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Lesson 8 - 1
Distribution of the Sample Mean
Objectives
• Understand the concept of a sampling distribution
• Describe the distribution of the sample mean for
sample obtained from normal populations
• Describe the distribution of the sample mean from
samples obtained from a population that is not
normal
Vocabulary
• Statistical inference – using information from a
sample to draw conclusions about a population
• Standard error of the mean – standard deviation of
the sampling distribution of x-bar
Conclusions regarding the sampling
distribution of X-bar
• Shape: normally distributed
• Center: mean equal to the mean of the population
• Spread: standard deviation less than the standard
deviation of the population
• Law of Large numbers tells us that as n increases
the difference between the sample mean, x-bar and
the population mean μ approaches zero
Central Limit Theorem
X or x-bar
Distribution
Regardless of the shape of the population, the sampling distribution of
x-bar becomes approximately normal as the sample size n increases.
Caution: only applies to shape and not to the mean or standard deviation
x
x
x
x
x
x
x
x
x
x
x
x
x
Random Samples Drawn from Population
Population Distribution
x
x
x
Summary of Distribution of x
Shape, Center and
Spread of Population
Distribution of the Sample Mean
Shape
Normal with mean, μ and
standard deviation, σ
Regardless of sample
size, n, distribution of
x-bar is normal
Population is not normal
with mean, μ and
standard deviation, σ
As sample size, n,
increases, the distribution
of x-bar becomes
approximately normal
Center
Spread
μx-bar = μ
σ
σx-bar = ------n
μx-bar = μ
σ
σx-bar = ------n
What Happens to Sample Spread
If the random variable X has a normal
distribution with a mean of 20 and a standard
deviation of 12
– If we choose samples of size n = 4, then the
sample mean will have a normal distribution with
a mean of 20 and a standard deviation of 6
– If we choose samples of size n = 9, then the
sample mean will have a normal distribution with
a mean of 20 and a standard deviation of 4
Example 1
The height of all 3-year-old females is approximately
normally distributed with μ = 38.72 inches and σ = 3.17
inches. Compute the probability that a simple random
sample of size n = 10 results in a sample mean greater
than 40 inches.
P(x-bar > 40)
μ = 38.72 σ = 3.17 n = 10
σx = 3.17 / 10 = 1.00244
x-μ
1.28
40 – 38.72
Z = ------------- = ----------------- = ----------------σx
1.00244
1.00244
= 1.277
normalcdf(1.277,E99) = 0.1008
normalcdf(40,E99,38.72,1.002) = 0.1007
a
Example 2
We’ve been told that the average weight of giraffes is
2400 pounds with a standard deviation of 300 pounds.
We’ve measured 50 giraffes and found that the sample
mean was 2600 pounds. Is our data consistent with
what we’ve been told?
P(x-bar > 2600)
μ = 2400 σ = 300 n = 50
σx = 300 / 50 = 42.4264
x-μ
200
2600 – 2400
Z = ------------- = ----------------- = ----------------σx
42.4264
42.4264
= 4.714
normalcdf(4.714,E99) = 0.000015
normalcdf(2600,E99,2400,42.4264) = 0.0000001
a
Summary and Homework
• Summary:
– The sample mean is a random variable with a
distribution called the sampling distribution
• If the sample size n is sufficiently large (30 or more is a
good rule of thumb), then this distribution is
approximately normal
• The mean of the sampling distribution is equal to the
mean of the population
• The standard deviation of the sampling distribution is
equal to σ / n
• Homework
– pg 431 – 433; 3, 4, 6, 7, 12, 13, 22, 29
Homework
• 3) standard error of the mean
• 4) zero
• 6) population is normal
• 7) four
• 12) μ=64, σ= 18, n=36
μx-bar =64, σx-bar= 18/36 = 18/6 = 3
• 13) μ=64, σ= 18, n=36
μx-bar =64, σx-bar= 18/36 = 18/6 = 3
• 22) μ=81.7, σ= 6.9
•
a) P(x < 75) = 0.1658
normalcdf(-E99,75,81.7,6.9)
•
b) n=5 P(x<75) =0.01496
normalcdf(-E99,75,81.7,6.9/5)
•
c) n=8 P(x<75) =0.00301
normalcdf(-E99,75,81.7,6.9/8)
•
d) very unlikely (or unusual)
• 29) P(x < 45) = 0.0078
normalcdf(-E99,45,50,16/60)