Download Chapter 5 Normal Probability Distributions

Chapter 5
Normal Probability Distributions
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
A. Central Limit Theorem basically says this:
1. If you have samples of 30 or more, the distribution of sample
means will be a normal distribution.
2. If the population is normally distributed, then the distribution
of sample means will be normal no matter what sample size
you use.
3. Whether the original population distribution is normal or not,
the sampling distribution of sample means has a mean equal
to the population mean.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
4. The standard deviation of sample means will be smaller than
the standard deviation of the population.
a. The standard deviation of the distribution of sample
means, σₓ, is also called the standard error of the mean.
b. The standard deviation of sample means is equal to the
population standard deviation divided by the square root
𝜎
of n. 𝜎𝑥 =
𝑛
1) The bigger n (the sample size) gets, the smaller the
sampling standard deviation will get.
a) The more times we take a sample of the same
population, the more tightly grouped the results
will be.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
C. Probability and the Central Limit Theorem
1. Using what we’ve learned in Section 5-2, and what we’ve been
told here in Section 5-4, we can find the probability that a
sample mean will fall in a given interval of the sampling
distribution.
a. We will still use normalcdf, but instead of using the
population standard deviation, we will divide by 𝑛 and use
that answer as our standard deviation.
1) Everything else is the same as what we did in 5-2.
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 4 Page 275
The graph at the right shows
the length of time people
spend driving each day. You
randomly select 50 drivers
ages 15 to 19. What is the
probability that the mean time
they spend driving each day is
between 24.7 and 25.5
minutes? Assume that 𝜎 = 1.5
minutes.
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 4 Page 275
𝑛 > 30; Central Limit Theorem
applies.
Standard error equals
standard deviation (1.5)
divided by 50, (𝑛 = 50)
1.5
= .212
50
From here, we go to the
calculator:
normalcdf(24.7,25.5,25,.212)
= .912.
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 5 Page 276
The mean room and board expense per year of four-year colleges is
$6803. You randomly select 9 four-year colleges. What is the
probability that the mean room and board is less than $7088? Assume
that the room and board expenses are normally distributed, with a
standard deviation of $1125.
Because the population is normally distributed, you can use the Central
Limit Theorem to conclude that the distribution of sample means is
normally distributed.
1125
1125
µ = $6803; σ = $375 because (
)=
= 375
9
3
From here, normalcdf(-1E99,7088,6803,375)=.776
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 6 Page 277
A bank auditor claims that credit card balances are normally
distributed, with a mean of $2870 and a standard deviation of $900.
You randomly select 25 credit card holders and find that their mean
credit card balance is $2499.03.
1) What is the probability that a randomly selected credit card holder
has a credit card balance less than $2500?
2) You randomly select 25 credit card holders. What is the probability
that their mean credit card balance is less than $2500?
3) Compare the probabilities from (1) and (2) and interpret your
answer in terms of the auditor’s claim.
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 6 Page 277
A bank auditor claims that credit card balances are normally
distributed, with a mean of $2870 and a standard deviation of $900.
You randomly select 25 credit card holders and find that their mean
credit card balance is $2499.03.
1) What is the probability that a randomly selected credit card holder
has a credit card balance less than $2500?
We are talking about ONE individual here, so we use the given
standard deviation. (dividing by the 𝑛 would mean dividing by 1,
which is pointless).
normalcdf(-1E99,2500,2870,900)=.340
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 6 Page 277
A bank auditor claims that credit card balances are normally
distributed, with a mean of $2870 and a standard deviation of $900.
You randomly select 25 credit card holders and find that their mean
credit card balance is $2499.03.
2) You randomly select 25 credit card holders. What is the probability
that their mean credit card balance is less than $2500?
Now we are talking about a sample, so we divide by 𝑛.
900
900
The standard deviation of the sample is
=
= 180
25
normalcdf(0,2500,2870,180)=.020
5
Chapter 5
Normal Probability Distributions
Section 5-4 –
Example 6 Page 277
A bank auditor claims that credit card balances are normally
distributed, with a mean of $2870 and a standard deviation of $900.
You randomly select 25 credit card holders and find that their mean
credit card balance is $2499.03.
3) Compare the probabilities from (1) and (2) and interpret your
answer in terms of the auditor’s claim.
The probability of a single card holder owing less than $2500 is
34%, but the probability of the average of 25 card holders
balances is less than $2500 is only 2%.
Either the auditor is wrong about the distribution being normal, or
your sample is unusual and needs to be done again, more
carefully.
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.
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.
We will use either binompdf or normalcdf, as
needed.
b. There is a good review chart with this information displayed on
page 288 of your text book.
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 (1 – p)
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 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
𝜇 = 𝑛𝑝 = 33.15 ≈ 33.2
𝜎 = 𝑛𝑝𝑞 = 65 ∗ .51 ∗ .49 = 16.2 = 4.03 ≈ 4.0
The mean is 33.2 and the standard deviation is 4.0.
All this means is that for every 65 people you survey, you would
expect 33.2 of them to say that they achieved their resolution, with
a standard deviation of 4.0.
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
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
We would use these in the normalcdf distribution.
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(-1E99,39.5,33.2,4) 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 ≈ 6.7
Correcting for continuity means that we subtract .5 from 70
(the low end of the interval) to get 69.5.
normalcdf(69.5,1E99,76,6.7) = .834
We have an 83.4% 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 binompdf!!
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(200,.86,176) = .0612
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 somewhere between a 5.83% and a 6.12% chance of
getting exactly 176 out of 200 people to say that they use
Internet Explorer as their browser.
YOUR ASSIGNMENTS TODAY ARE:
Classwork:
Pages 278-279 #1-8, and 11-16 All (Skip 9 & 10)
Page 291 #1-16 All
Homework:
Pages 279-282 #18-38 Evens
Pages 292-294 #17-26 All