Chapter 7: Statistics Exercise – 7.2

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	46	54	43	39

Compute the mean number of calls per interval.

Solution:

Let be assumed mean (A) = 3

No. of calls xi	No. of intervals fi	u1 = xi − A = xi = 3	fiui
0	15	- 3	- 45
1	24	- 2	- 48
2	29	- 1	- 29
3	46	0	0
4	54	1	54
5	43	2	86
6	39	3	117
	N = 250		Sum = 135

Mean number of cells = 3 + 135/250
= 885/250
= 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

Solution:

Let the assumed mean (A) = 2

No. of heads per toss xi	No of intervals fi	ui = Ai –x = Ai - 2	fiui
0	38	- 2	- 76
1	144	- 1	- 144
2	342	0	0
3	287	1	287
4	164	2	328
5	25	3	75
	N = 1000		Sum = 470

Mean number of per toss = 2 + 470/1000 
= 2 + 0.47 
= 2.47 

 

Question: 3

The following table gives 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

Calculate the average number of branches per plant.

Solution:

Let the assumed mean (A) = 4

No of branches xi	No of plants fi	ui = xi − A = xi − 4	fiui
2	49	- 2	- 98
3	43	- 1	- 43
4	57	0	0
5	38	1	38
6	13	2	26
	N = 200		Sum = – 77

Average number of branches per plant = 4 + (-77/200) 
= 4 -77/200 
= (800 -77)/200
= 3.615

 

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.

Solution:

Let the assumed mean (A) = 2

No of children xi	No of families fi	ui = xi − A = xi − 2	fiui
0	10	- 2	- 20
1	21	- 1	- 21
3	42	1	42
4	15	2	30
5	7	5	35
	N = 20		Sum = 52

Average number of children for family = 2 + 52/150
= (300 +52)/150 
= 352/150 
= 2.35 (approx) 

 

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.

Solution:

Marks xi	Frequency fi	ui = xi  − A = xi − 2	fiui
15	5	- 10	- 50
20	8	- 5	- 40
22	8	- 3	- 24
24	20	- 1	- 20
25	23	0	0
30	18	5	90
33	13	8	104
38	3	12	36
45	3	20	60
	N = 122		Sum = 110

Question: 6

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

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.

Solution:

Let mean assumed mean (A) = 3

No of students absent xi	No of days fi	ui = xi − A = xi − 3	fiui
3	1	- 3	- 3
1	4	- 2	- 8
2	10	- 1	- 10
3	50	0	0
4	34	1	24
5	15	2	30
6	4	3	12
7	2	4	8
	N = 120		Sum =63

Mean number of students absent per day = 3 + 63/120
= (360 + 63)/120
= 423/120
= 3.53

 

Question: 7

In the first proof of 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	96	36	9	5	1

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

No of misprints per page xi	No of days fi	ui = xi − A = xi − 3	fiui
0	154	- 2	- 308
1	95	- 1	- 95
2	36	0	0
3	9	1	9
4	5	2	1
5	1	3	3
	N = 300		Sum = – 381

Average number of misprints per day = 2 + (- 381/300)
= (600-381)/300
= 219/300
= 0.73

 

Question: 8

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

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.

Solution:

Let the assumed mean (A) = 2

No of accidents	No of workers fi	ui = xi − A = xi − 3	fiui
0	70	- 2	- 140
1	52	- 1	- 52
2	34	0	0
3	3	1	3
4	1	2	2
	N = 100		Sum = – 187

Average no of accidents per day workers ⟹ x + (-187/160)
= 133/160
= 0.83

 

Question: 9

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

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

Solution:

Let the assumed mean (A) = 25

Marks xi	No of students fi	ui = xi − A = xi − 3	fiui<
5	15	-20	-300
10	50	-15	-750
15	80	-10	-800
20	76	-5	-380
25	72	0	0
30	45	5	225
35	39	10	390
40	9	15	135
45	8	20	160
50	6	25	150
	N = 400		Sum = -1170

Mean = 25 + (-1170)/400

= 22.075