Question

The variance of first 20 natural numbers is

Answer 1

Var[X]=E[(X-μ)²]

Expected value:
$E[X]=\sum_{i=1}^{20}p_iX_i$
For all numbers 1,...,20 probability is to equal 1/20.
$E[X]=\sum_{i=1}^{20}p_iX_i=\frac1{20}(1+2+...+19+20)=\frac1{20}\frac{(1+20)20}{2}=\bar{X_i}=10.5$

$Var[X]=E[(X-\mu)^2]=\frac1{20}[(1-10.5)^2+(2-10.5)^2+...+(20-10.5)^2]=calculated\ in\ Excel:\  33.25$
Other way:
$Var[X]=E[X^2]-(E[X])^2=\frac1{20}[(1^2+2^2+...+20^2)]-10.5^2=\\\frac1{20}\times2870-110.25=143.5-110.25=33.25$
The standard deviation:
$\sigma=\sqrt{Var[X]}\approx5.7663$
Answer:
Var[1, 2, 3, .... 19, 20] = 33.25

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Answer 2

Answer:

33.25

Step-by-step explanation:

First 20 natural numbers= 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

Mean = $\frac{\text{Sum of all observations}}{\text{Total no. of observations}}$

So, Mean = $\frac{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20}{20}$

Mean = $10.5$

Variance = $\sigma^2=\frac{ \sum(x_i-\bar{x})^2}{n}$

So, $\sigma^2=\frac{(1-10.5)^2+(2-10.5)^2+(3-10.5)^2+(4-\10.5)^2+(5-10.5)^2+(6-10.5)^2+(7-10.5)^2+(8-10.5)^2+(9-10.5)^2+(10-10.5)^2+(11-10.5)^2+(12-10.5)^2+(13-10.5)^2+(14-10.5)^2+(15-10.5)^2+(16-10.5)^2+(17-10.5)^2+(18-10.5)^2+(19-10.5)^2+(20-10.5)^2}{20}$

$\sigma^2=33.25$

Hence the variance of first 20 natural numbers is 33.25