# The mean and standard deviations of a group of 100 observations were found to be 20 and 3 respectively.

Question:

The mean and standard deviations of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations 21, 12 and 18 were incorrect. Find the mean and standard deviation if the incorrect observations were omitted.

Solution:

Given that number of observations (n) = 100

Incorrect Mean $(\bar{x})=20$

and Incorrect Standard deviation, $(\sigma)=3$

We know that,

$\overline{\mathrm{x}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}$

$\Rightarrow 20=\frac{1}{100} \sum x_{i}$

$\Rightarrow 20 \times 100=\sum x_{i}$

$\Rightarrow 2000=\sum x_{i}$

$\Rightarrow \sum x_{i}=2000$ ...........(i)

$\therefore$ Incorrect sum of observations $=2000$

Finding correct sum of observations, incorrect observations 21,12 and 18 are removed

So, Correct sum of observations = Incorrect Sum $-21-12-18$

$=2000-51$

$=1949$

Hence,

Correct Mean $=\frac{\text { Correct Sum of Observations }}{\text { Total number of observations }}$

$=\frac{1949}{100-3}$

$=\frac{1949}{97}$

$=20.09$

Now, Incorrect Standard Deviation ( $\sigma$ )

$=\frac{1}{\mathrm{~N}} \sqrt{\mathrm{N} \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}{ }^{2}\right)-\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}\right)^{2}}$

$3=\frac{1}{100} \sqrt{100 \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}{ }^{2}\right)-(2000)^{2}}$

$3 \times 100=\sqrt{100 \times\left(\text { Incorrect } \sum x_{i}^{2}\right)-4000000}$

$300=\sqrt{100 \times\left(\text { Incorrect } \sum x_{i}^{2}\right)-4000000}$

Squaring both the sides, we get

$(300)^{2}=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-4000000$

$\Rightarrow 90000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-4000000$

$\Rightarrow 90000+4000000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow 4090000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow \frac{4090000}{100}=\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow 40900=\left(\right.$ Incorrect $\left.\sum x_{i}{ }^{2}\right)$

Since, 21, 12 and 18 are removed

So,

Correct $\sum x_{i}^{2}=40900-(21)^{2}-(12)^{2}-(18)^{2}$

$=40900-441-144-324$

$=40900-909$

$=39991$

Now,

Correct Standard Deviation

$=\sqrt{\frac{\left(\text { Correct } \sum x_{i}^{2}\right)}{N}-\left(\frac{\text { Correct } \sum x_{i}}{N}\right)^{2}}$

$=\sqrt{\frac{39991}{97}-(20.09)^{2}}$

$\left[\because \overline{\mathrm{x}}=\frac{\text { Correct } \sum \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=20.09\right]$

$=\sqrt{412.27-403.60}$

$=\sqrt{8.67}$

$=2.94$

Hence, Correct Mean $=20.09$

and Correct Standard Deviation $=2.94$