The mean and standard deviation of 18 observations are found to be 7 and 4 respectively.

Solution:

Given that number of observations (n) = 18

Incorrect Mean $(\bar{x})=7$

and Incorrect Standard deviation, $(\sigma)=4$

We know that,

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

$\Rightarrow 7=\frac{1}{18} \sum_{i=1}^{18} x_{i}$

$\Rightarrow 7 \times 18=\sum_{i=1}^{18} x_{i}$

$\Rightarrow 126=\sum_{i=1}^{18} x_{i}$

$\Rightarrow \sum_{\mathrm{i}=1}^{18} \mathrm{x}_{\mathrm{i}}=126$ …(i)

$\therefore$ Incorrect sum of observations $=126$

Finding correct sum of observations, 12 was misread as 21

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

$=126-21+12$

$=117$

Hence,

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

$=\frac{117}{18}$

= 6.5

Now, Incorrect Standard Deviation (σ)

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

$4=\frac{1}{18} \sqrt{18 \times\left(\text { Incorrect } \sum x_{\mathrm{i}}^{2}\right)-(126)^{2}}$

$4 \times 18=\sqrt{18 \times\left(\text { Incorrect } \sum x_{\mathrm{i}}^{2}\right)-(126)^{2}}$

$72=\sqrt{18 \times\left(\text { Incorrect } \sum x_{\mathrm{i}}^{2}\right)-(126)^{2}}$

Squaring both the sides, we get

$(72)^{2}=18 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-(126)^{2}$

$\Rightarrow 5184=18 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-15876$

$\Rightarrow 5184+15876=18 \times$ Incorrect $\sum x_{i}^{2}$

$\Rightarrow 21060=18 \times$ Incorrect $\sum x_{i}^{2}$

$\Rightarrow \frac{21060}{18}=$ Incorrect $\sum x_{i}{ }^{2}$

$\Rightarrow 1170=$ Incorrect $\sum x_{i}{ }^{2}$

Since, 12 was misread as 21

So,

Correct $\sum_{i=1}^{18} x_{i}^{2}=1170-(21)^{2}+(12)^{2}$

$=1170-441+144$

= 873

Now,

Correct Standard Deviation

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

$=\sqrt{\frac{873}{18}-(6.5)^{2}}\left[\because \overline{\mathrm{x}}=\frac{\text { Correct } \sum \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=6.5\right]$

$=\sqrt{48.5-42.25}$

$=\sqrt{6.25}$

$=2.5$

Hence, Correct Mean $=6.5$

and Correct Standard Deviation $=2.5$