Question

If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a-b)^2$ is equal to :

• A. 16
• B. 12
• C. 24
• D. 32

Hint:

When given mean and variance, you can set up two equations involving the unknown variables. Often, you don't need to solve for the variables individually. Look for algebraic identities like $(a-b)^2 = (a+b)^2 - 4ab$ or $(a-b)^2 = (a^2+b^2) - 2ab$ to find the required expression directly.

Answer 1

The Correct Option is A

The data set is $\{6, 10, 7, 13, a, 12, b, 12\}$. There are $n=8$ observations.
The mean ($\bar{x}$) is given as 9.
$\bar{x} = \frac{\sum x_i}{n} = \frac{6+10+7+13+a+12+b+12}{8} = 9$.
$\frac{60+a+b}{8} = 9$.
$60 + a + b = 72 \implies a+b = 12$. (Equation 1)
The variance ($\sigma^2$) is given as $\frac{37}{4}$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\frac{37}{4} = \frac{6^2+10^2+7^2+13^2+a^2+12^2+b^2+12^2}{8} - (9)^2$.
$\frac{37}{4} = \frac{36+100+49+169+a^2+144+b^2+144}{8} - 81$.
$\frac{37}{4} = \frac{642+a^2+b^2}{8} - 81$.
Add 81 to both sides: $\frac{37}{4} + 81 = \frac{642+a^2+b^2}{8}$.
$\frac{37 + 324}{4} = \frac{642+a^2+b^2}{8}$.
$\frac{361}{4} = \frac{642+a^2+b^2}{8}$.
Multiply both sides by 8: $2 \times 361 = 642+a^2+b^2$.
$722 = 642 + a^2 + b^2$.
$a^2 + b^2 = 722 - 642 = 80$. (Equation 2)
Now we have a system of two equations:
1) $a+b=12$
2) $a^2+b^2=80$
We want to find $(a-b)^2$.
We know that $(a+b)^2 = a^2+b^2+2ab$.
$12^2 = 80 + 2ab \implies 144 = 80 + 2ab \implies 2ab = 64 \implies ab = 32$.
Now use the identity for $(a-b)^2$:
$(a-b)^2 = a^2+b^2-2ab$.
$(a-b)^2 = 80 - 64 = 16$.