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:
Correct Option: 1
Solution:
Let other two numbers be $\mathrm{a},(21-\mathrm{a})$
Now,
$10.25=\frac{\left(4+16+25+49+a^{2}+(21-a)^{2}\right)}{6}-(6.5)^{2}$
(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.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.