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. Find the correct standard deviation.

• A. $4$
• B. $6$
• C. $3$
• D. $5$

Answer 1

The Correct Option is D

Also, 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}\sum\limits_{i=1}^{n} x_{i}^{2} - \left(\bar{x}\right)^{2}} $ i.e. $5.1 = \sqrt{\frac{1}{100}\times {\text{Incorrect}} \sum\limits _{i=1}^{n} x_{i}^{2} - \left(40\right)^{2}}$ or, $26.01 = \frac{1}{100} \times {\text{Incorrect}} \sum\limits _{i=1}^{n} x_{i}^{2} -1600$ Therefore, Incorrect $\sum\limits _{i=1}^{n} x_{i}^{2} = 100\left(26.01+1600\right) = 162601$ Now, Correct $\sum\limits _{i=1}^{n} x_{i}^{2} =$ Incorrect $\sum\limits _{i=1}^{n} x_{i}^{2} -\left(50\right)^{2} +\left(40\right)^{2}$ $= 162601 -2500+1600 = 161701 $ Therefore correct standard deviation $ = \sqrt{\frac{\text{Correct} \sum x_{i}^{2}}{n}-\left({\text{Correct mean}}\right)^{2}}$ $= \sqrt{\frac{161701}{100}-\left(39.9\right)^{2} } $ $ = \sqrt{1617.01 -1592.01} $ $= \sqrt{25}$ $ = 5$