Normal Probability Distributions

1

Chapter

5

Elementary Statistics

Larson Farber

N

Normal Probability Distributions

2

Introduction to Normal

Distributions

Section 5.1

3

Properties of a Normal Distribution

•The mean, median, and mode are equal

•Bell shaped and is symmetric about the mean

•The total area that lies under the curve is one or 100%

x

4

•As the curve extends farther and farther away from the mean, it gets closer and closer to the x- axis but never touches it.

•The points at which the curvature changes are called inflection points. The graph curves downward between the inflection points and curves upward past the inflection points to the left and to the

right

x

Inflection pointInflection point

Properties of a Normal Distribution

5

Means and Standard Deviations

2012 15 1810 11 13 14 16 17 19 21 229

12 15 1810 11 13 14 16 17 19 20

Curves with different means different standard deviations

Curves with different means, same standard deviation

6

Empirical Rule

About 95% of the area lies within 2 standard

deviations

About 99.7% of the area lies within 3 standard deviations of the mean

68%

About 68% of the area lies within 1 standard deviation of the mean

3 2 32

7

Determining Intervals

An instruction manual claims that the assembly time for a product is normally distributed with a mean of 4.2 hours

and standard deviation 0.3 hours. Determine the interval in which 95% of the assembly times fall.

3 2 1 0 1 2 34.2 4.5 4.8 5.13.93.63.3x

4.2 - 2 (0.3) = 3.6 and 4.2 +2 (0.3) = 4.8. 95% of the assembly times will be between 3.6 and 4.8 hrs.

4.2 .

0.3

hrs

hrs

95% of the data will fall within 2 standard deviations of the mean.

8

The Standard Normal

Distribution

Section 5.2

9

The Standard ScoreThe standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean.

x

zdeviation standard

mean - value

The test scores for a civil service exam are normally distributed with a mean of 152 and standard deviation of 7. Find the standard z-score for a person with a score of:(a) 161 (b) 148 (c) 152

(c)

07

152152

z

z

29.17

152161

z

z(a)

57.07

152148

z

z(b)

10

The Standard Normal DistributionThe standard normal distribution has a mean of 0 and a standard deviation of 1.

Using z- scores any normal distribution can be transformed into the standard normal distribution.

4 3 2 1 0 1 2 3 4 z

11

Cumulative Areas

• The cumulative area is close to 1 for z scores close to 3.49.

• The cumulative area is close to 0 for z-scores close to -3.49.

• The cumulative area for z = 0 is 0.5000

The total area

under the curve

is one.

0 1 2 3-1-2-3 z

12

Find the cumulative area for a z-score of -1.25.

0 1 2 3-1-2-3 z

Cumulative Areas

0.1056

Read down the z column on the left to z = -1.2 and across to the column under .05. The value in the cell is 0.1056, the cumulative

area.

The probability that z is at most -1.25 is 0.1056. P ( z -1.25) = 0.1056

13

Finding ProbabilitiesTo find the probability that z is less than a given value, read the

cumulative area in the table corresponding to that z-score.

0 1 2 3-1-2-3 z

Read down the z-column to -1.4 and across to .05. The cumulative area is 0.0735.

Find P( z < -1.45)

P ( z < -1.45) = 0.0735

14

Finding ProbabilitiesTo find the probability that z is greater than a given

value, subtract the cumulative area in the table from 1.

0 1 2 3-1-2-3 z

P( z > -1.24) = 0.8925

Required areaFind P( z > -1.24)

The cumulative area (area to the left) is 0.1075. So the area to the right is 1 - 0.1075 = 0.8925.

0.10750.8925

15

Finding ProbabilitiesTo find the probability z is between two given values, find the

cumulative areas for each and subtract the smaller area from the larger.

Find P( -1.25 < z < 1.17)

1. P(z < 1.17) = 0.8790 2. P(z < -1.25) =0.1056

3. P( -1.25 < z < 1.17) = 0.8790 - 0.1056 = 0.7734

0 1 2 3-1-2-3 z

160 1 2 3-1 -2-3 z

Summary To find the probability that z is less than a given value, read the corresponding cumulative area.

0 1 2 3-1-2-3 zTo find the probability is greater than a given value, subtract the cumulative area in the table from 1.

0 1 2 3-1-2-3 z

To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger.

17

Normal Distributions

Finding Probabilities

Section 5.3

18

Probabilities and Normal Distributions

B 4 3.99 1

115100

115

100115

z

If a random variable, x is normally distributed, the probability that x will fall within an interval is equal to the area under the curve in the interval.IQ scores are normally distributed with a mean of 100 and standard deviation of 15. Find the probability that a person selected at random will have an IQ score less than 115.

To find the area in this interval, first find the standard score equivalent to x = 115.

19

B 4 3.99 1

0 1

Probabilities and Normal Distributions

Find P(z < 1)

B 4 3.99 1

115100

Standard Normal Distribution

Find P(x < 115)

Normal Distribution

P( z < 1) = 0.8413, so P( x <115) = 0.8413

SA

ME

SA

ME

100

15

0

1

20

Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected. Find the probability it is between $80 and $115.

P(80 < x < 115)

Normal Distribution

67.112

10080

z

25.112

100115

z

P(-1.67 < z < 1.25)0.8944 - 0.0475 = 0.8469

The probability a utility bill is between $80 and $115 is 0.8469.

Application

100

12

21

Normal Distributions

Finding Values

Section 5.4

22

z

0.9803

From Areas to z-scores

Locate 0.9803 in the area portion of the table. Read the values at the beginning of the corresponding row and at the top of the

column. The z-score is 2.06.

Find the z-score corresponding to a cumulative area of 0.9803.

z = 2.06 corresponds roughly to the 98th percentile.

4 3 2 1 0 1 2 3 40.9803

23

Finding z-scores From Areas

Find the z-score corresponding to the 90th percentile.

z0

.90

The closest table area is .8997. The row heading is 1.2 and column heading .08. This corresponds to z = 1.28.

A z-score of 1.28 corresponds to the 90th percentile.

24


Find the z-score with an area of .60 falling to its right.

.60.40

0 zz

With .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is –0.2 and column heading is .05. The z-score is –0.25.

A z-score of –0.25 has an area of .60 to its right. It also corresponds to the 40th percentile

25


Find the z-score such that 45% of the area under the curve falls between –z and z.

0 z-z

The area remaining in the tails is .55. Half this area isin each tail, so since .55/2 =.275 is the cumulative area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is –0.60. The positive z score is 0.60.

.45.275.275

26

From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and standard deviation of 7. Find the test score for a person with a standard score of (a) 2.33 (b) -1.75 (c) 0

(a) x = 152 + (2.33)(7) = 168.31

(b) x = 152 + ( -1.75)(7) = 139.75

(c) x = 152 + (0)(7) = 152

To find the data value, x when given a standard score, z:

x z

27

Finding Percentiles or Cut-off valuesMonthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills?

t 1.28 1.29 4

10%90%

Find the cumulative area in the table that is closest to 0.9000 (the 90th percentile.) The area 0.8997 corresponds to a z-score of 1.28.

x = 100 + 1.28(12) = 115.36.

$115.36 is the smallest value for the top 10%.

z

To find the corresponding x-value, usex z

28

The Central Limit Theorem

Section 5.5

29

Sampling DistributionsA sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means.

Sample

Sample

Sample

SampleSampleSample

The sampling distribution consists of the values of the sample means, ,...,,,,, 654321 xxxxxx

1x

2x

3x

4x 5x6x

30

x

the sample means will have a normal distribution


x

with a mean

nx

and standard deviation

If a sample n 30 is taken from a population with any type distribution that has a mean = and standard deviation =

x

xx

x

xx

x

xx

xxx

xxxxxx

x

xxx

xx

31

x

xx

x

xx

x

xx

xxx

xxxxxx

x

xx

xx

the distribution of means of sample size n , will be normal with a mean

standard deviation

nx

x


x

If a sample of any size is taken from a population with a

normal distribution with mean = and standard deviation=

32

Application

69.2

x

xx

x

xx

x

xx

xxx

xxxxxx

x

xx

xx

Distribution of means of sample size 60 , will be normal.

69.2

2.9

The mean height of American men (ages 20-29) is inches. Random samples of 60 such men are selected. Find the mean and standard deviation (standard error) of the sampling distribution.

69.2 2.9and

3744.060

9.2

x

69.2x mean

Standard deviation

33

Interpreting the Central Limit Theorem

The mean height of American men (ages 20-29) is = 69.2”. If a random sample of 60 men in this age group is selected, what is the probability the mean height for the sample is greater than 70”? Assume the standard deviation is 2.9”.

Find the z-score for a sample mean of 70:

14.23744.0

2.6970

x

xz

3744.060

9.2xstandard deviation

2.69xmean

Since n > 30 the sampling distribution of will be normalx

34


t 1.87 1.88 4

2.14

P ( > 70)x

zThere is a 0.0162 probability that a sample of 60 men will have a mean height greater than 70”.

= P (z > 2.14)

= 1 - 0.9838

= 0.0162


35

Application Central Limit TheoremDuring a certain week the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for the sample of 38 gas stations in California is between $1.169 and $1.179? Assume the standard deviation = $0.049.

63.00079.0

164.1169.1

z 90.1

0079.0

164.1179.1

z

0079.038

049.0

nx

standard deviation

mean 164.1 xx

Calculate the standard z-score for sample values of $1.169 and $1.179.

Since n > 30 the sampling distribution of will be normalx

36

.63 1.90

z

Application Central Limit TheoremP( 0.63 < z < 1.90)

= 0.9713 - 0.7357

= 0.2356

The probability is 0.2356 that the mean for the sample is

between $1.169 and $1.179.

37

Normal Approximation to

the Binomial

Section 5.6

38

Binomial Distribution Characteristics

• There are a fixed number of independent trials. (n)

• Each trial has 2 outcomes, Success or Failure.

• The probability of success on a single trial is p and

the probability of failure is q. p + q = 1

• We can find the probability of exactly x successes out

of n trials. Where x = 0 or 1 or 2 … n.

• x is a discrete random variable representing a count

of the number of successes in n trials.

and =np npq

39

Application34% of Americans have type A+ blood. If 500 Americans are sampled at random, what is the probability at least 300 have type A+ blood?

Using techniques of chapter 4 you could calculate the probability that exactly 300, exactly 301…exactly 500 Americans have A+

blood type and add the probabilities.

Or…you could use the normal curve probabilities to approximate the binomial probabilities.

If np 5 and nq 5, the binomial random variable x is approximately normally distributed with mean and

npqnp

40

Why do we require np 5 and nq 5?

0 1 2 3 4 5

4

4

n = 5p = 0.25, q = .75np =1.25 nq = 3.75

n = 20p = 0.25np = 5 nq = 15

n = 50p = 0.25np = 12.5 nq = 37.5

0 10 20 30 40 50

41

Binomial ProbabilitiesThe binomial distribution is discrete with a probability histogram graph. The probability that a specific value of x will occur is equal to the area of the rectangle with midpoint at x.

If n = 50 and p = 0.25 find P (14 x 16)

Add the areas of the rectangles with midpoints at x = 14, x = 15, x = 16.

14 15 16

0.111 0.0890.065

0.111 + 0.089 + 0.065 = 0.265

P (14 x 16) = 0.265

42

14 15 16

Correction for Continuity

Check that np= 12.5 5 and nq= 37.5 5.

Use the normal approximation to the binomial to find P(14 x 16) if n = 50 and p = 0.25

Values for the binomial random variable x are 14, 15 and 16.

43

14 15 16

Correction for Continuity

Check that np= 12.5 5 and nq= 37.5 5.

Use the normal approximation to the binomial to find P(14 x 16) if n = 50 and p = 0.25

The interval of values under the normal curve is 13.5 x 16.5.

To ensure the boundaries of each rectangle are included in the interval, subtract 0.5 from a left-hand boundary and add 0.5 to a right-hand boundary.

44

Normal Approximation to the BinomialUse the normal approximation to the binomial to find P(14 x 16) if n = 50 and p = 0.25

Adjust the endpoints to correct for continuity P(13.5 x 16.5)

12.50.33

3.062

13.5z

12.5

1.313.062

16.5z

P(0.33 z 1.31) = 0.9049 - 0.6293 = 0.2756

Convert each endpoint to a standard score

5.12)25(.50 np50(.25)(.75) 3.062npq

Find the mean and standard deviation using binomial distribution formulas.

45

ApplicationA survey of Internet users found that 75% favored government regulations on “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation.

Since np = 150 5 and nq = 50 5 use the normal approximation to the binomial.

Use the correction for continuity to translate to the continuous variable in the interval (- , 139.5). Find P( x < 139.5 )

1237.6)25)(.75(.200 npq

150)75(.200 np

The binomial phrase of “fewer than 140” means0, 1, 2, 3…139.

46

ApplicationA survey of Internet users found that 75% favored government regulations on “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation.

Use the correction for continuity P(x < 139.5)

1501.71

6.1237

139.5z

P( z < -1.71) = 0.0436

The probability that fewer than 140 are in favor of government regulation is approximately 0.0436

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Normal Probability Distributions