160−165, lower limit (l) of modal class = 160, class size (h) = 5,

frequency (f1) of the modal class = 20,

frequency (f0) of class preceding the modal class = 8,

frequency (f2) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:

Hence, the mode is 163.

It represents that the height of maximum number of students is 163 cm.

Now, to find the mean let us put the data in the table given below:

 Height (in cm) Number of students (fi) Class mark (xi) fixi 150−155 15 152.5 2287.5 155−160 8 157.5 1260 160−165 20 162.5 3250 165−170 12 167.5 2010 170−175 5 172.5 862.5 Total ∑fi = 60 ∑fixi =  9670

Thus, mean of the given data is 161.17.

It represents that on an average, the height of a student is 161.17 cm.

#### Question 4:

Find the mode of the following distribution:

 Class interval 10-14 14-18 18-22 22-26 26-30 30-34 34-38 38-42 Frequency 8 6 11 20 25 22 10 4

As the class 26-30 has the maximum frequency, it is the modal class.

#### Question 5:

Given below is the distribution of total household expenditure of 200 manual workers in a city:

 Expenditure (in Rs) No. of manual workers 1000-1500 24 1500-2000 40 2000-2500 31 2500-3000 28 3000-3500 32 3500-4000 23 4000-4500 17 4500-5000 5
Find the expenditure done by maximum number of manual workers.

As the class 1500-2000 has the maximum frequency, it is the modal class.

Hence, mode = Rs 1820

#### Question 6:

Calculate the mode from the following data:

 Monthly salary (in Rs) No. of employees 0-5000 90 5000-1000 150 10000-15000 100 15000-20000 80 20000-25000 70 25000-30000 10

As the class 5000-10000 has the maximum frequency, it is the modal class.

Hence, mode = Rs 7727.27

#### Question 7:

Compute the mode from the following data:

 Age (in years) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 Number of patients 6 11 18 24 17 13 5

As the class 15-20 has the maximum frequency, it is the modal class.

Hence, mode=17.3 years

#### Question 8:

Compute the mode from the following series:

 Size 45-55 55-65 65-75 75-85 85-95 95-105 105-115 Frequency 7 12 17 30 32 6 10

As the class 85-95 has the maximum frequency, it is the modal class.

Hence, mode=85.71

#### Question 9:

Compute the mode from the following data:

 Class interval 1-5 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 46-50 Frequency 3 8 13 18 28 20 13 8 6 4