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Question

The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:
Class intervalFrequencyClass intervalFrequency01002500600201002005600700f2200300f170080093004001280090074005001790010004


Solution

Class interval Frequency Cumulative frequency
0 – 100 2 2
100 – 200 5 7
200 – 300 f1 7 + f1
300 – 400 12 19 + f1
400 – 500 17 36 + f­1 (F)
500 – 600 20 (f) 56 + f1
600 – 700 f2 56 + f1 + f2
700 – 800 9 65 + f1 + f2
800 – 900 7 72 + f1 + f2
900 – 1000 4 76 + f1 + f2
  N = 100  

Given

Median = 525

Then, median class = 500 – 600

L = 500, f = 20, F = 36 + f1, h = 600 – 500 = 100
Median space equals space L plus fraction numerator begin display style N over 2 end style minus F over denominator f end fraction cross times h 525 space equals space 500 plus fraction numerator 50 minus left parenthesis 36 plus f subscript 1 right parenthesis over denominator 20 end fraction cross times 100 525 space equals space 500 space plus fraction numerator 50 minus 36 minus f subscript 1 over denominator 20 end fraction cross times 100 25 space equals space left parenthesis 14 space � space f subscript 1 right parenthesis space cross times space 5
5 = 14 – f1
f1 = 14 – 5 = 9

Given

Sum of frequencies = 100

2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

85 + f2 = 100

f2 = 100 – 85 = 15

f1 = 9 and f2 = 15


Mathematics
RD Sharma
Standard X

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