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प्रश्न
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?
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उत्तर
\[n = 100 \]
\[\text{ Mean } = \bar{X} = 40 \]
\[\sigma = SD = 5 . 1\]
\[\frac{1}{n}\sum x_i = \bar{X} \]
\[ \Rightarrow \sum x_i = 100 \times 40 = 4000 \left( \text{ This is an incorrect reading due to misread values . } \right)\]
\[ \text{ Corrected sum } , \sum x_i = 4000 - 50 + 40 \]
\[ = 3990\]
\[ \Rightarrow \text{ Corrected mean }= \frac{\text{ Corrected sum } }{100}\]
\[ = \frac{3990}{100}\]
\[ = 39 . 9 . . . (1)\]
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
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