Determine mean and standard deviation

Given first $n$ terms of an A.P. whose first term is a and common difference is $d$ Now we have to find mean and standard deviation The given AP in tabular form is as shown below,

Here we have assumed a as mean.

Given the AP have $n$ terms. And we know the sum of all the terms of AP can be written as,

$\sum x_{i}=\frac{n}{2}[2 a+(n-1) d]$

Now we will calculate the actual mean,

$\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$

Substituting the corresponding values, we get

$\overline{\mathrm{x}}=\frac{\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]}{\mathrm{n}}$

The above equation can be written as

$\overline{\mathrm{x}}=\frac{[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]}{2}$

$\bar{x}=a+\frac{[(n-1) d]}{2}$

$\bar{x}=a+\frac{(n-1)}{2} d$

Now we other two columns sums, i.e.

$\sum \mathrm{d}_{\mathrm{i}}=\sum\left(\mathrm{x}_{\mathrm{i}}-\mathrm{a}\right)=\mathrm{d}[1+2+3+\cdots+(\mathrm{n}-1)]=\mathrm{d}\left(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\right)$

$\sum \mathrm{d}_{\mathrm{i}}^{2}=\sum\left(\mathrm{x}_{\mathrm{i}}-\mathrm{a}\right)^{2}=\mathrm{d}^{2}\left[1^{2}+2^{2}+3^{2}+\cdots+(\mathrm{n}-1)^{2}\right]=\mathrm{d}^{2}\left(\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right)$

Now we know standard deviation is given by,

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

Substituting the corresponding values, we get

$\sigma=\sqrt{\frac{\mathrm{d}^{2}\left(\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right)}{\mathrm{n}}-\left(\frac{\mathrm{d}\left(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\right)}{\mathrm{n}}\right)^{2}}$

$\sigma=\sqrt{\mathrm{d}^{2}\left(\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6 \mathrm{n}}\right)-\mathrm{d}^{2}\left(\frac{\mathrm{n}^{2}(\mathrm{n}-1)^{2}}{4 \mathrm{n}^{2}}\right)}$

Cancelling the like terms, we get

$\sigma=\sqrt{\mathrm{d}^{2}\left(\frac{(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right)-\mathrm{d}^{2}\left(\frac{(\mathrm{n}-1)^{2}}{4}\right)}$

Taking out common terms we get

$\sigma=\sqrt{\left(\frac{d^{2}(n-1)}{2}\right)\left(\frac{(2 n-1)}{3}-\frac{n-1}{2}\right)}$

By taking the LCM, we get

$\sigma=\sqrt{\left(\frac{\mathrm{d}^{2}(\mathrm{n}-1)}{2}\right)\left(\frac{2(2 \mathrm{n}-1)-3(\mathrm{n}-1)}{6}\right)}$

$\sigma=\sqrt{\left(\frac{\mathrm{d}^{2}(\mathrm{n}-1)}{2}\right)\left(\frac{4 \mathrm{n}-2-3 \mathrm{n}+3}{6}\right)}$

$\sigma=\sqrt{\left(\frac{\mathrm{d}^{2}(\mathrm{n}-1)}{2}\right)\left(\frac{\mathrm{n}+1}{6}\right)}$

$\sigma=d \sqrt{\frac{\left(n^{2}-1\right)}{12}}$

Hence the mean and standard deviation of the given $\mathrm{AP}$ is

$a+\frac{(n-1)}{2} d_{\text {and }} d \sqrt{\frac{\left(n^{2}-1\right)}{12}}$ respectively.