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JEE 2023 Mathematics PYQ — The mean and standard deviation of the marks of 10 students were … | Mathem Solvex | Mathem Solvex

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The mean and standard deviation of the marks of 10 students were found to be 50 and 12, respectively, Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50, respectively. Then, the correct variance is ______.

Choose the correct answer:

A.
269
(Correct Answer)
B.
270
C.
271
D.
272

Correct Answer:

269

Explanation

Mean $\overset{x}{ˉ} = 50 \Rightarrow \sum x_{i} = 500$
Variance $σ^{2} = \frac{\sum x _{i}^{2}}{10} - (\overset{x}{ˉ})^{2} = 1 2^{2}$
$\frac{\sum x _{i}^{2}}{10} - 2500 = 144$
$\sum x_{i}^{2} = (144 + 2500) \times 10 = 26440$

\text{Now, correct mean} = \frac{500 - 45 - 50 + 20 + 25}{10} = \frac{450}{10} = 45

\text{Also, correct } \sum x_i^2 = 26440 - (45)^2 - (50)^2 + (20)^2 + (25)^2 = 26440 - 70 \times 20 - 70 \times 30 = 22940

\therefore \text{Correct variance} = \frac{22940}{10} - (\text{Correct mean})^2 = 2294 - (45)^2 = 2294 - 2025 = 269

Explanation

\text{Now, correct mean} = \frac{500 - 45 - 50 + 20 + 25}{10} = \frac{450}{10} = 45

\text{Also, correct } \sum x_i^2 = 26440 - (45)^2 - (50)^2 + (20)^2 + (25)^2 = 26440 - 70 \times 20 - 70 \times 30 = 22940

\therefore \text{Correct variance} = \frac{22940}{10} - (\text{Correct mean})^2 = 2294 - (45)^2 = 2294 - 2025 = 269

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