For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the corrected mean and standard deviation corresponding to the corrected figures are

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

\( \mu' = \frac{3955}{100} = \boxed{39.55} \)

Corrected Standard Deviation:

\( \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} \)

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96