The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1,2 and 6, then the other two are

Options

A. 2 and 9
B. 3 and 8
C. 4 and 7
D. 5 and 6

Answer

C

Explanation

Let the two unknown items be x and y, then
Mean`=4 implies (1+2+6+x+y)/(5)=4`
`implies x+y=11`
and variance=5.2
`implies (1^(2)+2^(2)+6^(2)+x^(2)+y^(2))/(5)-("Mean")^(2)=5.2`
`41+x^(2)+y^(2)=5[5.2+(4)^(2)]`
`41+x^(2)+y^(2)=106`
`x^(2)+y^(2)=65`
Solving (1) and (2) for x and y, we get
x=4, y=7 or x=7, y=4