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Mean and Standard Deviation of 100 Observations Were Found to Be 40 and 10 Respectively. - Mathematics

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.      

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Solution

Given:
Number of observations, n = 100
Mean,

\[\bar{x} \] = 40
Standard deviation,
\[\sigma\]   = 10
We know that \[x = \frac{\sum_{} x_i}{100}\]
\[ \Rightarrow \frac{\sum_{} x_i}{100} = 40\]
\[ \Rightarrow \sum_{} x_i = 4000\]
∴ Correct
\[ \sum_{} x_i = 4000 - \left( 30 + 70 \right) + \left( 3 + 27 \right) = 3930\]
Correct mean = \[\frac{\text{ Correct }  \sum_{} x_i}{100} = \frac{3930}{100} = 39 . 3\]
Now,
Incorrect variance,
\[\sigma^2 = \frac{\sum_{} x_i^2}{100} - \left( 40 \right)^2\]
\[\Rightarrow \frac{\sum_{} x_i^2}{100} = 100 + 1600 = 1700\]
\[ \Rightarrow \sum_{} x_i^2 = 170000\]
Correct \[\sum_{} x_i^2 = 170000 - {30}^2 - {70}^2 + 3^2 + {27}^2 = 164939\]
 
∴ Correct standard deviation \[= \sqrt{\frac{164939}{100} - \left( 39 . 3 \right)^2}\]
\[ = \sqrt{1649 . 39 - 1544 . 49}\]
\[ = \sqrt{104 . 9}\]
\[ = 10 . 24\]
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.6 | Q 8 | Page 42
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