Question

Here are the historical probabilities p(x) = P(X = x) that a fixed length of cast-iron piping will be manufactured and be found, on inspection, to have various numbers X of potentially weakening defects:x 0 1 2 3 4p(x) .45 .21 .13 .12 ,09F(x)Calculate the expected number of defects in a pipe E(X).

Answer 1

`E(X) =\sum_(i=1)^n X_i P(X_i)`

`E(X) = 0*0.45 +1*0.21 +2*0.13 +3*0.12 +4*0.09 =1.19`

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:

`E(X) =\sum_(i=1)^n X_i P(X_i)`

And replacing we got:

`E(X) = 0*0.45 +1*0.21 +2*0.13 +3*0.12 +4*0.09 =1.19`