Question

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.

Answer 1

\[n = 20 \]

\[\text{ Mean }  = \bar{X} \]

\[SD = \sigma = 2 \]

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

\[ \therefore \frac{1}{20} \sum x_i = 10_{} \]

\[ \Rightarrow \sum x_i = 200 \left[ {\text{ This is incorrect due to misread values } .} \right] . . . (1)\]

\[ \Rightarrow \text{ Variance } = \sigma^2 = 4\]

\[{\frac{1}{n}} \sum_{} {x_i}^2 - \left( {\bar{X}} \right) {}^2 = 4\]

\[ \Rightarrow {\frac{1}{20}} \sum_{} {x_i}^2 - {10}^2 = 4\]

\[ \Rightarrow{\frac{1}{20}} \sum_{} {x_i}^2 = 104\]

\[ \Rightarrow \sum_{} {x_i}^2 = 104 \times 20 = 2080 \left[ {\text{ This is incorrect due to misread values } .} \right] . . . (2)\]

(i)     If  observation 8 is omitted, then total 19 observations are left.

Incorrected \[\sum_{} x_i = 200\]

\[\text{ Corrected } \sum_{} x_i + 8 = 200\]

\[\text{ Corrected } \sum^{}_{} x_i = 192\]

\[ \Rightarrow \left( {\text{ Corrected mean } } \right) = {\frac{\text{ Corrected } \sum^{}_{} x_i}{19}}\]

\[ = {\frac{192}{19}}\]

\[ = 10 . 10\]

\[\text{ Using equation (2), we get: } \]

\[\text{ Corrected } \sum^{}_{} {x_i}^2 + 8^2 = 2080\]

\[ \Rightarrow \text{ Corrected } \sum^{}_{} {x_i}^2 = 2080 - 64 \]

\[ = 2016\]

\[ \therefore{\frac{1}{19}}\text{ Corrected}  \sum^{}_{} {x_i}^2 - \left({\text{ Corrected mean} } \right)^2 = \text{ Corrected variance } \]

\[ \Rightarrow \text{ Corrected variance } = {\frac{1}{19}} \times 2016 - \left({\frac{192}{19}} \right)^2 \]

 

\[ \Rightarrow\text{  Corrected variance} = {\frac{\left( 2016 \times 19 \right) - \left( 192 \right)^2}{{19}^2}} \]

\[ \Rightarrow \text{ Corrected variance } ={\frac{38304 - 36864}{{19}^2}} \]

\[ \Rightarrow \text{ Corrected variance } ={\frac{1440}{{19}^2}} \]

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

\[ = \sqrt{{\frac{1440}{{19}^2}}} \]

\[ = {\frac{12\sqrt{10}}{19}}\]

\[ = 1 . 997\]

Thus, if  8 is omitted, then the mean is 10.10 and SD is 1.997.
(ii)  When incorrect observation 8 is replaced by 12:

\[\text{ From equation }  (1): \]

\[\text{ Incorrected } \sum^{}_{} x_i = 200\]

\[\text{ Corrected } \sum^{}_{} x_i = 200 - 8 + 12 = 204\]

\[ \text{ Corrected }  \bar{X} = \frac{204}{20} = 10 . 2\]

\[\text{ Incorrected}  \sum^{}_{} {x_i}^2 = 2080 \left[ {\text{ from }(2)} \right]\]

\[\text{ Corrected } \sum^{}_{} {x_i}^2 = 2080 - 8^2 + {12}^2 \]

\[ = 2160\]

\[\text{ Corrected variance } = {\frac{1}{20}} \times \text{ Corrected } \sum^{}_{} {x_i}^2 - \left( {\text{ Corrected } \bar{X}} \right)^2 \]

\[ ={\frac{1}{20}} \times 2160 - \left( {\frac{204}{20}} \right)^2 \]

\[ = {\frac{\left( 2160 \times 20 \right) - \left( 204 \right)^2}{{20}^2}}\]

\[ = {\frac{43200 - 41616}{400}}\]

\[ = {\frac{1584}{400}}\]

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

\[ = \sqrt{{\frac{1584}{400}}} \]

\[ = {\frac{\sqrt{396}}{10}}\]

\[ = {\frac{19 . 899}{10}}\]

\[ = 1 . 9899\]

If 8 is replaced by 12, then the mean  is 10.2 and SD is 1.9899.