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If the median of the following frequency distribution is 32.5. Find the values of f_1\: \: \: and\: \: \: f_2.

Class

0-10 10-20 20-30 30-40 40-50 50-60 60-70

Total

Frequency

f_1 5 9 12 f_2 3 2 40

 

 

 

 
 
 
 
 

Answers (1)

Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total
Frequency f_1 5 9 12 f_2 3 2 40
Cumulative frequency f_1 5+f_1 14+f_1 26+f_1 26+f_1+f_2 29+f_1+f_2 31+f_1+f_2  

 

N=40=31+f_1+f_2

Median = l+\frac{\left [ \frac{n}{2}-cF \right ]}{F}\times h

 From the table,

n/2 = 40/2 = 20

l=30

f=12

CF=(14+F_1)

h=10

Putting values in the formula

 

32.5=30+\frac{\left [ 20-(14+F_1) \right ]}{12}\times 10

2.5=\frac{\left [ (6-F_1) \right ]}{12}\times 10

2.5\times 12={\left [ (6-F_1) \right ]\times 10

30=60-10f_1

f_1=\frac{30}{10}

f_1=3

From the previous data

31+f_1+f_2=40

31+3+f_2=40

f_2=6

Hence the missing frequencies are 3 and 6.

Posted by

Safeer PP

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