Compute the median for each of the following data
(i)Marks No.of students (ii)Marks No.of students
Less than 10 0 More than 150 0
Less than 30 10 More than 140 12
Less than 50 25 More than 130 27
Less than 70 43 More than 120 60
Less than 90 65 More than 110 105
Less than 110 87 More than 100 124
Less than 130 96 More than 90 141
Less than 150 100 More than 80 150
Answers
Step-by-step explanation:
STATISTICS
Compute the median for each of the following data:
(i)Marks No. of students (ii)Marks No. of students
Less than 10 0 More than 150 0
Less than 30 10 More than 140 12
Less than 50 25 More than 130 27
Less than 70 43 More than 120 60
Less than 90 65 More than 110 105
Less than 110 87 More than 100 124
Less than 130 96 More than 90 141
Less than 150 100 More than 80 150
November 22, 2019avatar
Roopal Jinka
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ANSWER
(i)
Class interval Frequency Cumulative frequency
0−10 0 0
10−30 10 10
30−50 15 25
50−70 18 43
70−90 22 65
90−110 22 87
110−130 9 96
130−150 4 100
N=100
We have, N=100,
2
N
=50
The cumulative frequency just greater than
2
N
is 65 then median class is 70−90 such that,
l=70,h=90−70=20,f=22,cf=43
Median=l+
f
2
N
−cf
×h
⇒ Median=70+
22
50−43
×20
⇒ Median=70+
22
7×20
⇒ Median=70+6.36
∴ Median=76.36
(ii)
Class interval Frequency Cumulative Frequency
80−90 9 9
90−100 17 26
100−110 19 45
110−120 45 90
120−130 33 123
130−140 15 138
140−150 12 150
150−160 0 150
N=150
We have, N=150, then,
2
N
=
2
150
=75
The cumulative frequency just more than
2
N
is 90 then the median class is 110−120 such that
l=70,h=120−110=10,f=45,cf=45
Median=l+
f
2
N
−cf
×h
⇒ Median=110+
45
75−45
×10
⇒ Median=110+
45
30×10
⇒ Median=110+6.67
∴ Median=116.67