Question

In class test marks scored by students are given in the following frequency distribution.
Marks 0-6 6-12 12-18 18-24 24-30
students 1 4 9 3 3
Find mean median of the data​

Answer 1

Median=15.33

Mean=15.9

Step-by-step explanation:

Sum of frequencies=N=20

N is even then

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

Median observation=$\frac{10^{th}+11^{th}}{2}=\frac{14+14}{2}=14$

It lies in interval 12-18

Median class=12-18

l=12

h=6

f=9

c.f=5

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

Where N=Total number of observation

f=Frequency of  median class

c.f =Cumulative frequency of class preceding median class

h=Size of class

l=Lower limit of median class

Substitute the values then we get

Median=$12+\frac{\frac{20}{2}-5}{9}\times 6$

Median=$12+3.33=15.33$

Class mark=$x_i=\frac{lower\;limit+upper\;limit}{2}$

By direct method

Mean$(\bar x)=\frac{\sum f_ix_i}{\sum f_i}$

Mean$(\bar x)=\frac{318}{20}=15.9$

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Answer 2

Answer:

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Step-by-step explanation:

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