Statistics
Correction of mean and standard deviation
Grade Class 12
Question:
The mean and standard deviation of marks of 200 students were 40 and 15 respectively. Later it was discovered that a score of 40 was wrongly read as 50. The correct standard deviation is
39.95
14.98
224.5
None of these
Step-by-Step Solution
Key Concept: Use: corrected $\Sigma x = n\bar{x} - \text{wrong} + \text{correct}$. Corrected $\Sigma x^2 = n(\sigma^2+\bar{x}^2) - (\text{wrong})^2 + (\text{correct})^2$. Then recompute $\sigma$.
Corrected $\sigma=\sqrt{\frac{364100}{200}-(39.95)^2}=14.98$.
Correct Answer: 2