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Chapter 5
Normal Probability Distributions
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
A. A sampling distribution is the probability distribution of a sample
statistic that is formed when samples of size n are repeatedly taken
from a population.
1. If the sample statistic is the sample mean, then the distribution is
the sampling distribution of sample means.
a. Every sample statistic has a sampling distribution.
2. Remember that sample means can vary from one another and
can also vary from the population mean.
a. This type of variation is to be expected and is called sampling
error.
3. Properties of sampling distributions of sample means:
a. The mean of the sample means is equal to the population
mean.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
b. The standard deviation of the sample means is equal to the
population standard deviation divided by the square root of
n.
1) The standard deviation of the sampling distribution of the
sample means is called the standard error of the mean.
B. The Central Limit Theorem
1. The Central Limit Theorem forms the foundation for the
inferential branch of statistics.
a. It describes the relationship between the sampling
distribution of sample means and the population that the
samples are taken from.
b. It is an important tool that provides the information you’ll
need to use sample statistics to make inferences about a
population mean.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
2. The Central Limit Theorem says:
a. If samples of size n, where n ≥ 30, are drawn from any
population with a mean μ and a standard deviation σ, then
the sampling distribution of sample means approximates a
normal distribution.
1) The greater the sample size (the larger number n is), the
better the approximation.
b. If the population itself is normally distributed, the sampling
distribution of sample means is normally distributed for any
sample size n.
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
a. In real life words, this means that if we take the average of all
of the means, from all of the samples that are done on one
population, the mean of those averages will equal the mean
of the population.
4. The sampling distribution of sample means has a variance equal
to 1/n times the variance of the population.
a. The standard deviation of sample means will be smaller than
the standard deviation of the population.
5. The sampling distribution of sample means has a standard
deviation equal to the population standard deviation divided by
the square root of n.
a. The distribution of sample means has the same center as the
population, but it is not as spread out.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
1) The bigger n (the sample size) gets, the smaller the
standard deviation will get.
a) The more times we take a sample of the same
population, the more tightly grouped the results will
be.
b. The standard deviation of the sampling distribution of the
sample means, σₓ, is also called the standard error of the
mean.
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.
Chapter 5
Normal Probability Distributions
Section 5-4 – Sampling Distributions and the Central Limit Theorem
a. To find a z-score of a random variable x, we took the value
minus the mean and divided by the standard deviation.
b. To convert the sample mean to a z-score, we alter that
slightly.
1) Instead of dividing by the standard deviation, we divide by
the sample error.
a) Remember, this is the standard deviation of the
population divided by the square root of n (the sample
size).
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(0,7088,6803,375)=.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.
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.
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(0,2500,2870,900)=.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.
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.
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.
Your Assignments are:
Classwork:
Pages 278-279 #1-16 All
Homework:
Pages 279-282 #18--38 Evens