JEE Main 2025 Apr 3 Shift 2, Mathematics Q4: Measure of Dispersion

JEE Main2025Apr 3, Shift 2Mathematics

Let the Mean and Variance of five observations $x_{1} = 1$ , $x_{2} = 3$ , $x_{3} = a$ , $x_{4} = 7$ and $x_{5} = b$ , $a > b$ , be $5$ and $10$ respectively. Then the Variance of the observations $n + x_{n}$ , $n = 1, 2, \dots, 5$ is

A
$17$
B
$16.4$
C
$17.4$
D
$16$

Solution

From the mean condition: $1 + 3 + a + 7 + b = 25 \Rightarrow a + b = 14$ .

From the variance condition: $\frac{1 + 9 + a ^{2} + 49 + b ^{2}}{5} - 25 = 10 \Rightarrow a^{2} + b^{2} = 116$ .

Solving with $a > b$ : $(a, b) = (10, 4)$ .

Now compute $y_{n} = n + x_{n}$ for $n = 1, \dots, 5$ : $y_{1} = 2, y_{2} = 5, y_{3} = 13, y_{4} = 11, y_{5} = 9$ .

Mean of $y_{n}$ : $\overset{y}{ˉ} = \frac{2 + 5 + 13 + 11 + 9}{5} = 8$ .

$\sum y_{n}^{2} = 4 + 25 + 169 + 121 + 81 = 400$ .

Variance $= \frac{400}{5} - 64 = 80 - 64 = 16$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q4: Measure of Dispersion

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