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Chapter 9: The Normal Probability Distribution
X
will be a binomial random variable since we have a fixed number of trials
(
= 200); two possible outcomes (either the item is classified as a defective
or a non-defective item); the probability of a non-defective item (successful
outcome) is fixed from item to item (
= 0.95); items will be non-defective
independent from each other.
Figure 9-46:
Bar Chart with Defective Rates
Next we need to check whether the conditions for the normal approximation
to the binomial distribution hold. That is, we need to check whether
> 5
and
–
> 5.
Now,
= 200
0.95 = 190 and
–
= 200
(1 – 0.95) = 200
0.05 = 10.
Since both
> 5 and
–
> 5,.we can use the normal approximation
to the binomial distribution. Thus,
= 200
0.95 = 190 and
√
=
√
= 3.0822.
We need to compute
P
(
X
192) =
P
(
X
192 + 0.5) =
P
(
X
192.5)
P
(
X
< 192.5).