Transcript Chapter 5

Chapter 5
Section 5-5
Chapter 5
Normal Probability Distributions
Section 5-5 – Normal Approximations to Binomial Distributions
A. Properties of a Normal Approximation to a Binomial Distribution
1. If np ≥ 5, and nq ≥ 5, then the binomial random variable x is
approximately normally distributed, with a mean that equals np
and a standard deviation that equals 𝑛𝑝𝑞.
a. Again, if np ≥ 5, and nq ≥ 5, then μ = np and σ = 𝑛𝑝𝑞
b. We need to remember from Section 4-2 what the properties
of a binomial experiment are:
1) n independent trials (we know before we start how many
trials there are going to be).
2) Only two possible outcomes (success or failure).
3) Probability of success is p.
4) Probability of failure is 1 – p, which we call q.
5) p is constant for each trial (the trials have nothing to do
with each other).
Chapter 5
Normal Probability Distributions
Section 5-5 – Normal Approximations to Binomial Distributions
2. Correction for Continuity
a. Binomial distributions only work for discrete data points.
1) When we want to calculate the exact binomial
probabilities, we can find the probability of each value of x
occurring and add them together. We did this in Chapter 4.
b. To use a continuous normal distribution to approximate a
binomial probability, you need to move .5 unit to each side of
the midpoint to include all possible x-values in the interval.
1) This is called making a correction for continuity.
a) We simply subtract .5 units from the lowest value and
add .5 units to the highest value.
3. There is a good review chart with this information displayed on
page 288 of your text book.
Chapter 5
Normal Probability Distributions
Section 5-5 – Normal Approximations to Binomial Distributions
a. The steps to using the Normal Distribution to Approximate
Binomial Probabilities are:
1) Verify that the binomial distribution applies.
a) Specify n, p, and q.
2) Determine if you can use the normal distribution to
approximate x, the binomial variable.
a) Are np and nq both greater than or equal to 5?
3) Find the mean and standard deviation for the distribution.
a) μ = np and σ = 𝑛𝑝𝑞.
4) Apply the approximate continuity correction. Shade the
corresponding area under the normal curve.
a) Subtract .5 unit from lowest value, add .5 unit to
highest value.
Chapter 5
Normal Probability Distributions
Section 5-5 – Normal Approximations to Binomial Distributions
5) Find the corresponding z-score(s).
𝑥−𝜇
a) 𝑧 =
𝜎
6) Find the probability.
a) Use the calculator.
Example 1A (Page 286)
51% of adults in the US who resolved to exercise more in the new
year achieved their resolution. You randomly select 65 adults in
the US whose resolution was to exercise more and ask each if he
or she achieved their resolution.
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If you can,
find the mean and standard deviation. If you cannot, explain
why.
What are n, p and q?
n = 65
p = 0.51
q = 0.49
Example 1A (Page 286)
51% of adults in the US who resolved to exercise more in the new
year achieved their resolution. You randomly select 65 adults in
the US whose resolution was to exercise more and ask each if he
or she achieved their resolution.
n = 65
p = 0.51
q = 0.49
Are np and nq greater than or equal to 5?
(65)(.51) = 33.15 and (65)(.49) = 31.85
Since both of these are greater than 5, we CAN use the normal
distribution.
Example 1A (Page 286)
51% of adults in the US who resolved to exercise more in the new
year achieved their resolution. You randomly select 65 adults in
the US whose resolution was to exercise more and ask each if he
or she achieved their resolution.
n = 65
p = 0.51
q = 0.49
REMEMBER THE ROUND-OFF RULE!!!!
Mean, standard deviation and variance are rounded to one
decimal place more than the x-values.
Since we are talking about adults, the x-values are whole
numbers. Hence, we round the mean and standard
deviation to the nearest tenth.
Example 1B (Page 286)
15% of adults in the US do not make New Year’s resolutions. You
randomly select 15 adults in the US and ask each if he or she
made a New Year’s resolution.
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If you can,
find the mean and standard deviation. If you cannot, explain
why.
What are n, p and q?
n = 15
p = 0.15
q = 0.85
Example 1B (Page 286)
15% of adults in the US do not make New Year’s resolutions. You
randomly select 15 adults in the US and ask each if he or she
made a New Year’s resolution.
n = 15
p = 0.15
q = 0.85
Are np and nq greater than or equal to 5?
(15)(.15) = 2.25 and (15)(.85) = 12.75
Since np < 5, we CANNOT use the normal distribution to
approximate the distribution of x.
Example 2 (Page 287)
Use a 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.
Since we are dealing with whole numbers, we subtract .5 from
the low end and add .5 to the high end.
270 - .5 = 269.5
310 + .5 = 310.5
Our interval is 269.5 < x < 310.5
2. The probability of at least 158 successes.
Since 158 is the low end, our interval is x > 157.5.
3. The probability of getting less than 63 successes.
We want all numbers less than 63, which makes 62 the upper
end. We add .5 to the upper end to get x < 62.5.
Example 3 (Page 288)
51% of adults in the US who resolved to exercise more in the new
year achieved their resolution. You randomly select 65 adults in
the US whose resolution was to exercise more and ask each if he
or she achieved their resolution.
What is the probability that fewer than 40 of them respond yes?
We know from Example 1A that we can use the normal
distribution, with a mean of 33.2 and a standard deviation of
4.0
Correcting for continuity means that we use 39.5, since 39 is
the highest number less than 40, and it is at the high end of
the interval.
normalcdf(0,39.5,33.2,4) gives us .942.
We have a 94.2% probability that fewer than 40 people
will respond “Yes”.
Example 4 (Page 289)
38% of people in the US admit that they snoop in other people’s
medicine cabinets. You randomly select 200 people in the United
States and ask each if they snoop in other people’s medicine
cabinets.
What is the probability that at least 70 will say yes?
Can we use the normal distribution?
𝑛𝑝 = 200 .38 = 76 and 𝑛𝑞 = 200 .62 = 124
Since both of these are ≥ 5, we CAN use the normal
distribution with a mean of 76 (np)
The standard deviation = 200 .38 .62 = 6.686
Correcting for continuity means that we subtract .5 from 70
(the low end of the interval) to get 69.5.
normalcdf(69.5,10000,76,6.686) = .835
We have an 83.5% probability that at least 70 people
will respond “Yes”.
Example 5 (Page 290)
1. A survey reports that 86% of internet users use Windows
Internet Explorer as their browser. You randomly select 200
internet users and ask each whether he or she uses Internet
Explorer as his or her browser.
What is the probability that exactly 176 say yes?
SOLUTION: Can we use the normal distribution?
np = 200(.86) = 172
nq = 200(.14) = 28
This means that we can use the normal approximation.
The mean is 𝑛𝑝 = 172
The standard deviation is 200 .86 (.14) ≈ 4.9
Since we want exactly 176, we use normalpdf!!
Example 5 (Page 290)
1. A survey reports that 86% of internet users use Windows
Internet Explorer as their browser. You randomly select 200
internet users and ask each whether he or she uses Internet
Explorer as his or her browser.
What is the probability that exactly 176 say yes?
2nd VARS binompdf(176,172,4.9) = .0583
Alternatively, we can use the normalcdf to find the area
between 175.5 and 176.5 (176 corrected for continuity).
2nd VARS normalcdf(175.5,176.5,172,4.9) = .0583
We have about a 5.83% chance of getting exactly 176 out of
200 people to say that they use Internet Explorer as their
browser.
YOUR ASSIGNMENTS TODAY ARE:
Classwork:
Page 291 #1-16 All
Homework:
Page 292-294 #17-26 All