Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.2
Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:
Number of calls | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
Number of intervals | 15 | 24 | 29 | 46 | 54 | 43 | 39 |
Compute the mean number of calls per interval.
Solution:
Let the assumed mean(A) be =3 (Generally we choose the middle element to be the assumed mean, but it’s not mandatory),
hence, the table is,
Number of calls Number of intervals
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
hence, mean of the calls =
=
=
Therefore, mean number of calls per interval is 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.
Number of heads per toss | 0 | 1 | 2 | 3 | 4 | 5 |
Number of tosses | 38 | 144 | 342 | 287 | 164 | 25 |
Solution:
Let the assumed mean (A) be = 2
hence, the table is,
Number of heads per toss
Number of tosses
0
38
-2
-76
1
144
-1
-144
2
342
0
0
3
287
1
287
4
164
2
328
5
25
3
75
Mean number of head per toss =
=
= 2.47
Therefore, mean number of head per toss is 2.47
Question 3: The following table gives the number of branches and number of plants in the garden of a school.
Number of branches | 2 | 3 | 4 | 5 | 6 |
Number of plants | 49 | 43 | 57 | 38 | 13 |
Calculate the average number of branches per plant.
Solution:
Let the assumed mean(A) be = 4
hence, the table is,
Number of branches
Number of plants
2
49
-2
-98
3
43
-1
-43
4
57
0
0
5
38
1
38
6
13
2
26
Average Number of branches per plant =
=
= 3.615
Therefore, mean number of branches per plant is 3.615
Question 4: The following table gives the number of children of 150 families in a village
Number of children | 0 | 1 | 2 | 3 | 4 | 5 |
Number of families | 10 | 21 | 55 | 42 | 15 | 7 |
Find the average number of children per family.
Solution:
Let the assumed mean(A) be = 2
Hence, the table is,
Number of children
Number of families
0
10
-2
-20
1
21
-1
-21
2
55
0
0
3
42
1
42
4
15
2
30
5
7
3
21
Average number of children per family =
=
= 2.35 (approximately)
Therefore, the average number of children per family is 2.35 (approximately)
Question 5: The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
Marks | 15 | 20 | 22 | 24 | 25 | 30 | 33 | 38 | 45 |
Frequency | 5 | 8 | 11 | 20 | 23 | 18 | 13 | 3 | 1 |
Find the average number of marks.
Solution:
Let the assume mean (A) be = 25
hence, the table is,
Marks
Frequency
15
5
-10
-50
20
8
-5
-40
22
11
-3
-33
24
20
-1
-20
25
23
0
0
30
18
5
90
33
13
8
104
38
3
13
39
45
1
20
20
Average number of marks =
=
= 26.08 (approximately)
Therefore, average number of marks is 26.08 (approximately)
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
Number of students absent | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
Number of Days | 1 | 4 | 10 | 50 | 34 | 15 | 4 | 2 |
Find the mean number of students absent per day.
Solution:
Let mean assumed mean (A) be = 3
Number of students absent
Number of Days
0
1
-3
-3
1
4
-2
-8
2
10
-1
-10
3
50
0
0
4
34
1
34
5
15
2
30
6
4
3
12
7
2
4
8
Mean number of students absent per day =
=
= 3.525
Therefore, the mean number of students absent per day is 3.525
Question 7: In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:
Number of misprints per page | 0 | 1 | 2 | 3 | 4 | 5 |
Number of page | 154 | 95 | 36 | 9 | 5 | 1 |
Find the average number of misprints per page.
Solution:
Let the assumed mean (A) be = 2
Number of misprints per page
Number of page
0
154
-2
-308
1
95
-1
-95
2
36
0
0
3
9
1
9
4
5
2
10
5
1
3
3
Average number of misprints per day =
=
= 0.73
Therefore, the average number of misprints per day is 0.73
Question 8: Find the mean from the following frequency distribution of marks at a test in statistics:
Number of accidents | 0 | 1 | 2 | 3 | 4 |
Number of workers | 70 | 52 | 34 | 3 | 1 |
Find the average number of misprints per page.
Solution:
Let the assumed mean (A) = 2
Number of accidents
Number of workers
0
70
-2
-140
1
52
-1
-52
2
34
0
0
3
3
1
3
4
1
2
2
Average no of accidents per day workers =
=
= 0.83
Therefore, average no of accidents per day workers 0.83
Question 9: Find the mean from the following frequency distribution of marks at a test in statistics:
Marks | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
Number of students | 15 | 50 | 80 | 76 | 72 | 45 | 39 | 9 | 8 | 6 |
Solution:
Let the assumed mean (A) be = 25
Marks
Number of students
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
Mean =
=
= 22.075
Therefore, the mean is 22.075
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