Respuesta :
Answer:
[tex] E(X) = 1.1* \frac{4}{9} -1*\frac{5}{9}= -\frac{1}{15}[/tex]
[tex] Var(X) = Var(2.1 I-1) = Var(2.1I) = 2.1^2 Var(I)[/tex]
And if we find the variance for the indicator variable we got:
[tex] Var(X)= 2.1^2 \frac{20*25}{45^2} =\frac{49}{45}[/tex]
Step-by-step explanation:
For this case we can define an indicator variable I that is 1 when we win and 0 when we not win.
And we can define the random variable X as the amount of money that we can win like this:
[tex] X = 2.1I-1[/tex]
We can find the probability of win like this:
[tex] p_{win}= \frac{5*4}{10C2}= \frac{20}{45}= 4/9[/tex]
And the probability of no win would be:
[tex] p_{Nowin}= 1-\frac{4}{9}= \frac{5}{9}[/tex]
The two possible values for X are 1.1 when we win and -1 when we not win.
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
We can find the expected value with this formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And if we replace the values that we have we got:
[tex] E(X) = 1.1* \frac{4}{9} -1*\frac{5}{9}= -\frac{1}{15}[/tex]
Now in order to find the variance we need to find the second moment defined as:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And if we replace we got:
[tex] E(X^2) = (1.1)^2 \frac{4}{9} - 1^2*\frac{5}{9} =-\frac{4}{225}[/tex]
We can calculate the variance using this definition:
[tex] Var(X) = Var(2.1 I-1) = Var(2.1I) = 2.1^2 Var(I)[/tex]
And if we find the variance for the indicator variable we got:
[tex] Var(X)= 2.1^2 \frac{20*25}{45^2} =\frac{49}{45}[/tex]
