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AGENDA:
• DG 8 --- 15 minutes
• Begin Part 2 Unit 1 Lesson 11
Accel Precalc
Unit #1 Data Analysis
Lesson #11 Central Limit Theorem
EQ: How do the mean and standard deviation
of a sampling distribution compare to the mean
and standard deviation of a population?
Recall:
Suppose a teacher gave an 8-point quiz to a small
class of four students. The results of the quiz were
2, 6, 4, and 8. Assume that these students constitute
a population.
Create a histogram of the original quiz scores.
Describe the shape of
the distribution.
uniform
1
.75
P(X)
.5
.25
2 4 6 8
Quiz Score
Find the mean and the standard deviation of
the population quiz scores.
(2)(.25) + (4)(.25) + (6)(.25) + (8)(.25) = 5
 = ________________________
2  52 .25  4  52 .25  6  52 .25  8  52 .25 
 = ____________________________
5  2.236
Now choose 2 quiz scores (with replacement).
Create a table listing all possible samples of
size 2 then calculate the mean for each
sample.
Samples of size n = 2
2 , 2
2 , 4
.
.
.
8 , 8
Mean
(2 + 2) / 2 = 2
(2 + 4) / 2 = 3
(8 + 8) / 2 = 8
Create a histogram of the sample means.
2 3 4 5 6 7 8
Sample Means
Describe the shape of the distribution.
Symmetric, bell-shaped
New Notation:
x
mean of the sample means
standard deviation of the
sample means
•Calculate the mean and standard deviation
of the sample means from your table above.
1 2 3 4 3 2 1
2   3   4   5   6   7   8   5
 16   16   16   16   16   16   16 
 X  ________________________
“mu x-bar”
X 
2 1 
2 2 
2 2 
2 1 








2

5

3

5

...

7

5

8

5
 
 _________
 
 
__________
 16 
 1.5811
 16 
“standard error of the mean”
 16 
 16 

2.236

 1.581
n
2
They are equal.
Three Properties of a Sampling
Distribution:
1. The mean of the sample means will
always equal the mean of the population.
2. The standard deviation will always
be greater than the standard error of
the means.
3. Central Limit Theorem ---As the
sample size of a sampling distribution
gets larger, the shape of the
distribution will become approximately
NORMAL regardless of the shape of
the population distribution.
Original Distribution
Normal
Not Normal
Sampling Distribution
Normal
if n > 30, sampling dist
becomes approx normal
Recall: Standardizing Values
Individual Data
Sample Mean
z = obs – mean
standard deviation
z = obs – mean
standard error
Ex 1. A.C. Nielsen reported that children between the ages
of 2 and 5 watch an average of 25 hours of TV per week.
Assume the variable is normally distributed and the
standard deviation is 3 hours. If 20 children between the
ages of 2 and 5 are randomly selected, find the probability
that the mean number of hours they watch TV will be
greater than 26.3 hours.
  25   3 n  20
P( X  26.3)
1.94


26..3325
25
26

26..33)) PP zz 
PP((XX  26
33


20
20



94))  0.0262
  PP((zz 11..94



Ex 2. The average age of a vehicle registered in
the United States is 8 years or 96 months.
Assume the standard deviation is 16 months. If a
random sample of 36 vehicles is selected, find the
probability that the mean of their age is between
90 and 100 months.
  96   16 n  36
P(90  X  100)


9096
96
1009696
90
100
P(90  XX 100
100))PP

z

z
16
16
16
16


36
36
36
 36
-2.25
1.5


  PP
2.225.25 z z 1.51.5



 0.921
Ex 3. The average number of pounds of meat
that a person consumes a year is 218.4 pounds.
Assume that the standard deviation if 25 pounds
and the distribution is approximately normal.
a.
Find the probability that a person selected at
random consumes less than 224 pounds per year.
  218.4   25 n  1
P( X  224)
0.22
 218
 218
.4 .4 
  224224
( X224
 224
P( XP
)  P) zP z 
  P( zP( z0.220).22
 0) .5871
25 25  
 
Ex 3. The average number of pounds of meat
that a person consumes a year is 218.4 pounds.
Assume that the standard deviation if 25 pounds
and the distribution is approximately normal.
b.
If a sample of 40 individuals is selected, find
the probability that the mean of the sample will be
less than 224 pounds per year.
  218.4   25 n  40
P( X  224)


224  218.4

P( X  224)  P z 
25


40

1.42


  P( z  1.42)  0.922



Assignment:
Practice Worksheet: Central
Limit Theorem
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