The Bernoulli distribution is a distribution whose random variable can only take 0 or 1
The distribution is given as:
p(0) = 1 - p
p(1) = p
The expected value of x2, E(x2) is calculated as:
[tex]E(x^2) = \sum x^2 * P(x)[/tex]
So, we have:
[tex]E(x^2) = 0^2 * (1- p) + 1^2 * p[/tex]
Evaluate the exponents
[tex]E(x^2) = 0 * (1- p) + 1 * p[/tex]
Multiply
[tex]E(x^2) = 0 +p[/tex]
Add
[tex]E(x^2) = p[/tex]
Hence, the value of E(x2) is p
This is calculated as:
[tex]V(x) = E(x^2) - (E(x))^2[/tex]
Start by calculating E(x) using:
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
[tex]E(x) = 0 * (1- p) + 1 * p[/tex]
[tex]E(x) = p[/tex]
Recall that:
[tex]V(x) = E(x^2) - (E(x))^2[/tex]
So, we have:
[tex]V(x) = p - p^2[/tex]
Factor out p
[tex]V(x) = p(1 - p)[/tex]
Hence, the value of V(x) is p(1 - p)
The expected value of x79, E(x79) is calculated as:
[tex]E(x^{79}) = \sum x^{79} * P(x)[/tex]
So, we have:
[tex]E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p[/tex]
Evaluate the exponents
[tex]E(x^{79}) = 0 * (1- p) + 1 * p[/tex]
Multiply
[tex]E(x^{79}) = 0 + p[/tex]
Add
[tex]E(x^{79}) = p[/tex]
Hence, the value of E(x79) is p
Read more about probability distribution at:
https://brainly.com/question/15246027
