RD Sharma Solutions: Statistics (Exercise 14.4)

# Statistics (Exercise 14.4) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

``` Page 2

172 -174
We have
N = 420
N/2 = 420/ 2 = 120
171.5-174.5 18
N =420
420
The cumulative frequency just greater than N/2 is 275 then 165.5 - 168.5 is the median class such,
that
L = 165.5, f = 142, F = 133 and h = 168.5 - 165.5 = 3
li-F
Mean = L + N2-FfXhL + -
2
- x h
f
= 166.5 + 1.63 = 168.13
3. Following is the distribution of I.Q of 100 students. Find the median I.Q.
I.Q:
No of
students:
Soln:
Class interval
(exclusive)
55- 64
65- 74
75- 84
85- 94
95-104
55-64 65-74 75-84 85-94 95-104 105-
1
114 115-124 125-134 135-144
2
22
Class interval
(inclusive)
54.5-64-5
64.5-74.5
74.5-84.5
84.5-94.5
94.5-104.5
9
8
22
Class interval
frequency
1
2
9
22
33 (f)
33
2 1
Cumulative
frequency
1
3
12
34 (f)
67
Page 3

172 -174
We have
N = 420
N/2 = 420/ 2 = 120
171.5-174.5 18
N =420
420
The cumulative frequency just greater than N/2 is 275 then 165.5 - 168.5 is the median class such,
that
L = 165.5, f = 142, F = 133 and h = 168.5 - 165.5 = 3
li-F
Mean = L + N2-FfXhL + -
2
- x h
f
= 166.5 + 1.63 = 168.13
3. Following is the distribution of I.Q of 100 students. Find the median I.Q.
I.Q:
No of
students:
Soln:
Class interval
(exclusive)
55- 64
65- 74
75- 84
85- 94
95-104
55-64 65-74 75-84 85-94 95-104 105-
1
114 115-124 125-134 135-144
2
22
Class interval
(inclusive)
54.5-64-5
64.5-74.5
74.5-84.5
84.5-94.5
94.5-104.5
9
8
22
Class interval
frequency
1
2
9
22
33 (f)
33
2 1
Cumulative
frequency
1
3
12
34 (f)
67
Page 4

172 -174
We have
N = 420
N/2 = 420/ 2 = 120
171.5-174.5 18
N =420
420
The cumulative frequency just greater than N/2 is 275 then 165.5 - 168.5 is the median class such,
that
L = 165.5, f = 142, F = 133 and h = 168.5 - 165.5 = 3
li-F
Mean = L + N2-FfXhL + -
2
- x h
f
= 166.5 + 1.63 = 168.13
3. Following is the distribution of I.Q of 100 students. Find the median I.Q.
I.Q:
No of
students:
Soln:
Class interval
(exclusive)
55- 64
65- 74
75- 84
85- 94
95-104
55-64 65-74 75-84 85-94 95-104 105-
1
114 115-124 125-134 135-144
2
22
Class interval
(inclusive)
54.5-64-5
64.5-74.5
74.5-84.5
84.5-94.5
94.5-104.5
9
8
22
Class interval
frequency
1
2
9
22
33 (f)
33
2 1
Cumulative
frequency
1
3
12
34 (f)
67
Page 5

172 -174
We have
N = 420
N/2 = 420/ 2 = 120
171.5-174.5 18
N =420
420
The cumulative frequency just greater than N/2 is 275 then 165.5 - 168.5 is the median class such,
that
L = 165.5, f = 142, F = 133 and h = 168.5 - 165.5 = 3
li-F
Mean = L + N2-FfXhL + -
2
- x h
f
= 166.5 + 1.63 = 168.13
3. Following is the distribution of I.Q of 100 students. Find the median I.Q.
I.Q:
No of
students:
Soln:
Class interval
(exclusive)
55- 64
65- 74
75- 84
85- 94
95-104
55-64 65-74 75-84 85-94 95-104 105-
1
114 115-124 125-134 135-144
2
22
Class interval
(inclusive)
54.5-64-5
64.5-74.5
74.5-84.5
84.5-94.5
94.5-104.5
9
8
22
Class interval
frequency
1
2
9
22
33 (f)
33
2 1
Cumulative
frequency
1
3
12
34 (f)
67
```

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

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