In a batch of pedometers, are believed to be defective. A quality-control engineer randomly selects units to test. Let random variable Xthe number of defective units that are among the units tested. The probability mass function f(x)P(Xx) is given below. f(x){(0,), (1,), (2,), (3,)} Recall that the mean of a discrete random variable X with probability mass function f(x)P(Xx) is given by . Find for the probability mass function above. What does this number represent?

In a batch of 28 pedometers, 3 are believed to be defective. A quality-control engineer randomly selects 6 units to test. Let random variable X = the number of defective units that are among the units tested.

The probability mass function f (x) is:

f (x) = {(0, 0.47009), (1, 0.42308), (2, 0.10073), (3, 0.00611)}

Solution:

The mean of a discrete random variable X with probability mass function f (x) is:

[tex]\mu=\sum\limits^{n}_{x=1}{x\times f(x)}[/tex]

Compute the mean for the probability mass function above as follows:

[tex]\mu=\sum\limits^{n}_{x=1}{x\times f(x)}[/tex]

[tex]=(0\times 0.47009)+(1\times 0.42308)+(2\times 0.10073)+(3\times 0.00611)\\=0+0.42308+0.20146+0.01833\\=0.64287[/tex]

Thus, the mean for the probability mass function is 0.64287.

This values represents the expected number of defective units for every 6 units tested.