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 are the correct mean and standard deviation?

Answer 1

\[n = 100 \]

\[\text{ Mean } = \bar{X} = 40 \]

\[\sigma = SD = 5 . 1\]

\[\frac{1}{n}\sum x_i = \bar{X} \]

\[ \Rightarrow \sum x_i = 100 \times 40 = 4000 \left( \text{ This is an incorrect reading due to misread values . } \right)\]

\[ \text{ Corrected sum  } , \sum x_i = 4000 - 50 + 40 \]

\[ = 3990\]

\[ \Rightarrow \text{ Corrected mean }= \frac{\text{ Corrected sum } }{100}\]

\[ = \frac{3990}{100}\]

\[ = 39 . 9 . . . (1)\]

To find the corrected SD:

\[\sqrt{\text{ Variance } } = \sigma \]

\[ \Rightarrow \sigma^2 = \left( 5 . 1 \right)^2 = \text{ Variance } \]

\[\text{ According to the formula } , \]

\[\frac{1}{n} \sum_{} {x_i}^2 - \left( \bar{X} \right)^2 = \text{ Variance} \]

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 - \left( 40 \right)^2 = 26 . 01\]

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 - 1600 = 26 . 01\]

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 = 1626 . 01\]

\[ \Rightarrow \sum_{} {x_i}^2 = 162601 \left( \text{ But, this is incorrect due t o misread values } \right)\]

\[ \Rightarrow \text{ Corrected } \sum_{} {x_i}^2 = 162601 - {50}^2 + {40}^2 \]

\[ = 161701 . . . . (2)\]

\[\text{ Corrected variance } = \frac{1}{100}\text{ Corrected } \sum_{} {x_i}^2 - \left( \text{ Corrected mean }  \right)^2 \]

\[ = \frac{161701}{100} - \left( 39 . 9 \right)^2 \left[\text{  using equations (1) and (2) }  \right]\]

\[ = 1617 . 01 - 1592 . 01\]

\[ = 25\]

\[ \text{ Corrected SD }  = \sqrt{{\text{ Corrected variance} }}\]

\[ = \sqrt{{25}} \]

\[ = 5\]

 Corrected mean = 39.9
 Corrected standard deviation = 5