The mean of 6 distinct observations is 6.5 and their variance is 10.25.

Question

The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are

Accepted Answer

Let other two numbers be a, (21 - a)

Now,

$$10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}$$

(Using formula for variance)

$$ \Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}$$

$$ \Rightarrow {a^2} + {(21 - a)^2} = 221$$

$$\therefore$$ a = 10 and (21 - a) = 21 - 10 = 11

So, remaining two observations are 10, 11.

$$\Rightarrow$$ Option (1) is correct.

Solutions · 1