Mathematics Solutions Solutions for Class 8 Maths Chapter 12 - Statistics

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Page No 69:

Question 1:

The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

No. of saplings (Scores) x_i	No. of students (frequncy) f_i	f_i×x_i
1	4	4
2	6	$0$
3	12	$0$
4	8	$0$
	N = $0$	∑f_ix_i= $0$

Mean

x

\frac{0}{N}

\frac{0}{0}

0

∴ The average number of trees planted

0

Answer:

No. of saplings (Scores) x_i	No. of students (frequncy) f_i	f_i×x_i
1	4	4
2	6	= 2 $\times$ 6 = $12$
3	12	= 3 $\times$ 12 = $36$
4	8	= 4 $\times$ 8 = $32$
	N = $30$	∑f_ix_i= $84$

Mean

x

\frac{\sum_{} f_{i} x_{x}}{N}

\frac{84}{30}

2.8

∴ The average number of trees planted

2.8

Page No 70:

Question 2:

The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

Electricity used (units) x_i	No. of Families (frequency) f_i	f_i×x_i
30	7	.......
45	2	.......
60	8	.......
75	5	.......
90	3	.......
	N = .......	∑f_ix_i =.......

(1) How many families use 45 units electricity?
(2) State the score, the frequency of which is 5.
(3) Find N, and ∑f_ix_i
(4) Find the mean of electricity used by each family in the month of May.

Answer:

The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

Electricity used (units) x_i	No. of Families (frequency) f_i	f_i×x_i
30	7	30 × 7 = 210
45	2	45 × 2 = 90
60	8	60 × 8 = 480
75	5	75 × 5 = 375
90	3	90 × 3 = 270
	N = 25	∑f_ix_i = 1425

(1) 2 families use 45 units electricity.

(2) The score with a frequency of 5 is 75.

(3) N = 25, and ∑f_ix_i = 1425

(4) Since,

Mean = \frac{\sum_{} f_{i} x_{i}}{N} = \frac{1425}{25} = 57

So, the mean of electricity used by each family in the month of May is 57.

Page No 70:

Question 3:

The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2, 3, 5, 6, 4, 2. Prepare a frequency table and find the mean of members of 40 families.

Answer:

Number of members (x_i)	Number of families (f_i)	f_ix_i
1	2	2
2	8	16
3	7	21
4	8	32
5	7	35
6	6	36
7	2	14
	N = 40	$\sum_{}$ f_ix_i = 156

Since, mean =

\frac{\sum_{} f_{i} x_{i}}{N} = \frac{156}{40} = 3.9

So, the mean of members of 40 families is 3.9.

Page No 70:

Question 4:

The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exibition is:
2 3 4 1 2 3 1 5 4 2 3 1 3 5 4 3 2 2 3 2. Prepare a frequency table and find the mean of the data.

Answer:

The frequency table of the data is follows:

Number of projects (x_i)	Frequency (f_i)	f_ix_i
1	3	1 $\times$ 3 = 3
2	6	2 $\times$ 6 = 12
3	6	3 $\times$ 6 = 18
4	3	4 $\times$ 3 = 12
5	2	5 $\times$ 2 = 10
	N = 20	$\sum_{} f_{i} x_{i}$ = 55

Since, mean of the data =

\frac{\sum_{} f_{i} x_{i}}{N} = \frac{55}{20} = 2.75

Hence, the mean of the data is 2.75.

Page No 71:

Question 1:

Observe the following graph and answer the question.

(1) State the type of the graph.
(2) How much is the savings of Vaishali in the month of April?
(3) How much is the total savings of Saroj in the months March and April?
(4) How much more is the total savings of Savita than the total savings of Megha?
(5) Whose savings in the month of April is the least?

Answer:

(1) This is a bar graph.

(2) The savings of Vaishali in the month of April is Rs 600.

(3) The total savings of Saroj in the months March and April is Rs 800.

(4) Since, the total savings of Savita = Rs 1000, and

The total savings of Megha = Rs 500

Now, 1000 $-$ 500 = 500

So, the total savings of Savita is Rs 500 more than the total savings of Megha.

(5) The savings of Megha in the month of April is the least i.e. Rs 200.

Page No 72:

Question 2:

The number of boys and girls, in std 5 to std 8 in a Z.P. school is given in the table. Draw a subdivided bar graph to show the data.
(Scale : On Y axis, 1 cm = 10 students)

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20

Answer:

We have,

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20

The subdivided bar graph of the given data is as follows:

Page No 72:

Question 3:

In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

Town →	Karjat	Wadgoan	Shivapur	Khandala
Year â†´	Karjat	Wadgoan	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150

Answer:

We have,

Town →	Karjat	Wadgoan	Shivapur	Khandala
Year â†´	Karjat	Wadgoan	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150

The subdivided bar graph of the given data is as follows:

Page No 72:

Question 4:

In the following table, data of the transport means used by students in 8^th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale : On Y axis : 1 cm = 500 students)

Town →	Paithan	Yeola	Shahapur
Mean of Communication â†´	Paithan	Yeola	Shahapur
cycle	3250	1500	1250
Bus and Auto	750	500	500
On foot	1000	1000	500

Answer:

We have, the following table that shows the data of the transport means used by students in 8^th standard for commutation between home and school is given.

Town →	Paithan	Yeola	Shahapur
Mean of Communication â†´	Paithan	Yeola	Shahapur
cycle	3250	1500	1250
Bus and Auto	750	500	500
On foot	1000	1000	500

The subdivided bar diagram of the given data is as follows:

Page No 73:

Question 1:

Show the following information by percentage bar graph.

Division of standard 8	A	B	C	D
Number of students securring grade A	45	33	10	15
Total number of students	60	55	40	75

Answer:

We have,

The following information:

Division of standard 8	A	B	C	D
Number of students securring grade A	45	33	10	15
Total number of students	60	55	40	75
Percentage of students securring grade A	$\frac{45}{60} \times 100 = 75$	$\frac{33}{55} \times 100 = 60$	$\frac{10}{40} \times 100 = 25$	$\frac{15}{75} \times 100 = 20$

The percentage bar graph of the given data is as follows:

Page No 74:

Question 2:

Observe the following graph and answer the questions.

(1) State the type of the bar graph.
(2) How much percent is the Tur production to total production in Ajita'a farm?
(3) Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
(4) Whose percentage production fo Tur is the least?
(5) State production percentages of Tur and gram in Sudha's farm

Answer:

(1) This is a percentage bar graph.

(2) The Tur production is 60% of the total production in Ajita'a farm.

(3) The production of gram in the farm of Yash is 50%, whereas

The production of gram in the farm of Ravi is 30%.

Since, 50 $-$ 30 = 20

So, the production of Gram in the farms of Yash is more than that of Yash by 20%.

(4) The production fo Tur in the farm of Sudha is the least, i.e. 40%.

(5) The production percentages of Tur and gram in Sudha's farm are 40% and 60%, repectively.

Page No 74:

Question 3:

The following data is collected in a survey of some students of 10^thstandard from some schools. Draw the percentage bar graph of the data.

School	1^st	2^nd	3^rd	4^th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24

Answer:

We have,

School	1^st	2^nd	3^rd	4^th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24
Total number of students	150	80	50	40
Percentage of students whose inclination towards science stream	$\frac{90}{150} \times 100 = 60$	$\frac{60}{80} \times 100 = 75$	$\frac{25}{50} \times 100 = 50$	$\frac{16}{40} \times 100 = 40$
Percentage of students whose inclination towards commerce stream	$100 - 60 = 40$	$100 - 75 = 25$	$100 - 50 = 50$	$100 - 40 = 60$

The percentage bar graph of the given data is as follows:

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