JEE Main 2023 — Statistics Question with Solution

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:

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

Therefore, the correct variance is 269.