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Section 7.5

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Section 7.5. The Central Limit Theorem. Theorem 7.1 for a Normal Probability Distribution. Let x be a random variable with a normal distribution with mean m and standard deviation s . Let x be the sample mean corresponding to random samples of size n taken from the distribution. - PowerPoint PPT Presentation
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Section 7.5 The Central Limit Theorem 7.5 / 1
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Page 1: Section 7.5

Section 7.5

The Central Limit Theorem

7.5 / 1

Page 2: Section 7.5

Theorem 7.1for a Normal Probability Distribution

(a) The x distribution is a normal distribution. (b) The mean of the x distribution is m (the same

mean as the original distribution).(c) The standard deviation of the x distribution is s/

(the standard deviation of the original distribution, divided by the square root of the sample size).

n

Let x be a random variable with a normal distribution with mean m and standard deviation s. Let x be the sample mean corresponding to random samples of size n taken from the distribution.Then the following are true:

7.5 / 2

Page 3: Section 7.5

We can use this theorem to draw conclusions about means of samples

taken from normal distributions.

If the original distribution is normal, then the sampling distribution will be

normal.

7.5 / 3

Page 4: Section 7.5

mm x

7.5 / 4

The mean of the sampling distribution is equal to the mean of the original distribution.

The Mean of the Sampling Distribution xm

The Standard Deviation of the Sampling Distribution xs

The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

x nss

Page 5: Section 7.5

To Calculate z Scores

7.5 / 5

nxxz

x

x

sm

s

m

Page 6: Section 7.5

The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2 minutes.

1. Find the probability that a trip between the cities takes more than 15 minutes.

2. Find the probability that mean time of nine trips between the cities is more than 15 minutes.

7.5 / 6

Page 7: Section 7.5

7

• Find the probability that a trip between the cities takes more than 15 minutes.

3264.06736.01)45.0z(P

45.02.21415z

Mean = 14 minutes, standard deviation = 2.2 minutes

Page 8: Section 7.5

Mean = 14 minutes, standard deviation = 2.2 minutes

• Find the probability that mean time of nine trips between the cities is more than 15 minutes.

7.5 / 8

73.092.2

n

14

x

x

s

s

mm

15 14 1.370.73

( 1.37) 1 0.9147 0.0851

x

x

xz

P z

ms

Page 9: Section 7.5

Standard Error of the Mean

• The standard error of the mean is the standard deviation of the sampling distribution

n

mean the oferror standard the on,distributi sampling x theFor

x ss

Page 10: Section 7.5

What if the Original Distribution Is Not Normal?

• Use the Central Limit Theorem.

7.5 / 10

Page 11: Section 7.5

Theorem 7.2Central Limit Theorem for Any Probability

Distribution

If x has any distribution with mean m and standard deviation s, then the sample mean based on a random sample of

size n will have a distribution that approaches the normal distribution of the normal random variable with mean m and standard deviation as n increases without limit.

7.5 / 11

x

ns

Page 12: Section 7.5

How large should the sample size be to permit the application of the Central Limit Theorem?

• In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.

7.5 / 12

For most x distributions, if we use a sample size of 30 or larger, the x distribution will be approximately normal.

Page 13: Section 7.5

Use the central limit theorem to convert the distribution to the standard normal distribution

The mean of the sampling distribution is the same as the mean of the original distribution.

The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

7.5 / 13

xm m x nss

/x

x

x xzn

m ms s

Page 14: Section 7.5

Guided Exercise 10Central limit theorem(a) Suppose that x has a normal distribution

with n = 18 and standard deviation σ = 3. If you draw random samples of size 5 from the x distribution and represents the sample mean, what can you say about distribution? How can you standardize the distribution?

7.5 / 14

x

xx

Page 15: Section 7.5

Guided Exercise 10 cont.

Since the x distribution is given to be normal, the distribution also will be normal even though the sample size is much less that 30.

The mean is:

The standard deviation is:

We could standardize as follows:

7.5 / 15

x

x

18xm m

/ 3 / 5 1.3x ns s

181.3/

x xznm

s

Page 16: Section 7.5

Guided Exercise 10 cont.

(b) Suppose you know that the x distribution has mean μ = 75 and standard deviation σ = 12,but you have no information as to whether or not

the x distribution is normal. If you draw samples of size 30 from the x distribution and represents the sample mean,

what you can say about the distribution? How can you standardize the distribution?

7.5 / 16

x

x

x

Page 17: Section 7.5

Guided Exercise 10 cont.

Since the sample size is large enough, the distribution will be approximately a normal distribution.

The mean of the distribution is:

The standard deviation of the distribution is:

We could standardize as follows:

7.5 / 17

x

x

x

75xm m

/ 12 / 30 2.2x ns s

752.2/

x xznm

s

Page 18: Section 7.5

Guided Exercise 10 cont.Suppose you did not know that the x had a normal

distribution. Would you be justified in saying that the distribution is approximately normal if the sample size were n = 8?

No, the sample size should be 30 or larger if we don’t know that x has a normal distribution.

7.5 / 18

x

Page 19: Section 7.5

How to find Probabilities regarding Given a probability distribution of x values where

n = sample sizeμ = mean of the x distributionσ = standard deviation of the x distribution

1. If the x distribution is normal, then the distribution is normal.2. Even if the x distribution is not normal, if the sample size is

then by the central limit theorem, the distribution is approximately normal.

3. Convert to z using the formula4. Use the standard normal distribution

to find the corresponding probability of events regarding

7.5 / 19

x

x30n x

/x

x

x xzn

m ms s

x

x

x

Page 20: Section 7.5

Guided Exercise 11Probability regardingIn mountain country, major highways sometimes use

tunnels instead of long, winding roads over high passes. However, too many vehicles in a tunnel at the same time can cause a hazardous situation. Traffic engineers are studying a long tunnel in Colorado. If x represents the time for the vehicle to go through the tunnel, it is known that the x distribution has mean μ = 12.1 minutes and standard deviation σ = 3.8 minutes under ordinary traffic conditions. From a histogram of x values, it was found that the x distribution is mound-shaped with some symmetry about the mean.

7.5 / 20

x

Page 21: Section 7.5

Guided Exercise 11Engineers have calculated that, on average, vehicles should

spend from 11 to 13 minutes in the tunnel. If the time is less than 11 minutes, traffic is moving too fast for safe travel in the tunnel. If the time is more than 13 minutes, there is a problem of bad air (too much carbon monoxide and other pollutants).

Under ordinary conditions, there are about 50 vehicles in the tunnel at one time.

What is the probability that the mean time for 50 vehicles to go through the tunnel will be from 11 to 13minutes.

We will answer this question in steps.7.5 / 21

Page 22: Section 7.5

Guided Exercise 11(a) Let represent the sample mean based on

samples of size 50. Describe the distribution.

From the central limit theorem, we expect the distribution to be approximately normal with mean and standard deviation

7.5 / 22

xx

x

12.1xm m 3.8 0.5450x n

ss

Page 23: Section 7.5

Guided Exercise 11(b) Find P(11< <13).

We convert the interval P(11< <13) to a standard z interval and use the standard probability table to find our answer. Since

converts to converts to Therefore, P(11< <13) = P(-2.04 < z < 1.67) = 0.6525 – 0.0207 = 0.9318

(c) Comment on your answer for part (b)It seems that 93% of the time there should be no safety hazard for average traffic flow 7.5 / 23

xx

12.10.54/

x xznm

s

11x

13x

11 12.1 2.040.54

z and

13 12.1 1.670.54

z

x

x

Page 24: Section 7.5

24

Application of the Central Limit Theorem

Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be:

a. more than 530 pounds?b. less than 500 pounds?c. between 495 and 515 pounds?

Page 25: Section 7.5

Are we authorized to use the Normal Distribution?

• Yes, we are attempting to draw conclusions about means of large samples.

7.5 / 25

Page 26: Section 7.5

26

Applying the Central Limit Theorem

What is the probability that the mean weight will be more than 530 pounds?Consider the distribution of sample means:

P( x > 530): z = 530 – 510 = 20 = 2.22 9 9

P(z > 2.22) = _______.0132

9100/90,510 xx sm

Page 27: Section 7.5

27

Applying the Central Limit Theorem

What is the probability that their mean weight will be less than 500 pounds?

P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9

P(z < – 1.11) = _______.1335

Assignment 18

Page 28: Section 7.5

28

Applying the Central Limit Theorem

What is the probability that their mean weight will be between 495 and 515 pounds?

P(495 < x < 515) :

for 495: z = 495 – 510 = - 15 = - 1.67 9 9

for 515: z = 515 – 510 = 5 = 0.56 9 9

P(–1.67 < z < 0.56) = ______.6648


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