Download 38178 .42.19 )96)(.04)(. 500( ) 1( ≈ = = − = p np )62.2 ( 38178 .4

SECTION 11.4 SUMMARY
The following directions will appear at the top of your third test and final exam.
Directions: For problems involving a binomial experiment,
if n < 10 use the formula for binomial probabilities.
If it is binomial and n ≥ 10, then
either A) use the normal approximation if it is valid
or B) if the normal approximation is not valid, write "NOT VALID", do not solve the
problem, and go on to the next problem.
Here are some examples: (Pictures and bold must be shown.)
Example 1) Suarez Office Supplies manufactures pencils with, on the average, 4% of the pencils
defective. A random sample of 500 pencils is selected from each day's production and tested. What
is the probability that in the sample there are 32 or more defective pencils?
Discussion of example 1: This problem is binomial with
n = 500
success: pencil is defective
p = .04
x = number of defective pencils
Since n ≥ 10, we have to test to see if the normal approximation can be used. That requires that
we compute µ ± 3σ.
µ = np = (500)( .04 ) = 20
σ =
np (1 − p ) =
( 500 )(. 04 )(. 96 ) =
19 . 2 ≈ 4 . 38178
µ ± 3σ = 20 ± 3(4.4) = 20 ± 13. 2 = 6.8 or 33.2
Since both of these numbers are between 0 and n (that is, between 0 and 500), we can use the normal
approximation. To convert from binomial to normal, we draw a histogram and shade the rectangles
corresponding to the desired probability.
We see that the shaded region begins at 31.5.
So P(X ≥ 32), where X is a binomial random
variable, is approximately equal to P(Y≥31.5) where
Y is a normal random variable. This is because the
sum of the areas of the rectangles is almost the same
as the area under the bell-shaped curve. We write:
P(X ≥ 32) ≈ P(Y ≥ 31.5)
Now that we have a normal random variable, we
need to convert it to standard normal so that we can use the table in the back of the text. To do this,
we subtract the mean and divide by the standard deviation.
 Y − 20 31.5 − 20 
P( X ≥ 32) ≈ P(Y ≥ 31.5) = P
≥
 = P(Z ≥ 2.62)
 4.38178 4.38178 
Now we draw our second picture of the problem and use the
standard normal table in the back of the text to get: .5-.4956 = .0044
Page 2
Example 2) IQ scores are normally distributed with a mean of 100 and a standard deviation of
15. Find the probability that a randomly selected person has an IQ score below 110.
Discussion of Example 2: The thing to note here is that unlike example 1, this problem is NOT
binomial. In fact, the first line tells us that our random variable is normal. This means that you
do not have to convert it to normal (that is, we do NOT make the one-half adjustment that was
necessary in the last example). If we let X = the IQ score of a randomly selected person, then we
need to find P ( X < 110 ) , where X is a normal random variable. To convert to standard
normal, we subtract the mean and divide by the standard deviation:
 X −100 110−100
P( X < 110) = P
<
 = P(Z < .67)
15 
 15
Now looking up .67 in the table in the back of the text, and adding
.5 to represent the area to the left of zero, we get
.5 + .2486 = .7486
Example 3) 90% of all Gator football players have cheated on a test. If five Gator football
players are selected at random, what is the probability that all five have cheated on a test?
Discussion of Example 3: This problem is binomial with
n=5
success: player has cheated on a test
p = .9
X = the number of Gator football players who have cheated on a test
Since n < 10 we must use our memorized formula for binomial probabilities.
P( X = 5) =
( )(.9) (.1)
5
5
5
0
= (1)(.59049)(1) = .59049
Example 4) A new drug produces undesirable side-effects in 1% of the patients who use it. If 50
patients use this drug, what is the probability that no more than five suffer undesirable sideeffects?
Discussion of Example 4: This problem is binomial with
n = 50
success: patient has undesirable side-effects
p = .01
X = the number of patients having undesirable side-effects
Since n ≥10 we have to check to see if the normal approximation is valid. That is, we check to
see if µ ± 3 σ are between 0 and n.
µ = np = (50)( .01 ) = .5
σ = np(1 − p) = (50)(.01)(.99) ≈ .7035624
µ ± 3σ = .5 ± 3(.7) = .5 ± 2.1 = -1.6 or 2.6
Since -1.6 is not between 0 and 50, we write NOT VALID, and go on to the next problem. This
means that the method of the normal approximation does not yield a good approximation.