The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took 50 instead of 40 for one observation. What is the correct mean and standard deviation?

Correct option is C

Solution:

​$$Mean = 40 \Rightarrow \frac{\Sigma_{i=1}^{100} x_i}{100} = 40$$​​

​$$\Rightarrow \Sigma_{i=1}^{100} x_i = 4000$$​​

​$$\text{SD} = 5.1 \Rightarrow \sigma^2 = (5.1)^2\\\ \\\Rightarrow \frac{\Sigma_{i=1}^{100} x_i^2}{100} - (40)^2 = 26.01\\\ \\\Rightarrow \frac{\Sigma_{i=1}^{100} x_i^2}{100} = 162601\\\ \\\text{Thus, incorrect} \Sigma_{i=1}^{100} x_i = 4000 \text{ and incorrect} \Sigma_{i=1}^{100} x_i^2 = 162601\\\ \\\text{Now, incorrect} \Sigma_{i=1}^{100} x_i = 4000\\\ \\\text{Correct} \Sigma_{i=1}^{100} x_i^2 = \{162601 - (50)^2 + (40)^2\} = 161701\\\ \\\text{Correct variance} = \frac{\text{Correct} \Sigma_{i=1}^{100} x_i^2}{100} - (\text{Correct Mean})^2\\\ \\= \frac{161701}{100} - (39.9)^2\\\ \\= (1617.01) - (40 - 0.1)^2\\\ \\= (1617.01) - (1600 + 0.01 \times 8)\\\ \\= (1617.01 - 1592.01) = 25\\\ \\\text{Correct SD} = \sqrt{25} = 5 \\\ \\\text{Hence, correct mean = 39.9 and correct SD = 5.}$$​​