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A Student Obtained the Mean and Standard Deviation of 100 Observations as 40 and 5.1 Respectively. It Was Later Found that One Observation Was Wrongly Copied as 50, the Correct Figure Being 40 - Mathematics

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प्रश्न

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.

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उत्तर

\[\bar{x}_o = 40, \sigma_o = 5 . 1, \text{ and } n = 100\]
\[ \Rightarrow \Sigma x_o = 4000\]
\[\text{ Corrected sum of observations,}  \Sigma x_n = 4000 - 50 + 40 = 3990\]
\[n = 100\]
\[ \Rightarrow \bar{x}_n = \frac{\Sigma x_n}{n} = 39 . 90\]
\[\text{ Also, SD, } \sigma_o = 5 . 1\]
\[ \Rightarrow \Sigma \left( x_i - \bar{x}_o \right)^2 = 2601\]
\[ \Rightarrow \Sigma \left( x_i - \bar{x}_n \right)^2 = 2601 - \left( 50 - 40 \right)^2 + \left( 40 - 39 . 90 \right)^2 = 2601 - 100 + 0 . 01 = 2501 . 01\]
\[ \Rightarrow \sigma_n = \sqrt{\frac{\Sigma \left( x_i - \bar{x}_n \right)^2}{n}} = \sqrt{\frac{2501 . 01}{100}} = 5\]

 

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अध्याय 32: Statistics - Exercise 32.6 [पृष्ठ ४२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 32 Statistics
Exercise 32.6 | Q 4 | पृष्ठ ४२

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