JEE Main 2025 Apr 7 Shift 1, Mathematics Q12: Measure of Dispersion

JEE Main2025Apr 7, Shift 1Mathematics

The mean and standard deviation of 100 observations are 40 and 5.1, respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to

A
$445$
B
$451$
C
$447$
D
$449$

Solution

Correct mean. Original sum $= 100 \times 40 = 4000$ . Replace the wrong value 50 with the correct 40:

$μ = \frac{4000 - 50 + 40}{100} = \frac{3990}{100} = 39.9$

Correct standard deviation. Use $σ^{2} = \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2}$ .

Incorrect $\sum x_{i}^{2} = 100 \cdot [(5.1)^{2} + 4 0^{2}] = 100 (26.01 + 1600) = 162601$ .

Correct $\sum x_{i}^{2} = 162601 - 5 0^{2} + 4 0^{2} = 162601 - 2500 + 1600 = 161701$ .

$σ^{2} = \frac{161701}{100} - (39.9)^{2} = 1617.01 - 1592.01 = 25 \Rightarrow σ = 5$ .

Compute the required expression.

$10 (μ + σ) = 10 (39.9 + 5) = 10 \times 44.9 = 449$

$∴$ Option (4).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q12: Measure of Dispersion

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