Transcript 5.5 Normal Approximations to Binomial Distributions

5.5 Normal Approximations to
Binomial Distributions
Statistics
Mrs. Spitz
Fall 2008
Objectives/Assignment
• How to decide when the normal distribution can
approximate the binomial distribution
• How to find the correction for continuity
• How to use the normal distribution to
approximate binomial probabilities
• Assignment: pp. 241-243 #1-18 all
Approximating a Binomial Distribution
• In Section 4.2, you learned how to find binomial
probabilities. For instance, a surgical procedure has an
85% chance of success and a doctor performs the
procedure on 10 patients, it is easy to find the probability
of exactly two successful surgeries.
• But what if the doctor performs the surgical procedure on
150 patients and you want to find the probability of fewer
than 100 successful surgeries?
• To do this using the techniques described in 4.2, you
would have to use the binomial formula 100 times and
find the sum of the resulting probabilities. This is not
practical and a better approach is to use a normal
distribution to approximate the binomial distribution.
Normal Approximation to a Binomial
Distribution
To see why this result is valid, look at the following slide
and binomial distributions for p = 0.25 and n = 4, 10, 25
and 50. Notice that as n increases, the histogram
approaches a normal curve.
Study Tip
• Properties of a binomial experiment
– n independent trials
– Two possible outcomes: success or failure
– Probability of success is p; probability of a
failure is 1 – p = q
– p is constant for each trial
Ex. 1: Approximating the Binomial
Distribution
• Two binomial experiments are listed. Decide
whether you can use the normal distribution to
approximate x, the number of people who reply
yes. If so, find the mean and standard deviation.
If not, explain why.
1. Thirty-seven percent of Americans say they
always fly an American flag on the Fourth of
July. You randomly select 15 Americans and ask
each if he or she always flies an American flag
on the Fourth of July.
Ex. 1: Approximating the Binomial
Distribution - Solution
1. Thirty-seven percent of Americans say they always fly
an American flag on the Fourth of July. You randomly
select 15 Americans and ask each if he or she always
flies an American flag on the Fourth of July.
In this binomial experiment, n = 15, p = 0.37 and q =
0.63, so:
np = 15(0.37) = 5.55 and nq = 15(0.63) = 9.45
Because np ≥ 5 and nq ≥ 5, you can use the normal
distribution with  = 5.55 and
  npq  15  0.37  0.63  1.87
to approximate the distribution of x.
Ex. 1b: Approximating the Binomial
Distribution - Solution
2. Ninety-three percent of Americans want the national
anthem to remain the same. You randomly select 65
Americans and ask each if he or she want the national
anthem to remain the same.
In this binomial experiment, n = 65, p = 0.93 and q =
0.07, so:
np = 65(0.93) = 60.45 and nq = 65(0.07) = 4.55
Because nq  5, you cannot use the normal distribution
to approximate the distribution of x.
Note for TI users
• The TI’s cannot calculate the cumulative
binomial probability for n = 10,000, p = 0.4
and x = 9000. The probability can be
calculated using a normal approximation.
There are issues with memory limitation
for your calculator for the binomial
distribution.
Correction for Continuity
• The binomial distribution is discrete and can be
represented by a probability histogram. To calculate an
exact binomial probability, you can use the binomial
formula for each value of x and add the results.
Geometrically, this corresponds to adding the areas of
bars in the probability histogram. When you do this,
remember that each bar has a width of one unit and x is
the midpoint of the interval.
• When you use a continuous normal distribution to
approximate a binomial probability, you need to move
0.5 units to the left and right of the midpoint to include all
possible x-values in the interval. When you do this, you
are making a correction for continuity.
Correction for Continuity
Ex. 2: Using a Correction for Continuity
Use the correction for continuity to convert each of the following
binomial intervals to a normal distribution interval.
1. The probability of getting between 270 and 310 successes, inclusive
2. The probability of getting more than 157 and less than 420
successes
3. The probability of getting less than 63 successes.
SOLUTION:
1. The probability of getting between 270 and 310 successes, inclusive
The midpoint values are 270, 271, . . . 310. The boundaries for the
normal distribution are 269.5 < x < 310.5
Ex. 2: Using a Correction for Continuity
Use the correction for continuity to convert each of the following
binomial intervals to a normal distribution interval.
2. The probability of getting more than 157 and less than 420
successes
SOLUTION:
2. The probability of getting more than 157 and less than 420
successes
The midpoint values are 158, 159, . . . 419. The boundaries for the
normal distribution are 157.5 < x < 419.5 (less than means but not
equal to
Ex. 2: Using a Correction for Continuity
Use the correction for continuity to convert each of the following
binomial intervals to a normal distribution interval.
3. The probability of getting less than 63 successes.
SOLUTION:
3. The probability of getting less than 63 successes.
The midpoint values are . . . 60, 61, 62. The boundary for the normal
distribution is x < 62.5 (less than means but not equal to
Ex. 3: Approximating a Binomial Probability
Thirty-seven percent of Americans say they always fly an
American flag on the Fourth of July. You randomly select
15 Americans and ask each if he or she flies an
American flag on the Fourth of July. What is the
probability that fewer than eight of them reply yes?
SOLUTION: From Example 1, you know that you can
use a normal distribution with  = 5.55 and  ≈1.87 to
approximate the binomial distribution. By applying the
continuity correction, you can rewrite the discrete
probability P(x < 8) as P (x < 7.5). The graph on the
next slide shows a normal curve with  = 5.55 and 
≈1.87 and a shaded area to the left of 7.5. The z-score
that corresponds to x = 7.5 is
Continued . . .
z
x

7.5  5.55

 1.04
1.87
Using the Standard Normal
Table,
P (z<1.04) = 0.8508
So, the probability that fewer than eight people respond yes is
0.8508
Ex. 4: Approximating a Binomial Probability
Twenty-nine percent of Americans say they are confident
that passenger trips to the moon will occur during their
lifetime. You randomly select 200 Americans and ask if
he or she thinks passenger trips to the moon will occur in
his or her lifetime. What is the probability that at least 50
will say yes?
SOLUTION: Because np = 200 ● 0.29 = 58 and nq =
200 ● 0.71 = 142, the binomial variable x is
approximately normally distributed with
  np  58
and
  npq  200  0.29  0.71  6.42
Ex. 4 Continued
Using the correction for continuity, you can rewrite
the discrete probability P (x ≥ 50) as the
continuous probability P ( x ≥ 49.5). The graph
shows a normal curve with  = 58 and  = 6.42,
and a shaded area to the right of 49.5.
Ex. 4 Continued
The z-score that corresponds to 49.5 is
z
x

49.5  58

 1.32
6.42
So, the probability that at least 50 will say yes is:
P(x ≥ 49.5) = 1 – P(z  -1.32)
= 1 – 0.0934
= 0.9066
Study Tip
• In a discrete distribution, there is a
difference between P (x ≥ c) and P( x > c).
This is true because the probability that x
is exactly c is not zero. IN a continuous
distribution, however, there is no
difference between P (x ≥ c) and P (x >c)
because the probability that x is exactly c
is zero.
Ex. 5: Approximating a Binomial Probability
A survey reports that 48% of Internet users use
Netscape as their browser. You randomly select
125 Internet users and ask each whether he or
she uses Netscape as his or her browser. What
is the probability that exactly 63 will say yes?
SOLUTION: Because np = 125 ● 0.48 = 60 and nq =
125 ● 0.52 = 65, the binomial variable x is approximately
normally distributed with
  np  60
and
  npq  125  0.48  0.52  5.59
Ex. 5 Continued
Using the correction for continuity, you can rewrite
the discrete probability P (x ≥ 63) as the
continuous probability P ( 62.5 < x < 63.5). The
graph shows a normal curve with  = 60 and  =
5.59, and a shaded area between 62.5 and 63.5.
Ex. 5 Continued
The z-scores that corresponds to 62.5 and 63.5
are:
z
z
x

x

62.5  60

 0.45
5.59
63.5  60

 0.63
5.59
Ex. 5 Continued
So, the probability that at least 50 will say yes is:
P(62.5 < x < 63.5) = P (0.45 < z < 0.63)
= P(z < 0.63) – P(z < 0.45)
= 0.7357 – 0.6736
= 0.0621
There is a probability of about 0.06 that exactly
63 of the Internet users will say they use
Netscape.
Reminders for the next couple of weeks
Friday and Monday– Work in Class on 5.5 and
Review Chapter 5.
Tuesday – Chapter 5 Test/Binder Grade
Wednesday – Review for Semester Exam – old
exams
Thursday– Semester Exam
Friday – 6.1Confidence Intervals for the Mean
(Large Samples)
Tuesday –6.2 Confidence Intervals for the Mean
(Small Samples)
Thursday – 6.3 Confidence Intervals for
Population Proportions