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Chapter 6:. Section 6.5: Central Limit Theorem. Properties of the Distribution of Sample Means. 1. The mean of the sample means will be the same as the population mean. 2. The standard deviation of the sample means will be smaller than the standard deviation of

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Chapter 6

Chapter 6:

Section 6.5: Central Limit Theorem


Chapter 6

Properties of the Distribution of Sample Means

1. The mean of the sample means will be the

same as the population mean

2. The standard deviation of the sample means

will be smaller than the standard deviation of

the population; and standard deviation of the mean =

σ

√n


Chapter 6

In the previous lesson, we learned that when

solving for z:

X – µ

σ

z =

Now, since we know the standard deviation of

The sample means is:

σ

√n

σ

√n

We simply plug in for the standard deviation

In the same equation and use X instead of X:

_

_

X – µ

z =

σ

√n


Chapter 6

**HINT: When using this equation in the

Calculator, be sure to put parenthesis around

BOTH the numerator and the denominator

_

(X – µ)

z =

( )

σ

√n


Chapter 6

Difference between the z = equations:

X – µ

σ

1. For individual data:

z =

_

(X – µ)

2. For sample means:

z =

σ

√n


Chapter 6

EXAMPLE 1: The mean time to complete a psychology exam is

34 minutes with a standard deviation of 8. If a class of 30

students takes the exam, find the probability that the sample

mean will be more than 30 minutes.

Are we dealing with individual data or sample means?

_

(X – µ)

Sample means, so we use the equation z =

σ

√n

µ =

34

σ=

8

n =

30

_

More than 30 minutes:

X =

30

_

(30 – 34)

(X – µ)

=

= -2.74

z =

σ

√n

8

√30

( )


Chapter 6

EXAMPLE 1: The mean time to complete a psychology exam is

34 minutes with a standard deviation of 8. If a class of 30

students takes the exam, find the probability that the sample

mean will be more than 30 minutes.

µ =

34

σ=

8

n =

30

z = -2.74

_

More than 30 minutes:

X =

30

-2.74

0

z = -2.74

Area = 0.4969

Since the questions asks for MORE THAN 30, shade to the right

Area = 0.4969 + 0.5 = 0.9969


Chapter 6

EXAMPLE 2: The average hourly wage of fast-food workers

employed by a nationwide chain is $5.55. The standard

deviation is $1.15. If a sample of 50 workers is selected, find

the probability that the mean of the sample will be between

$5.25 and $5.90?

Are we dealing with individual data or sample means?

_

(X – µ)

Sample means, so we use the equation z =

σ

√n

µ =

5.55

σ=

1.15

n =

50

_

Between $5.25 and $5.90:

X =

5.25, 5.90

(5.90 – 5.55)

(5.25 – 5.55)

z =

z =

= 2.15

= -1.84

1.15

√50

1.15

√50

( )

( )


Chapter 6

EXAMPLE 2: The average hourly wage of fast-food workers

employed by a nationwide chain is $5.55. The standard

deviation is $1.15. If a sample of 50 workers is selected, find

the probability that the mean of the sample will be between

$5.25 and $5.90?

z = -1.84, 2.15

µ =

5.55

σ=

1.15

n =

50

_

Between $5.25 and $5.90:

X =

5.25, 5.90

-1.84

0

2.15

z = -1.84

Area = 0.4671

z = 2.15

Area = 0.4842

Since the questions asks for BETWEEN, shade between and add

Area = 0.4671 + 0.4842 = 0.9513


Chapter 6

EXAMPLE 3: The mean grade point average of the engineering

majors at a large university is 3.23, with a standard deviation

of 0.72. In a class of 48 students, find the probability that the

mean grade point average of the student is less than 3.15.

Are we dealing with individual data or sample means?

_

(X – µ)

Sample means, so we use the equation z =

σ

√n

µ =

3.23

σ=

0.72

n =

48

_

Less than 3.15:

X =

3.15

_

(3.15 – 3.23)

(X – µ)

=

z =

= -0.77

σ

√n

0.72

√48

( )


Chapter 6

EXAMPLE 3: The mean grade point average of the engineering

majors at a large university is 3.23, with a standard deviation

of 0.72. In a class of 48 students, find the probability that the

mean grade point average of the student is less than 3.15.

z = -0.77

µ =

3.23

σ=

0.72

n =

48

_

Less than 3.15:

X =

3.15

-0.77

0

z = -0.77

Area = 0.2794

Since the questions asks for LESS THAN, shade to the left

Area = 0.5 – 0.2794 = 0.2206


Chapter 6

EXAMPLE 4: At a large university, the mean age of graduate

students who are majoring in psychology is 32.6 years, and the

standard deviation is 3 years. Assume normally distributed.

If an individual from the department is randomly selected,

find the probability that his/her age is less than 33.2 years.

Are we dealing with individual data or sample means?

X – µ

σ

Individual data, so we use the equation z =

µ =

32.6

σ=

3

Less than 33.2:

X =

33.2

X – µ

σ

33.2 – 32.6

3

= 0.20

=

z =

z = 0.20

Area = 0.0793

Area = 0.5 + 0.0793 = 0.5793

0 0.20


Chapter 6

EXAMPLE 4: At a large university, the mean age of graduate students

who are majoring in psychology is 32.6 years, and the standard

deviation is 3 years. Assume normally distributed.

If a random sample of 15 individuals is selected, find the

probability that the mean age of the students in the sample

Will be less than 33.2 years.

_

Are we dealing with individual data or sample means?

(X – µ)

Sample means, so we use the equation z =

σ

√n

Less than 33.2:

_

µ =

32.6

σ=

3

n =

15

_

X =

33.2

(X – µ)

(33.2 – 32.6)

z =

=

= 0.77

σ

√n

( )

_3

√15

z = 0.77

Area = 0.2794

Area = 0.5 + 0.2794 = 0.7794

0 0.77


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