Question

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

• A. 1
• B. 3
• C. 7
• D. 5

Answer 1

Answer: (C)

7

$\Rightarrow$ Let other two observations are $x,y$

$\therefore$ $Mean=\frac{3+4+4+x+y}{5}$

$\therefore$ $4=\frac{11+x+y}{5}$

$\therefore$ $20=11+x+y$

$\therefore$ $x+y=9$ ------ ( 1 )

$Variance=\frac{1}{5}\left[\right(3-4{)}^2+(4-4{)}^2+(4-4{)}^2+(x-4{)}^2+(y-4{)}^2]$

$5.20=\frac{1}{5}[1+0+0+x^2-8x+16+y^2-8y+16]$

$26=33+x^2+y^2-8(x+y)$

$26=33+x^2+y^2-8\left(9\right)$ [ From ( 1 ) ]

$\therefore$ $x^2+y^2=65$ ----- ( 2 )

$x+y=9$ [ From ( 1 )]

$x^2+y^2+2xy=81$ [ Squaring both sides ]

$65+2xy=81$ [ From ( 2 ) ]

$2xy=16$

$xy=8$

$\therefore$ $x=\frac{8}{y}$ -- ( 3 )

Substituting ( 3 ) in ( 1 ),

$\frac{8}{y}+y=9$

$y^2+8=9y$

$y^2-9y+8=0$

$\therefore$ $y=8$ and $y=1$

For $y=8$

$x+8=9$

$\therefore$ $x=1$

$\therefore$ We get $x=1$ and $y=8$

$\therefore$ Difference between other two observation $=8-1=7$