The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4, then absolute value of the difference of the other two observations is

Options

A. 1
B. 7
C. 5
D. 3

Answer

B

Explanation

Given mean `bar(x) = 4`
variance `sigma^(2)=5.20`
and number of observation n = 5
Let `x_(1)=3, x_(2)=4, x_(3)=4 and x_(4), x_(5)` be the five observations
So, `sum_(i=1)^(5)x_(i)=5*bar(x)=5xx4=20`
`rArr x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=20`
`rArr 3+4+4+x_(4)+x_(5)=20`
`rArr x_(4)+x_(5)=9" ...(i)" `
Now, variance `sigma^(2)=(sum_(i=1)^(5)x_(i)^(2))/(5)-(bar(x))^(2)`
`rArr (x_(1)^(2)+x_(2)^(2)+x_(3)^(2)+x_(4)^(2)+x_(5)^(2))/(5)-(4)^(2)=5.20`
`rArr (9+16+16+x_(4)^(2)+x_(5)^(2))/(5)=16+5.20`
`rArr 41+x_(4)^(2)+x_(5)^(2)=5xx21.20`
`rArr x_(4)^(2)+x_(5)^(2)=106-41`
`rArr x_(4)^(2)+x_(5)^(2)=65 " ...(ii)" `
`because (x_(4)+x_(5))^(2)=x_(4)^(2)+x_(5)^(2)+2x_(4)x_(5)`
`therefore 81=65+2x_(4)x_(5) " " `[from Eqs. (i) and (ii)]
`rArr 16=2x_(4)x_(5)`
`rArr x_(4)x_(5)=8 " ...(iii)" `
Now, `(|x_(4)-x_(5)|)^(2)=x_(4)^(2)+x_(5)^(2)-2x_(4)x_(5)`
`=65-16 " " `[from Eqs. (ii) and (iii)]
`=49`
`rArr |x_(4)-x_(5)|=7`