Transcript Chapter 7 The Normal Probability Distribution

Chapter 7
The Normal Probability
Distribution
7.5
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 x ?”
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.
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 “10minute 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 7
The Normal Probability
Distribution
7.6
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
BINOMIAL
P(X = a)
APPROXIMATE
NORMAL
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?