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Class intervalless than 140less than 150less thao 155lesston 160less than 165frequency 47181165solve graphically ​

Answer»

Answer:

Where CF= cumulative frequency

Median =1+

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

h×

F

(

2

N

−cf)

⎭

⎪

⎪

⎪

⎬

⎪

⎪

⎪

⎫

Here

N=100

N/2=50

cf just GREATER than 50 is 56

Median class =150−155

So, I=150, h=5, f=22 and

cf=cf of preceding class i.e 34

Substitute all the VALUE in the above formula, we get

Median =150+{5×(50−34)/22}

=150+3.64

=153.64

solution


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