Question

The means 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

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

Answer 1

Answer: (C)

4 and 7

Answer 2

Let the two unknown items be x and y, then

Mean $= 4 \Rightarrow \frac{1 + 2 + 6 + x + y}{5} = 4$

⇒ $x + y = 11$ .....(i)

and variance = 5. 2

⇒ $\frac{1^{2} + 2^{2} + 6^{2} + x^{2} + y^{2}}{5} - (\text{mean})^{2} = 5.2$

$41 + x^{2} + y^{2} = 5\lbrack 5.2 + (4)^{2}\rbrack$

$41 + x^{2} + y^{2} = 106$

$x^{2} + y^{2} = 65$ .....(ii)

Solving (i) and (ii) for $x$ and y, we get

$x = 4,y = 7$ or $x = 7,y = 4$.