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if x(1),x(2)….x(10) are 10 observations,...

if `x_(1),x_(2)….x_(10)` are 10 observations, in which mean of `x_(1),x_(2),x_(3),x_(4)` is 11 while mean of `x_(5),x_(6)….x_(10)` is 16. also `x_(1)^(2)+x_(2)^(2)+….x_(10)^(2)=2000` then value of standard devition is

A

`2sqrt(2)`

B

2

C

4

D

`sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
key idea Formula of standard deviation (sigma), for n observations
`=sqrt((Sigmax_(1)^(2))/(n)-((Sigmax_(1))/(n))^(2))`
Given 10 observations are `x_(1),x_(2),x_(3), … , x_(10)`
` therefore (x_(1)+x_(2)+x_(3)+x_(4))/(4) =11`
`rArr x_(1)+x_(2)+x_(3)+x_(4)=44 " ...(i)" `
`and (x_(5)+x_(6)+x_(7)+x_(8)+x_(9)+x_(10))/(6)=16`
` implies x_(5)+x_(6)+x_(7)+x_(8)+x_(9)+x_(10)=96 " ...(ii)" `
So, mean of given 10 observations `=(44+96)/(10)=(140)/(10)=14`
Since, the sum of squares of all the observations = 2000
`x_(1)^(2)+x_(2)^(2)+x_(3)^(2)+ ... +x_(10)^(2)=2000 " ...(iii)'' `
Now, `sigma^(2)=("standard deviation")^(2)=(Sigma x_(i)^(2))/(10)-((Sigma x_(i)^(2))/(10))^(2)`
`=(2000)/(10)-(14)^(2)=200-196=4`
So, `sigma =2`
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