Chapter 7The Normal Probability

Distribution7.5

Sampling Distributions;

The Central Limit Theorem

Illustrating Sampling Distributions

Step 1: Obtain a simple random sample of size n.

Illustrating Sampling Distributions

Step 1: Obtain a simple random sample of size n.Step 2: Compute the sample mean.

Illustrating Sampling Distributions

Step 1: Obtain a simple random sample of size n.Step 2: Compute the sample mean.Step 3: Assuming we are sampling from a finite population, repeat Steps 1 and 2 until all simple random samples of size n have been obtained.

EXAMPLE Illustrating a Sampling Distribution

The following data represents the ages of 6 individuals that are members of a weekend golf club.

23, 25, 49, 32, 38, 43

Treat these 7 individuals as a population.

(a) Compute the population mean.

(b) List all possible samples of size n = 2 and determine the sample mean age.

(c) Construct a relative frequency distribution of the sample means. This distribution represents the sampling distribution of the sample mean.

EXAMPLE Illustrating a Sampling Distribution

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 5 from this population.

The data on the following slide represent the sample means for the 200 simple random samples of size n = 5.

For example, the first sample of n = 5 had the following data:

2.493 2.466 2.473 2.492 2.471

The mean of the sample means is 2.46, the same as the mean of the population.

The standard deviation of the sample means is 0.0086, which is smaller than the standard deviation of the population.

The next slide shows the histogram of the sample means.

“What role does n, the sample size, play in

sampling distribution of ?”

Suppose the sample mean is computed for samples of size n = 1 through n = 200. That is, the sample mean is recomputed each time an additional individual is added to the sample. The sample mean is then plotted against the sample size.

x

EXAMPLE Finding the Area Under a Normal Curve

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams Approximate the sampling distribution of the sample mean by obtaining 200 simple random samples of size n = 15 from this population of pennies minted after 1982.

The mean of the 200 sample means is 2.46 (the mean of the population).

The standard deviation of the 200 sample means is 0.0049.

Notice that the standard deviation of the sample means is smaller for the larger sample size.

EXAMPLE Describing the Distribution of the Sample Mean

The weights of pennies minted after 1982 are approximately normally distributed with mean 2.46 grams and standard deviation 0.02 grams.

What is the probability that in a simple random sample of 10 pennies minted after 1982, we obtain a sample mean of at least 2.465 grams?

EXAMPLE Sampling from a Non-normal Population

The following distribution represents the number of people living in a household for all homes in the United States in 2000. Obtain 200 simple random samples of size n = 4; n = 10 and n = 30. Draw the histogram of the sampling distribution of the sample mean.

EXAMPLE Using the Central Limit Theorem

Suppose that the mean time for an oil change at a “10-minute oil change joint” is 11.4 minutes with a standard deviation of 3.2 minutes.

(a) If a random sample of n = 35 oil changes is selected, describe the sampling distribution of the sample mean.

(b) If a random sample of n = 35 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?

EXAMPLE Using the Central Limit Theorem

(c ) If a random sample of n = 50 oil changes is selected, what is the probability the mean oil change time is less than 11 minutes?

(d) What effect did increasing the sample size have on the probability?

Chapter 7The Normal Probability

Distribution7.6

The Normal Approximation to the Binomial Probability

Criteria for a Binomial Probability Experiment

An experiment is said to be a binomial experiment provided

1. The experiment is performed a fixed number of times. Each repetition of the experiment is called a trial.

2. The trials are independent. This means the outcome of one trial will not affect the outcome of the other trials.

3. For each trial, there are two mutually exclusive outcomes, success or failure.

4. The probability of success is fixed for each trial of the experiment.

Notation Used in the Binomial Probability Distribution

• There are n independent trials of the experiment

• Let p denote the probability of success so that 1 – p is the probability of failure.

• Let x denote the number of successes in n independent trials of the experiment. So, 0 < x < n.

As the number of trials n in a binomial experiment increase, the probability distribution of the random variable X becomes symmetric and bell-shaped. As a general rule of thumb, if np(1 – p) > 10, then the probability distribution will be approximately symmetric and bell-shaped.

P(X = 18) ≈ P(17.5 < X < 18.5)

P(X < 18) ≈ P(X < 18.5)

Summary

EXACT APPROXIMATE BINOMIAL NORMALP(X = a) P(a - 0.5 < X < a + 0.5)

P(X < a) P(X < a + 0.5)

P(X > a) P(X > a - 0.5)

P(a < X < b) P(a - 0.5 < X < b + 0.5)

NOTE: For P(X < a) = P(X < a - 1), so rewrite P(X < 5) as P(X < 4).

EXAMPLE Using the Binomial Probability Distribution Function

According to the United States Census Bureau, 18.3% of all households have 3 or more cars.

(a) In a random sample of 200 households, what is the probability that at least 30 5 have 3 or more cars?

(b) In a random sample of 200 households, what is the probability that less than 15 have 3 or more cars?

(c) Suppose in a random sample of 500 households, it is determined that 110 have 3 or more cars. Is this result unusual? What might you conclude?