Question

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 1

Correct Answer: 269

Let the total sum of the marks of the 10 students be \( \Sigma x_i \). From the given mean of 50: \[ \frac{\Sigma x_i}{10} = 50 \quad \Rightarrow \quad \Sigma x_i = 500. \] Now, correcting the sum of the marks: \[ \Sigma x_i = 500 - 45 - 50 + 20 + 25 = 450. \] The corrected sum of squares \( \Sigma x_i^2 \) is: \[ \sigma^2 = \frac{\Sigma x_i^2}{10} - \left( \frac{\Sigma x_i}{10} \right)^2. \] Using the given standard deviation of 12: \[ \sigma^2 = 144 \quad \Rightarrow \quad \frac{\Sigma x_i^2}{10} - 2500 = 144. \] Solving: \[ \Sigma x_i^2 = 26440. \] Correcting the squares for the wrongly recorded marks: \[ \Sigma x_i^2 = 26440 - (45^2) - (50^2) + (20^2) + (25^2) = 26440 - 2025 - 2500 + 400 + 625 = 22940. \] Now, calculating the corrected variance: \[ \sigma^2 = \frac{22940}{10} - \left( \frac{450}{10} \right)^2 = 2294 - 2025 = 269. \] Thus, the correct variance is 269.