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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
 
Average number of marks = 25 + 110/102
= (2550 + 110)/102 
= 2660/102
= 26.08 (Approx)
 

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


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