Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Statistics (Exercise 14A)
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Page 1
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
Page 2
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Page 3
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
Page 4
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
=
i i
i
i
i
f x
f
=
= 103.05
Thus, the mean of the following data is 103.05.
8. If the mean of the following frequency distribution is 24, find the value of p.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 3 4 P 3 2
Sol:
The given data is shown as follows:
Class Frequency (fi) Mid values (x i) (fi xi)
0 10 3 5 15
10 20 4 15 60
20 30 p 25 25p
30 40 3 35 105
40 50 2 45 90
Total
fi = 12 + p fi xi = 270 + 25p
The mean of the given data is given by,
=
i i
i
i
i
f x
f
24 =
24 (12 + p) = 270 + 25p
288 + 24p = 270 + 25p
25p 24p = 288 270
p = 18
Hence, the value of p is 18.
9. The following distribution shows the daily pocket allowance of children of a locality. If the
mean pocket allowance is 18 , find the missing frequency f.
Daily
pocket
allowance
11 13 13 15 15 17 17 19 19 - 21 21 23 23 25
Page 5
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
=
i i
i
i
i
f x
f
=
= 103.05
Thus, the mean of the following data is 103.05.
8. If the mean of the following frequency distribution is 24, find the value of p.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 3 4 P 3 2
Sol:
The given data is shown as follows:
Class Frequency (fi) Mid values (x i) (fi xi)
0 10 3 5 15
10 20 4 15 60
20 30 p 25 25p
30 40 3 35 105
40 50 2 45 90
Total
fi = 12 + p fi xi = 270 + 25p
The mean of the given data is given by,
=
i i
i
i
i
f x
f
24 =
24 (12 + p) = 270 + 25p
288 + 24p = 270 + 25p
25p 24p = 288 270
p = 18
Hence, the value of p is 18.
9. The following distribution shows the daily pocket allowance of children of a locality. If the
mean pocket allowance is 18 , find the missing frequency f.
Daily
pocket
allowance
11 13 13 15 15 17 17 19 19 - 21 21 23 23 25
Number of
children
7 6 9 13 f 5 4
Sol:
The given data is shown as follows:
Daily pocket Number of
children (fi)
Class mark (xi) fi xi
11 13 7 12 84
13 15 6 14 84
15 17 9 16 144
17 19 13 18 234
19 21 f 20 20f
21 23 5 22 110
23 25 4 24 96
Total
fi = 44 + f fi xi = 752 + 20f
The mean of the given data is given by,
=
i i
i
i
i
f x
f
18 =
18 (44 + f) = 752 + 20f
792 + 18f = 752 + 20f
20f 18f = 792 752
2f = 40
f = 20
Hence, the value of f is 20.
10. The mean of following frequency distribution is 54. Find the value of p.
Class 0 20 20 40 40 60 60 80 80 100
Frequency 7 p 10 9 13
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
0 20 7 10 70
20 40 p 30 30p
40 60 10 50 500
60 80 9 70 630
80 100 13 90 1170
Total
fi = 39 + p fi xi = 2370 + 30p
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Document Description: RD Sharma Solutions: Statistics (Exercise 14A) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation.
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