If for a distribution ∑ (x −5)=3,

Given for a distribution ∑ (x −5) = 3, ∑ (x −5)2 = 43 and the total number of item is 18

Now we have to find the mean and standard deviation.

As per given criteria,

Number of items, n=18

And given ∑(x – 5) = 3,

And also given, ∑(x −5)2 = 43

But we know mean can be written as,

$\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum(\mathrm{x}-5)}{\mathrm{n}}$

Here assumed mean is 5, so substituting the corresponding values in above equation, we get

$\bar{x}=5+\frac{3}{18}=\frac{18 \times 5+3}{18}=\frac{93}{18}=5.17$

And we know the standard deviation can be written as,

$\sigma=\sqrt{\frac{\sum(\mathrm{x}-5)^{2}}{\mathrm{n}}-\left(\frac{\sum(\mathrm{x}-5)}{\mathrm{n}}\right)^{2}}$

Substituting the corresponding values, we get

$\sigma=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}$

$\sigma=\sqrt{2.39-(0.166)^{2}}$

$\sigma=\sqrt{2.39-0.027}=\sqrt{2.363}$

Hence $\sigma=1.54$

So the mean and standard deviation of given items is $5.17$ and $1.54$ respectively.