Question

Suppose you have a 6-face unfair dice with numbers 1,2,3,4,5,6 on each of its faces. If the probability distribution of throwing a number up is as follows: 1 with probability 1/12 2 with probability 1/12 3 with probability 1/12 4 with probability 1/12 5 with probability 1/12 6 with probability 7/12 What are the mathematical expectation and variance of the dice based on the above probability distribution

Answer 1

Answer:

Expectation is equal to 4.75

Variance is equal to 3.02

Step-by-step explanation:

Consider the attached image.

Let $x_i$ denotes each of the outcome.

Expectation $=E(X)$ = ∑$x_if(x_i)$

$=1(\frac{1}{12}) +2(\frac{1}{12}) +3(\frac{1}{12}) +4(\frac{1}{12}) +5(\frac{1}{12}) +6(\frac{7}{12})$

$=\frac{1}{12}+\frac{2}{12}+\frac{3}{12}+\frac{4}{12}+\frac{5}{12}+\frac{42}{12}\\\\=\frac{57}{12}\\\\=4.75$

$E(X^2)=$ ∑ $x_i^2f(x_i)$

$=1^2(\frac{1}{12}) +2^2(\frac{1}{12}) +3^2(\frac{1}{12}) +4^2(\frac{1}{12}) +5^2(\frac{1}{12}) +6^2(\frac{7}{12})$

$=(\frac{1}{12}) +4(\frac{1}{12}) +9(\frac{1}{12}) +16(\frac{1}{12}) +25(\frac{1}{12}) +36(\frac{7}{12})\\\\=\frac{307}{12}$

Variance  $=E(X^2)-[E(X)]^2$

$=(\frac{307}{12})-(\frac{57}{12})^2\\\\ =\frac{307}{12}-\frac{3249}{144}\\\\ =\frac{3684-3249}{144}\\\\ =\frac{435}{144}\\\\ =3.02$

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