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

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 $=14.995$