JEE Main 2024 — Statistics Question with Solution

To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry. The original erroneous observation was 8, and the correct observation is 12. The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.

The sum of all 20 original observations ($S_{\text{original}}$) can be calculated from the mean formula:

$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$10=\frac{S_{\text{original}}}{20}$

$S_{\text{original}}=10\times 20=200$

The corrected sum of observations ($S_{\text{correct}}$) will replace the incorrect observation (8) with the correct one (12):

$S_{\text{correct}}=S_{\text{original}}-8+12=200-8+12=204$

The corrected mean (${\mu }_{\text{correct}}$) is:

${\mu }_{\text{correct}}=\frac{S_{\text{correct}}}{20}=\frac{204}{20}=10.2$

To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data. The original sum of squares ($S{S}_{\text{original}}$) is calculated from the original standard deviation formula, where $\sigma =2$:

${\sigma }^2=\frac{SS}{n}$

$4=\frac{S{S}_{\text{original}}}{20}$

$S{S}_{\text{original}}=4\times 20=80$

To calculate the corrected sum of squares ($S{S}_{\text{correct}}$), we need to adjust $S{S}_{\text{original}}$ by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation:

$S{S}_{\text{correct}}=S{S}_{\text{original}}-(8-10{)}^2+(12-10.2{)}^2$

$S{S}_{\text{correct}}=80-(-2{)}^2+(1.8{)}^2$

$S{S}_{\text{correct}}=80-4+3.24=79.24$

Finally, the corrected standard deviation (${\sigma }_{\text{correct}}$) is:

${\sigma }_{\text{correct}}=\sqrt{\frac{S{S}_{\text{correct}}}{20}}$

${\sigma }_{\text{correct}}=\sqrt{\frac{79.24}{20}}$

${\sigma }_{\text{correct}}=\sqrt{3.962}$

The corrected standard deviation is closest to the value given in Option B, which is

$\sqrt{3.96}$