Question

If the values $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},....\frac{1}{n}$ occur at frequencies $1,2,3,4,5,\mathrm{6...},n$ respectively, in a frequency distribution, then the mean is

• A. $1$
• B. $n$
• C. $\frac{1}{n}$
• D. $\frac{2}{n+1}$

Answer 1

The correct option is D $\frac{2}{n+1}$

Given, the values $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},....\frac{1}{n}$ occur at frequencies $1,2,3,4,5,\mathrm{6...},n$ respectively, in a frequency distribution.

Then mean of the distribution will be,

$=\frac{1.1+2.\frac{1}{2}+3.\frac{1}{3}+......+n.\frac{1}{n}}{1+2+3+.....+n}$

$=\frac{n}{\frac{n(n+1)}{2}}$ [ Since $1+2+3+...+n=\frac{n(n+1)}{2}$]

$=\frac{2}{n+1}$.