QJEEM23Ma5314: JEE Main 2023 Mathematics - Class 11 JEEM23Ma5314 Solution: The mean and standard deviation of the marks of 10 students…

1. The initial mean is given by:

$\bar{x}=50$

So, the total sum of the marks initially was:

$\sum x_i = \bar{x} \times n = 50 \times 10 = 500$

1. We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:

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

1. The initial variance is given as:

$\sigma^2 = 144$

We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:

$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2 = 144 + 50^2 = 2644$

Then, the sum of the squares of the initial marks is:

$\sum x_i^2 = n \times \frac{\sum x_i^2}{n} = 10 \times 2594 = 26440$

1. The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:

$\sum x_{i{\text{correct}}}^2 = \sum x_i^2 - 45^2 - 50^2 + 20^2 + 25^2 = 26400 - 45^2 - 50^2 + 20^2 + 25^2 = 22940$

1. Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:

$\sigma_{\text{correct}}^2 = \frac{\sum x_{i{\text{correct}}}^2}{n} - \left(\frac{\sum x{i_{\text{correct}}}}{n}\right)^2 = \frac{22940}{10} - \left(\frac{450}{10}\right)^2 = 2294 - 45^2 = 2294 - 2025 = 269$

Therefore, the correct variance is 269.