Find the median wage from the following data:Wages(in Rs) 800 - 820 820 - 840 840 - 860 860 - 880 880 - 900 900 - 920 920 - 940Number of workers 7 14 19 25 20 10 5
Answers
Solution :-
wages(Rs.)----------- Fi ---------- CF
800 - 820 ----------- 7 ----------- 7
820 - 840 ----------- 14 ----------- 21
840 - 860 ----------- 19 ----------- 40
860 - 880 ----------- 25 ----------- 65
880 - 900 ----------- 20 ----------- 85
900 - 920 ----------- 10 ----------- 95
920 - 940 ----------- 5 ----------- 100
Here , n = 100 .
So,
→ (n/2) = 50 .
Then, Cumulative frequency greater than 50 is 65, corresponds to the class 860 - 880.
Therefore,
→ Class 860 - 880 is the median class.
Now,
- Median = l + [{(n/2) - cf} / f] * h
from data we have :-
- l = lower limit of median class = 860
- n = total frequency = 100
- cf = Cumulative frequency of class before median class = 40 .
- f = frequency of median class = 25 .
- h = size of class = 20 .
Putting all value we get :-
→ Median = 860 + [(50 - 40)/25] * 20
→ Median = 860 + (10/25) * 20
→ Median = 860 + 8
→ Median = 868 (Ans.)
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Class interval Frequency
0−100 2
100−200 5
200−300 x
300−400 12
400−500 17
500−600 20
600−700 y
700−800 9
800−900 7
900−1000 4
Medium
Correct option is D)
Computation of Median
Class interval Frequency (f) Cumulative frequency (cf)
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19+x
400-500 17 36+x
500-600 20 56+x
600-700 y 56+x+y
700-800 9 65+x+y
800-900 7 72+x + y
900-1000 4 76+x + y
Total = 100
We have,
N=∑f
i
=100
⇒76+x+y=100⇒x+y=24
It is given that the median is 525. Clearly, it lies in the class 500−600
∴l=500,h=100,f=20,F=36+x and N=100
Now,
Median=i+
f
2
N
−F
×h
⇒525=500+
20
50−(36+x)
×100
⇒525−500=(14−x)×5
⇒25=70−5x⇒5x=45⇒x=9
Putting x=9 inx+y=24, we get y=15.
Hence, x=9and y=15.