05.02 Sample Means

Practice #4 The number of bags of chips sold each day at a vending machine at a theme park follows a Normal distribution, with a mean of 50 and a standard deviation of 15. A random sample of the number of bags of chips sold 20 days during the year is taken. What is the probability that the mean number of bags of chips sold will be less than 45?

0.0680

Practice #3 Akron Cinema sells an average of 500 tickets on Mondays, with a standard deviation of 50 tickets. If a simple random sample is taken of the mean amount of ticket sales from 30 Mondays in a year, what is the probability that the mean will be greater than 510?

0.1366

Normal distribution application example part 1 So it's finally time to figure out whether that pesky soda company was lying about its drinks! The true amount of soda in a can should be 12 ounces, with a standard deviation of 0.5 ounce. We purchase a 36-pack of soda and find the mean amount of soda per can is only 11.92 ounces. 1. Describe the graph of the sampling distribution of the mean for samples of a 36-pack of soda. 2. What are the mean and standard deviation of the sampling distribution of line over x?

1. Because the sample size of 36 is greater than 30, the CLT tells us the graph of the sampling distribution is approximately Normal. 2. µ line over x = µ = 12 σx = σ/sqrt(n) = 0.5/sqrt(36) = 0.5/6 = 0.0833 The mean of the sampling distribution is 12 ounces and the standard deviation is 0.0833 ounce (because there are at least 10 • 36 = 360 cans of soda in the population), as shown in the graph.

To use the CLT you must check two conditions:

1. Independence: The sample values must be independent of one another. 2. Sample size: The sample size, n, must be sufficiently large (n ≥ 30).

Practice 1 The cost for a cup of coffee at Starbucks follows a Normal distribution, with a mean of $3.25 and a standard deviation of $0.25. 1. What is the probability that a randomly selected person who goes to Starbucks will spend more than $3.50? 2. Find the probability that the mean amount of money spent at Starbucks for a simple random sample of 10 people is more than $3.50. 3. Find the probability that the mean amount of money spent at Starbucks for a simple random sample of 10 people is between $3.15 and $3.30. 4. Ninety-nine percent of the time, 10 randomly selected people who go to Starbucks will have a mean purchase of at most how much?

1. Let X be the cost of a cup of coffee at Starbucks for a randomly selected person. X follows N(3.25, 0.25). Because the conditions for a Normal distribution have been met, you can state that the Normal cumulative distribution function with a lower bound of 3.50, an upper bound of 9999, a mean of 3.25, and a standard deviation of 0.25 will result in a probability of 0.1587. Using a calculator, the command is normalcdf(3.50, 9999, 3.25, 0.25) = 0.1587. You could also calculate the z-score and use the Standard Normal Distribution table. z = 3.50−3.25/0.25 = 1 P(X > 3.50) = P(z > 1) = 1 - P(z ≤ 1) = 1 - 0.8413 = 0.1587. The probability that a randomly selected person who goes to Starbucks will spend more than $3.50 is 0.1587, or about 15.87%. 2. Because this sample is taken from a Normal population, the sample meets the Normal criteria as well. For a simple random sample of 10 people, the sampling distribution of the sample mean amount of money spent has the mean µx = µ = 3.25. The 10% condition is met because there are at least 10(10) = 100 Starbucks customers. The standard deviation is σx = σ/sqrt(n) = 0.25/sqrt(10) = 0.0791. Because the population distribution is Normal, the values will follow an N(3.25, 0.0791) distribution. To find the probability that x>3.50, we can use normalcdf(3.50, 9999, 3.25, 0.0791) = 7.874 • 10-4 = 0. There is an approximately zero probability that the mean amount of money spent at Starbucks for a simple random sample of 10 people is more than $3.50. 3. Method 1 To find the probability that 3.15 < x< 3.30, we can use normalcdf(3.15, 3.30, 3.25, 0.0791) = 0.6333. Because the conditions for a Normal distribution have been met, you can state that the Normal cumulative distribution function with a lower bound of 3.15, an upper bound of 3.30, a mean of 3.25, and a standard deviation of 0.0791 will result in a probability of 0.6333. There is a 0.6333, or about a 63.33%, probability that the mean amount of money spent at Starbucks for a simple random sample of 10 people is between $3.15 and $3.30. Method 2 Calculate each z-score and use the Standard Normal Table (Table A) to find the probability for each z-score. Then, subtract the probabilities to find the area between the two values: z = 3.30−3.25/0.0791 = 0.6321 and z = 3.15−3.25/0.0791 = −1.2642 P(3.15<X<3.30) = P(z<0.63)−P(z<−1.26) = 0.7357−0.1038 = 0.6319 4. With N(3.25, 0.0791), you are interested in finding the area under the curve up to 0.99, so you should use invNorm(0.99, 3.25, 0.0791) on a calculator, which yields 3.43. Ninety-nine percent of the time, 10 randomly selected people who go to Starbucks will have a mean purchase of at most $3.43.

Normal distribution application example part 2 3a. What is the probability of getting a mean of 11.92 ounces or less in a 36-pack of soda?

Because we know the 36-pack of soda has a sampling distribution of N(12, 0.0833), we can use Normal probability calculations: normalcdf(0, 11.92, 12, 0.0833) = 0.1684 There is a 0.1684, or approximately 16.84%, chance of getting a mean of 11.92 ounces or less in a 36-pack of soda. Or... You could also calculate the z-score and then use the Standard Normal Probability table provided in the AP Resource Packet: z = 11.92−12/0.0833 = −0.9604 P(X < 11.92) = P(z ≤ -0.9604) = 0.1685

Sampling distribution of the sample means example part 1 Was the soda company lying or not? Each 36-pack of soda has a certain mean amount of soda per can (line above x). Each of these means is probably close to, but not necessarily equal to, 12 ounces (given the company is telling the truth). Let's examine the distribution of all these means from 100,000 cans of soda. 1. Notice that the population distribution of 100,000 cans of soda is slightly skewed to the right, whereas the distribution of the sample means is approximately Normal.

Looking at these histograms, you can see that the sampling distribution of the sample means is centered at the population mean, μ, and is less spread out than the population distribution. So you still don't know whether the soda company is lying. There's more to learn first!

Sampling distribution of the sample means example part 2 Was the soda company lying or not? Each 36-pack of soda has a certain mean amount of soda per can (line above x). Each of these means is probably close to, but not necessarily equal to, 12 ounces (given the company is telling the truth). Let's examine the distribution of all these means from 100,000 cans of soda. 2. Also note that the sampling distribution of the sample means has less variability than the population distribution of 100,000 cans.

Looking at these histograms, you can see that the sampling distribution of the sample means is centered at the population mean, μ, and is less spread out than the population distribution. So you still don't know whether the soda company is lying. There's more to learn first!

CLT in Action #3

Repeat the same process again with n = 10. Here is the sampling distribution for that sample size. This certainly looks more Normal, and if you repeat this process one more time for n = 30, you see an even more Normal-looking result.

CLT in Action #2

Suppose that you gather 1,000 samples of three from the previous population. For each sample, you can compute its average. If you do that, you will have 1,000 averages. This set of 1,000 means is called a "sampling distribution" and, according to the CLT, the sampling distribution will approach a Normal distribution as the sample size n, used to produce it, increases. Here is what your sample distribution looks like for n = 3. As you can see, it certainly looks unimodal, although not necessarily Normal. If you repeat the same process with a larger sample size, you should see the sampling distribution start to become more Normal.

CLT in Action #1

The height in this histogram denotes the frequency of the number in the population. As you can see, the distribution isn't Normal, uniform, or any other commonly known pattern. To sample from this distribution, you need to define a sample size, n, which is the number of observations you sample at a time. Suppose you choose n = 3. This means you will sample in groups of three. For the population in the histogram, you might sample groups of three using a random number generator.

CLT in Action #4

The histograms demonstrate that as the sample size n is increased, the resultant sample mean distribution becomes more Normal. Furthermore, the distribution variance decreases.

When finding cumulative probabilities for values of x, you have three possible options

What It Sounds Like How It Looks What It Means (X < x) or (X ≤ x) normalcdf(−999999, x, x,σ x) Lower limit, upper limit, mean, standard deviation (X > x) or (X ≥ x) normalcdf(x, 999999, x,σ x) Lower limit, upper limit, mean, standard deviation x1 < X < x2 normalcdf(x1,x2,x,σ x) Lower limit, upper limit, mean, standard deviation Remember, always use the standard deviation of the sample, not the population!

Practice #2 What does the Central Limit Theorem State?

When n is sufficiently large, the sampling distribution of line above x is roughly approximated by the Normal curve.

AP Practice Question Consider the sampling distribution of a sample mean obtained by a random sampling from an infinite population. This population has a distribution that is highly skewed toward the larger values. a. How is the mean of the sampling distribution related to the mean of the population? b. How is the standard deviation of the sampling distribution related to the standard deviation of the population? c. How is the shape of the sampling distribution affected by the sample size?

a. The mean of the sampling distribution is equal to the mean of the population. b. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size. c. The sampling distribution is skewed for small sample sizes. The shape of the sampling distribution gets more and more Normal-like (bell shaped) as the sample size increases.

The amount of variation of a measure in a sample, from a population or subpopulation, depends on...

both the amount of variation in the population or subpopulation and on the size of the sample.

Practice #1 The mean of a population is 70 with a standard deviation of 10. A sample size of 40 is used. What are the mean and the standard deviation for this sample?

mean = 70 standard deviation = 10/sqrt(40)

line above x symbol represents

mean of a sample

µ symbol represents

mean of the entire population

When discussing quantitative variables, you are interested in the...

sampling distribution of the sample mean.

Central Limit Theorem (CLT)

the sampling distribution of the mean of any independent, random variable will be Normal or nearly Normal if the sample size is large enough

When you check the conditions for Normality and know: 1. Either the population is Normal, so the sampling distribution of x is Normal for any sample size or 2. The population distribution is not Normal or it is unknown, but the sample size is large (n ≥ 30), so by the CLT you know the sampling distribution is well approximated by a Normal curve then...

you know you can use Normal approximations for the mean and standard deviation.

Formula #1

µ line over x = µ µ line over x = mean of the sampling distribution of the means µ = mean of the population values

Formula #2

σx = σ/sqrt(n) σx = standard deviation of the sampling distribution of the means σ/sqrt(n) = standard deviation of the population divided by the square root of the sample size