SPA 3e_ Teachers Edition _ Ch 6

L E S S O N 6.5 • The Sampling Distribution of a Sample Mean 435
3. Click on “Clear lower 3” to start clean. Then click
on the “100,000” button under “Sample:” to simulate
taking 100,000 SRSs of size n 5 5 from the
population. Answer these questions:
• Does the simulated sampling distribution
(blue bars) have a recognizable shape? Click
the box next to “Fit normal.”
• To the left of each distribution is a set of
summary statistics. Compare the mean of
the simulated sampling distribution with the
mean of the population.
Describing a Sampling Distribution of a Sample Mean
When Sampling from a Normal Population
• How is the standard deviation of the simulated
sampling distribution related to the
standard deviation of the population?
4. Click “Clear lower 3.” Use the drop-down menus
to set up the bottom graph to display the mean
for samples of size n 5 20. Then sample 100,000
times. How do the two distributions of x compare:
shape, center, and variability?
5. What have you learned about the shape of the
sampling distribution of x when the population
has a normal shape?
As the activity demonstrates, if the population distribution is normal, so is the
sampling distribution of x. This is true no matter what the sample size is.
Suppose that a population is normally distributed with mean m and standard deviation s. Then
the sampling distribution of x for SRSs of size n has a normal distribution with mean m and
standard deviation s "n.
FYI
The exact formula for the standard
deviation of x when sampling without
replacement is s x = s # N − n
"n Å N − 1 .
When n is small relative to N (less than
N − n
10%), the value of Å N − 1 is
approximately 1 and therefore doesn’t
s
change the value of much. The
"n
N − n
factor is sometimes called the
Å N − 1
finite population correction factor.
Lesson 6.5
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In the next lesson, you will learn what happens when sampling from a non-normal
population.
Finding Probabilities Involving x
Now we have enough information to calculate probabilities involving x when the
population distribution is normal.
Are those peanuts underweight?
Probabilities involving x
PROBLEM: At the P. Nutty Peanut Company, dry-roasted, shelled peanuts
are placed in jars labeled “16 ounces” by a machine. The distribution of
weights in the jars is approximately normal with a mean of 16.1 ounces and
a standard deviation of 0.15 ounces. Find the probability that the mean
weight of 10 randomly selected jars is less than the advertised weight of
16 ounces.
e XAMPLe
SOLUTION:
Let x 5 the sample mean weight of 10 randomly
• Mean: m x - 5 16.1 ounces
selected jars. To find P(x ≤ 16), we have to know the
• SD: s x - = 0.15
mean, standard deviation, and shape of the sampling
"10 = 0.047 ounces distribution of x. Recall that m x = m and s x = s "n .
Charles Nesbit/Getty
Images
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Alternate Example
How hungry are the hounds?
Probabilities involving x
PROBLEM: The local SPCA (Society for the
Prevention of Cruelty to Animals) feeds the
animals it shelters, but the amount of food
needed per day varies. The distribution of
the weight of dry dog food used per day
is approximately normal, with a mean of
30 pounds and standard deviation of 5.1
pounds. Find the probability that the mean
weight of dry dog food for 6 randomly
selected days is greater than 33 pounds.
SOLUTION:
Let x 5 the sample mean weight of dog
food on 6 randomly selected days.
• Mean: m x = 30 pounds
• SD: s x = 5.1 = 2.08 pounds
!6
• Shape: Approximately normal
because the population distribution is
approximately normal
33
23.76 25.84 27.92 30.00 32.08 34.16 36.24
Sample mean weight of dry dog food
33 − 30
Using Table A: z = = 1.44 and
2.08
P(Z ≥ 1.44) 5 1 2 0.9251 5 0.0749
Using technology: Applet/normalcdf(lower:
33, upper: 10000, mean: 30, SD: 2.08) 5
0.0746
L E S S O N 6.5 • The Sampling Distribution of a Sample Mean 435
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