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, then the correct standard deviation is

• A. $4$
• B. $6$
• C. $3$
• D. $5$

Answer 1

Answer: (D)

Standard deviation $\left(\sigma \right)=\sqrt{\frac{1}{n} \displaystyle \sum _{i = 1}^{n} x_{i}^{2} - \frac{1}{n^{2}} \left(\displaystyle \sum _{i = 1}^{n} x_{i}\right)^{2}}$
$=\sqrt{\frac{1}{n} \displaystyle \sum _{i = 1}^{n} x_{i}^{2} - \left(\overset{-}{x}\right)^{2}}$
$\Rightarrow 5.1=\sqrt{\frac{1}{100} \times \text{Incorrect} \displaystyle \sum _{i = 1}^{n} x_{i}^{2} - \left(40\right)^{2}}$
Or, $26.01=\frac{1}{100}\times $ Incorrect $\displaystyle \sum _{i = 1}^{n} x_{i}^{2}-1600$
Therefore,
Incorrect $\displaystyle \sum _{i = 1}^{n}x_{i}^{2}=100\left(\right.26.01+1600\left.\right)=162601$
Correct mean, $\bar{x }=\frac{3990}{100}=39.9$
Now, Correct $\displaystyle \sum _{i = 1}^{n} x_{i}^{2}=$ Incorrect $\displaystyle \sum _{i = 1}^{n} x_{i}^{2}-\left(50\right)^{2}+\left(40\right)^{2}$
$=162601-2500+1600=161701$
Therefore, correct standard deviation
$=\sqrt{\frac{C o r r e c t \displaystyle \sum x_{i}^{2}}{n} - \left(C o r r e c t m e a n\right)^{2}}$
$=\sqrt{\frac{161701}{100} - \left(39.9\right)^{2}}=\sqrt{1617.01 - 1592.01}=\sqrt{25}=5$