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Answers
Solution :-
Height ------- Frequency ------- LB --------- CF
149.5-154.5 ------- 14 ------------149.5 ---------14
154.5-159.5 -------11 ------------154.5 ----------25
159.5-164.5 -------11 -------------159.5 -------- 36
164.5-169.5 -------10 ------------164.5 -------- 46
169.5-174.5 ------ 8 -------------169.5 -------- 54
-----------------⅀ F = 54
Here , n = 54 .
So,
→ (n/2) = 27.
Then, Cumulative frequency greater than 27 is 36, corresponds to the class 159.5 - 164.5
Therefore,
→ Class 159.5 - 164.5 is the median class.
Now,
- Median = l + [{(n/2) - cf} / f] * h
from data we have :-
- l = Lower limit of median class = 159.5
- n = Total frequency = 54
- cf = Cumulative frequency of class before median class = 25.
- f = Frequency of median class = 11 .
h = Size of class = 5
Putting all value we get :-
→ Median = 159.5 + [(27 - 25)/11] * 5
→ Median = 159.5 + (2/11) * 5
→ Median = 159.5 + 0.91
→ Median = 160.41 (Ans.)
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Given : Distribution of height
To Find : Median of Data
Solution:
As interval are continues , please convert into continuous interval and arrange in ascending order
Class Interval f cf
149.5 - 154.5 14 14
154.5 - 159.5 11 25
159.5 - 164.5 11 36
164.5 - 169.5 10 46
169.5 - 174.5 8 54
∑f = 54
54/2 = 27
27 frequency lies in Class Interval 159.5 - 164.5
Median = 159.5 + 5 * ( 27 - 25 ) / 11
= 159.5 + 0.91
= 160.41
Median of Data is 160.41
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