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

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berno berno · Answer 1 · 2020-10-23T00:54:28+02:00

Answer:

Expectation is equal to 4.75

Variance is equal to 3.02

Step-by-step explanation:

Consider the attached image.

Let [tex]x_i[/tex] denotes each of the outcome.

Expectation [tex]=E(X)[/tex] = ∑[tex]x_if(x_i)[/tex]

[tex]=1(\frac{1}{12}) +2(\frac{1}{12}) +3(\frac{1}{12}) +4(\frac{1}{12}) +5(\frac{1}{12}) +6(\frac{7}{12})[/tex]

[tex]=\frac{1}{12}+\frac{2}{12}+\frac{3}{12}+\frac{4}{12}+\frac{5}{12}+\frac{42}{12}\\\\=\frac{57}{12}\\\\=4.75[/tex]

[tex]E(X^2)=[/tex] ∑ [tex]x_i^2f(x_i)[/tex]

[tex]=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})[/tex]

[tex]=(\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}[/tex]

Variance [tex]=E(X^2)-[E(X)]^2[/tex]

[tex]=(\frac{307}{12})-(\frac{57}{12})^2\\\\ =\frac{307}{12}-\frac{3249}{144}\\\\ =\frac{3684-3249}{144}\\\\ =\frac{435}{144}\\\\ =3.02[/tex]