# RD Sharma Solutions Class 10 Statistics Exercise 7.2

### RD Sharma Class 10 Solutions Chapter 7 Ex 7.2 PDF Free Download

#### Exercise 7.2

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.

Soln: Let be assumed mean (A) = 3

 No. of calls $x_{i}$ No. of intervals $f_{i}$ $u_{1} = x_{i} -A =x_{i}= 3$ $f_{i}u_{i}$ 0 15 -3 -45 1 24 -2 -47 2 29 -1 -39 3 46 0 0 4 54 54 5 43 2 43(2) = 86 6 39 3 47 N= 250 Sum = 135

Mean number of cells = $3 + frac{135}{250}$ = $frac{885}{250}$ = 3.54

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

Soln:Let the assumed mean (A) = 2

 No. of heads per toss $x_{i}$ No of intervals $f_{i}$ $u_{i}=A_{i}-x =A_{i}- 2$ $f_{i}u_{i}$ 0 38 -2 -7 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

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.

Soln:

Let the assumed mean (A) = 4

 No of branches $x_{i}$ No of plants $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 4$ $f_{i}u_{i}$ 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

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.

Soln: Let the assumed mean (A) = 2

 No of children $x_{i}$ No of families $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 2$ $f_{i}u_{i}$ 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)

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.

Soln: Let the assumed mean (A) = 25

 Marks $x_{i}$ Frequency $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 2$ $f_{i}u_{i}$ 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)

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.

Soln:Let mean assumed mean (A) = 3

 No of students absent $x_{i}$ No of days $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 3$ $f_{i}u_{i}$ 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

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.

Soln: Let the assumed mean (A) = 2

 No of misprints per page $x_{i}$ No of days $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 3$ $f_{i}u_{i}$ 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)

= 2 – 381/300

= (600-381)/300

= 219/300

= 0.73

8. The following distribution gives the number of accidents met by 160 workers in a factory during a month.

 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.

Soln: Let the assumed mean (A) = 2

 No of accidents No of workers $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 3$ $f_{i}u_{i}$ 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

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

 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

Soln:Let the assumed mean (A) = 25

 Marks $x_{i}$ No of students $f_{i}$ $u_{i}=x_{i}-A =x_{i}- 3$ $f_{i}u_{i}$< 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