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

 

 

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