Statistics – Mean and Variance – JEE Main 02 April 2025 Shift 2

Solution:

Formula for mean:
$$\bar{x} = \frac{\sum x_i}{N}$$

$$9 = \frac{6+4+a+8+b+12+10+13}{8}$$
$$72 = 53+a+b$$
$$a+b = 19$$

Formula for variance:
$$\sigma^2 = \frac{\sum x_i^2}{N} – (\bar{x})^2$$

$$9.25 = \frac{6^2+4^2+a^2+8^2+b^2+12^2+10^2+13^2}{8} – (9)^2$$
$$\frac{37}{4} = \frac{36+16+a^2+64+b^2+144+100+169}{8} – 81$$
$$\frac{37}{4} + 81 = \frac{529+a^2+b^2}{8}$$
$$\frac{37+324}{4} = \frac{529+a^2+b^2}{8}$$
$$\frac{361}{4} = \frac{529+a^2+b^2}{8}$$
$$361 \times 2 = 529+a^2+b^2$$
$$722 = 529+a^2+b^2$$
$$a^2+b^2 = 193$$

Formula for square of sum:
$$(a+b)^2 = a^2+b^2+2ab$$

$$(19)^2 = 193+2ab$$
$$361 = 193+2ab$$
$$2ab = 168$$
$$ab = 84$$

$$a+b+ab = 19+84 = 103$$

Ans. (2)