Question

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

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (D)

Given, $n=100$, Incorrect mean, $\bar{x}=40$, Incorrect standard deviation, $\sigma =5.1$
Now, $\bar{x}=\frac{\Sigma x_i}{n}$
$\Rightarrow 40=\frac{\Sigma x_i}{100}$
$\Rightarrow \Sigma x_i=4000$
Correct $\Sigma x_i=$ Incorrect $\Sigma x_i-50+40$
$=4000-50+40=3990$
$\therefore$ Correct $\bar{x}=\frac{\text{ Correct }\Sigma x_i}{n}=\frac{3990}{100}=39.9$
Also, $\sigma =\sqrt{\frac{\sum x_i^2}{n}-(\bar{x}{)}^2}$
$\Rightarrow 5.1=\sqrt{\frac{\sum x_i^2}{100}-(40{)}^2}$
$\Rightarrow (5.1{)}^2=\frac{\sum x_i^2}{100}-1600$
$\Rightarrow 26.01+1600=\frac{\sum x_i^2}{100}$
$\Rightarrow \sum x_i^2=162601$
Correct $$\begin{gathered} \sum x_i^2=\text{ Incorret }\sum x_i^2-(50{)}^2+(40{)}^2 \\ =162601-(50{)}^2+(40{)}^2=161701 \end{gathered}$$
$\therefore$ Correct standard deviation $=\sqrt{\frac{\text{ Correct }\sum x_i^2}{n}-(\text{ Correct }\bar{x}{)}^2}=\sqrt{\frac{161701}{100}-(39.9{)}^2}$
$=\sqrt{1617.01-1592.01}=\sqrt{25}=5$