Respuesta :
Answer:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) [/tex]
And if we replace we got:
[tex] E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8[/tex]
So we expect about 0.8 defective computes in a batch of 4 selected.
Step-by-step explanation:
Previous concepts
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.
Solution to the problem
For this case we have the following distribution given:
X 0 1 2 3 4
P(X) 0.4096 0.4096 0.1536 0.0256 0.0016
And we satisfy that [tex] P(X_i) \geq 0[/tex] and [tex] \sum P(X_i) =1[/tex] so we have a probability distribution. And we can find the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) [/tex]
And if we replace we got:
[tex] E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8[/tex]
So we expect about 0.8 defective computes in a batch of 4 selected.
