# 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 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.

Solution:

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

$\bar{x}=\frac{1}{n} \sum_{j=1}^{20} x_{i}$

$10=\frac{1}{20} \sum_{i=1}^{20} x_{i}$

$\Rightarrow \sum_{i=1}^{20} x_{i}=200$

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

$\therefore$ Correct mean $=\frac{\text { Correct sum }}{19}=\frac{192}{19}=10.1$

Standard deviation, $\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}{ }^{2}-\frac{1}{n^{2}}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}$

$\Rightarrow 2=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)^{2}}$

$\Rightarrow 2=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-(\bar{x})^{2}} \quad\left[\right.$ as, $\left.\frac{1}{n} \sum_{i=1}^{n} x=\bar{x}\right]$

$\Rightarrow 2=\sqrt{\frac{1}{20} \times \text { Incorrect } \sum_{i=1}^{n} x_{i}{ }^{2}-(10)^{2}}$

$\Rightarrow 4=\frac{1}{20} \times$ Incorrect $\sum_{i=1}^{n} x_{i}^{2}-100$

$\Rightarrow 4=\frac{1}{20} \times$ Incorrect $\sum_{i=1}^{n} x_{i}{ }^{2}-100$

$\Rightarrow \frac{1}{20} \times$ Incorrect $\sum_{i=1}^{n} x_{i}^{2}=104$

$\Rightarrow$ Incorrect $\sum_{i=1}^{n} x_{i}{ }^{2}=2080$

Now, correct $\sum_{i=1}^{n} x_{i}{ }^{2}=$ Incorrect $\sum_{i=1}^{n} x_{i}{ }^{2}-(8)^{2}$

$\Rightarrow$ correct $\sum_{i=1}^{n} x_{i}{ }^{2}=2080-64=2016$

$\therefore$ correct Standard Deviation $=\sqrt{\frac{1}{n} \text { correct } \sum_{i=1}^{n} x_{i}^{2}-(\text { correct mean })^{2}}$

$\Rightarrow$ correct Standard Deviation $=\sqrt{\frac{1}{19} \times 2016-\left(\frac{192}{19}\right)^{2}}$

$\Rightarrow$ correct Standard Deviation $=\sqrt{\frac{2016}{19}-\left(\frac{192}{19}\right)^{2}}$

$\Rightarrow$ correct Standard Deviation $=\frac{\sqrt{1440}}{19}=\frac{12 \sqrt{10}}{19}$

$\Rightarrow$ correct Standard Deviation $=\frac{12 \times 3.162}{19}=1.997$

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

$\therefore$ Correct sum of observations $=200-8+12=204$

$\therefore$ Correct mean $=\frac{\text { Correct sum }}{20}=\frac{204}{20}=10.2$

Standard deviation $\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{0} x_{i}^{2}-\frac{1}{n^{2}}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-(\bar{x})^{2}}$

$\Rightarrow 2=\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^{n} x_{i}{ }^{2}-(10)^{2}}$

$\Rightarrow 4=\frac{1}{20}$ Incorrect $\sum_{i=1}^{n} x_{i}^{2}-100$

$\Rightarrow$ Incorrect $\sum_{i=1}^{n} x_{i}^{2}=2080$

$\therefore$ Correct $\sum_{i=1}^{n} x_{i}{ }^{2}=$ Incorrect $\sum_{i=1}^{n} x_{i}{ }^{2}-(8)^{2}+(12)^{2}$

$=2080-64+144$

$=2160$

$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$

$=\sqrt{\frac{2160}{20}-(10.2)^{2}}$

$=\sqrt{108-104.04}$

$=\sqrt{3.96}$

$=1.98$