The mean and standard deviation of 20 observations are found to be 10 and 2 respectively

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 by mistake was taken 8 instead of 12. The correct standard deviation is

Smart Achievers · Accepted Answer

(3)

$\frac{\sum x_{i}}{20} = 10, \sum x_{i} = 10 \times 20 = 200$

$If 8 replaced by 12 then \sum x_{i} = 200 - 8 + 12 = 204$

$Now, correct mean (\bar{x}) = \frac{\sum x_{i}}{20} = \frac{204}{20} = 10.2$

$Standard deviation = 2$

$So, variance = {(S . D)}^{2} = 2^{2} = 4$

$By definition,$ $\Rightarrow \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = 4 \Rightarrow \frac{\sum x_{i}^{2}}{20} - {(10)}^{2} = 4$

$\Rightarrow \sum x_{i}^{2} = 2080$

$\Rightarrow V_{a r} = \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = \frac{2160}{20} - {(10.2)}^{2}$

Solution

Q.
The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

1 $\sqrt{3.86}$

2 $1.8$

3 $\sqrt{3.96}$

4 $1.94$

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Solution

Q. The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

1 3.86

2 1.8

3 3.96

4 1.94