Question

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.

Answer 1

\[n = 100 \]

\[\text{ Mean } = \bar{X} = 20 \]

\[SD = \sigma = 3 \]

\[\text{ Misread values are 21, 21 and } 18 . \]

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

\[ \Rightarrow{\frac{1}{100}} \sum x_i = 20_{} \]

\[ \Rightarrow \sum x_i = 20 \times 100 = 2000_{} \left[ {\text{ This sum is incorrect due to misread values }  .} \right] . . . . (1) \]

\[\text{ If three misread values are to be omitted, the total number of enteries will be 97 }  . \]

\[\text{ Also, } \sum x_i = 2000 - \left( {21 - 21 - 18} \right) = 1940 \]

\[\text{ Corrected }  \bar{X} ={\frac{1940}{97}} = 20 . . . . (2)\]

\[\sigma = 3 \]

\[ \Rightarrow \text{ Variance } = \sigma^2 = 9\]

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

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 - {20}^2 = 9\]

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 = 9 + 400 \]

\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 = 409\]

\[ \Rightarrow \sum_{} {x_i}^2 = 409 \times 100 = 40900 \left( \text{ This is an incorrect sum due to misread values }. \right) . . . (2)\]

\[\text{ Corrected } \sum_{} {x_i}^2 = 40900 - \left( {21}^2 + {21}^2 + {18}^2 \right)\]

\[ = 40900 - 441 - 441 - 324\]

\[ = 39694 . . . . (3) \]

\[\text{ From equations (2) and (3), we get: }  \]

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

\[ = \frac{1}{97} \times 39694 - \left( 20 \right)^2 \]

\[ = 409 . 216 - 400 \]

\[ = 9 . 216\]

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

\[ = \sqrt{{9 . 216}} \]

\[ = 3 . 0357 \]

 Thus, after omitting three values, the mean would be 20 and SD would be 3.0357.