# 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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