Answers

2016-04-21T03:56:26+05:30
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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