Answer:
[tex] E(X) =\sum_{i=1}^n X_i P(X_i)[/tex]
[tex]E(X) = 0*0.45 +1*0.21 +2*0.13 +3*0.12 +4*0.09 =1.19[/tex]
Step-by-step explanation:
For this case we have the probability density function given:
X 0 1 2 3 4
P(X) 0.45 0.21 0.13 0.12 0.09
The cdf of F(X) would be given by:
X 0 1 2 3 4
F(X) 0.45 0.66 0.79 0.91 1
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.
And we can use the following formula in order to calculate the expected value:
[tex] E(X) =\sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex]E(X) = 0*0.45 +1*0.21 +2*0.13 +3*0.12 +4*0.09 =1.19[/tex]