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प्रश्न
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.
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उत्तर
\[n = 100 \]
\[\text{ Mean } = \bar{X} = 20 \]
\[SD = \sigma = 3 \]
\[\text{ Misread values are 21, 21 and } 18 . \]
\[ \frac{1}{n}\sum_{} x_i = \bar{X} \]
\[ \Rightarrow{\frac{1}{100}} \sum x_i = 20_{} \]
\[ \Rightarrow \sum x_i = 20 \times 100 = 2000_{} \left[ {\text{ This sum is incorrect due to misread values } .} \right] . . . . (1) \]
\[\text{ If three misread values are to be omitted, the total number of enteries will be 97 } . \]
\[\text{ Also, } \sum x_i = 2000 - \left( {21 - 21 - 18} \right) = 1940 \]
\[\text{ Corrected } \bar{X} ={\frac{1940}{97}} = 20 . . . . (2)\]
\[\sigma = 3 \]
\[ \Rightarrow \text{ Variance } = \sigma^2 = 9\]
\[ = \text{ Variance } = \frac{1}{n} \sum_{} {x_i}^2 - \left( \bar{X} \right)^2 \]
\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 - {20}^2 = 9\]
\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 = 9 + 400 \]
\[ \Rightarrow \frac{1}{100} \sum_{} {x_i}^2 = 409\]
\[ \Rightarrow \sum_{} {x_i}^2 = 409 \times 100 = 40900 \left( \text{ This is an incorrect sum due to misread values }. \right) . . . (2)\]
\[\text{ Corrected } \sum_{} {x_i}^2 = 40900 - \left( {21}^2 + {21}^2 + {18}^2 \right)\]
\[ = 40900 - 441 - 441 - 324\]
\[ = 39694 . . . . (3) \]
\[\text{ From equations (2) and (3), we get: } \]
\[\text{ Corrected variance } = {\frac{1}{n}} \sum_{} {x_i}^2 - \left( {\bar{X}} \right)^2 \]
\[ = \frac{1}{97} \times 39694 - \left( 20 \right)^2 \]
\[ = 409 . 216 - 400 \]
\[ = 9 . 216\]
\[ \text{ Corrected SD } = \sqrt{{\text{ Corrected variance} }} \]
\[ = \sqrt{{9 . 216}} \]
\[ = 3 . 0357 \]
Thus, after omitting three values, the mean would be 20 and SD would be 3.0357.
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