# For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively.

Question:

For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the score of 43 was misread as 34. Find the correct mean and standard deviation.

Solution:

Given that number of observations (n) = 200

Incorrect Mean $(\overline{\mathrm{x}})=40$

and Incorrect Standard deviation, $(\sigma)=15$

We know that,

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

$\Rightarrow 40=\frac{1}{200} \sum x_{i}$

$\Rightarrow 40 \times 200=\sum \mathrm{x}_{\mathrm{i}}$

$\Rightarrow 8000=\sum \mathrm{x}_{\mathrm{i}}$

$\Rightarrow \sum x_{i}=8000$ ............(i)

$\therefore$ Incorrect sum of observations $=8000$

Finding correct sum of observations, 43 was misread as 34

So, Correct sum of observations $=$ Incorrect Sum $-34+43$

$=8000-34+43$

$=8009$

Hence

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

$=\frac{8009}{200}$

$=40.045$

Now, Incorrect Standard Deviation ( $\sigma$ )

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

$15=\frac{1}{200} \sqrt{200 \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}^{2}\right)-(8000)^{2}}$

$15 \times 200=\sqrt{200 \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}^{2}\right)-64000000}$

$3000=\sqrt{200 \times\left(\text { Incorrect } \sum x_{i}^{2}\right)-64000000}$

Squaring both the sides, we get

$(3000)^{2}=200 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-64000000$

$\Rightarrow 9000000=200 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-64000000$

$\Rightarrow 9000000+64000000=200 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow 73000000=200 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow \frac{73000000}{200}=\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

$\Rightarrow 365000=\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$

Since, 43 was misread as 34

So,

Correct $\sum x_{i}^{2}=365000-(34)^{2}+(43)^{2}$

$=365000-1156+1849$

$=125000+693$

$=365693$

Now

Correct Standard Deviation

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

$=\sqrt{\frac{365693}{200}-(40.045)^{2}}$

$\left[\because \overline{\mathrm{x}}=\frac{\text { Correct } \sum \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=40.045\right]$

$=\sqrt{1828.465-1603.602025}$

$=\sqrt{224.862975}$

$=14.995$

Hence, Correct Mean $=40.045$

and Correct Standard Deviation $=14.995$