Plastic parts manufactured using an injection molding process may exhibit one or more defects, including sinks, scratches, black spots, and son on. Let x represent the number of defects on a single part, and suppose the distribution of x is as follows: x: 0 1 2 3 4 p(x): 0.79 0.09 0.05 0.04 0.03 What is the expected value, E(X), for the number of defects on a randomly selected plastic part

The expected value for the number of defects on a randomly selected plastic part is 0.43.

Step-by-step explanation:

We have the following probability distribution:

P(X = 0) = 0.79

P(X = 1) = 0.09

P(X = 2) = 0.05

P(X = 3) = 0.04

P(X = 4) = 0.03

What is the expected value, E(X), for the number of defects on a randomly selected plastic part?

Each value multiplied by its probability. So

The expected value for the number of defects on a randomly selected plastic part is 0.43.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

3 is greater then both of those numbers