Question

the median of the following data is 49.2 . Find the values of f1 and de , if £fi =90

marks – 20-30, 30-40,40-50, 50-60, 60-70, 70-80,80-90

Frequency – f1 , 15,25, 20, f2, 8, 10 respectively ​

Answer 1

the median of the following data is 49.2 . The values of f1 and f2 are 7 and 5 respectively , if £fi =90  marks – 20-30, 30-40,40-50, 50-60, 60-70, 70-80,80-90  Frequency – f1 , 15,25, 20, f2, 8, 10  ​

 Step-by-step explanation: Given that median=49.2 From the table (shown below), we get f1+15+25+20+f2+8+10=90. That is, f1+f2=12. Formula for  $median=L+\frac{\frac{n}{2} -cf}{f}\times i$ where, L is the lower limit of the median class f is the frequency of the median class cf is the cumulative frequency of the class preceding to the median class i is the width of the median class N is the sum of all frequencies The median class being the class which contains the (N/2)th observation. Here the median class is 40-50, since the median is 49.2 so that  L=40, N=90, cf=f1+15, f=25 and i=10 putting these values in the equation of median we get, $49.2=40+\frac{45-(f1+15)}{25}\times10\\ 9.2=\frac{300-f1}{25}\\ 230=300-10f1\\10f1=70\\f1=7\\therefore, f2=12-7=5$

Answer 2

$f_1=7,f_2=5$

Step-by-step explanation:

N=$\sum f_i=90$

$78+f_1+f_2=90$

$f_1+f_2=90-78=12$

Median=49.2

N is even , therefore

Median class=$\frac{\frac{N}{2}^{th}+(\frac{N}{2}+1)^{th}}{2})$

Mean of 45th and 46th observation lie in class 40-50

Median class=40-50

Median=$l+\frac{\frac{N}{2}-cf}{f}\times h$

Where l=Lower limit of median class

f=Frequency of median class

c.f=Cumulative frequency of preceding median class

h=Size of class

We have f=25

h=10

l=40

c.f=$f_1+15$

Substitute the values in the given formula

$49.2=40+\frac{45-f_1-15}{25}\times 10$

$49.2-40=\frac{30-f_1}{25}\times 10$

$9.2\times \frac{25}{10}=30-f_1$

$23=30-f_1$

$f_1=30-23=7$

Substitute the value then we get

$7+f_2=12$

$f_2=12-7=5$

Hence, $f_1=7,f_2=5$

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