JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Statistics

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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 ______.

Answer: 269

Solution

Given:

\overset{ˉ}{X} = 50

σ = 12

, so

σ^{2} = 144

.

Original sum:

\sum x_{i} = 10 \times 50 = 500

Corrected sum:

\sum x_{i, correct} = 500 - 45 - 50 + 20 + 25 = 450

From

σ^{2} = \frac{\sum x _{i}^{2}}{n} - \overset{ˉ}{X}^{2}

144 = \frac{\sum x _{i}^{2}}{10} - 2500 ⟹ \sum x_{i}^{2} = 26440

Corrected sum of squares:

\sum x_{i, correct}^{2} = 26440 - 4 5^{2} - 5 0^{2} + 2 0^{2} + 2 5^{2} = 26440 - 2025 - 2500 + 400 + 625 = 22940

Correct variance:

σ_{correct}^{2} = \frac{22940}{10} - (\frac{450}{10})^{2} = 2294 - 2025 = 269

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer

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