Question

The mean of the values 1, 2, 3, ..n with respective frequencies x, 2x, 3x… nx is:

Answer 1

Answer:

$\mathbf{\dfrac{2n+1}{3}}$

Step-by-step explanation:

Given :

• Values are 1, 2, 3, ..n
• There frequencies are x, 2x, 3x… nx

To find :

Mean of the given values

Solution :

We know that, mean of grouped frequency if a, b, c, d, ... are its values and A, B, C, D ... are its respective frequencies then,

$\boxed {\mathrm{Mean = \overline{X} = \dfrac{aA + bB + cC + dD + ...}{A + B + C + D + ...}}}$

Here, values are, 1, 2, 3, ...n and their frequencies are x, 2x, 3x… nx

So,

$\implies \mathrm{\overline X = \dfrac{1.x + 2. 2x + 3.3x + ... n.nx}{x + 2x + 3x + ...nx } }$

$\implies \mathrm{\overline X = \dfrac{(1^2 + 2^2 + 3^2 + ... n^2)x}{(1 + 2 + 3 + ...n)x}}$

$\implies \mathrm{\overline X = \dfrac{(1^2 + 2^2 + 3^2 + ... n^2)\bcancel{x}}{(1 + 2 + 3 + ...n)\bcancel{x}}}$

We know that,

$\longmapsto \boxed {\text{Sum of n numbers }= \mathrm{\dfrac{n(n+1)}{2}}}$

$\longmapsto \boxed {\text{Sum of n^2 numbers }= \mathrm{\dfrac{n(n+1)(2n+1)}{6}}}$

So, we get,

$\implies \mathrm{\overline X = \dfrac{\bigg(\dfrac{n(n+1)(2n+1)}{6} \bigg)}{\bigg(\dfrac{n(n+1)}{2}\bigg)}}$

$\implies \mathrm{\overline X = \dfrac{\bigg(\dfrac{\bcancel{n}\bcancel{(n+1)}(2n+1)}{\bcancel{6}^3} \bigg)}{\bigg(\dfrac{\bcancel{n}\bcancel{(n+1)}}{\bcancel{2}}\bigg)}}$

$\implies \mathrm{\overline X = \dfrac{2n+1}{3} }$

$\therefore \underline {\boxed{ \mathrm{Mean = \dfrac{2n+1}{3}}}}$

Hope it helps!!