2 Normal Distribution

1

Taburan Normal Chapter

5

Statistik AsasPn Norizah

2

x

Sifar-sifat Taburan Distribution

μ

Inflection pointInflection point

•The mean, median, and mode are equal

•Bell shaped and symmetric about the mean

•The total area under the curve is one or 100%•The curve approaches but never touches the x- axis as

it extends farther away from the mean.

•The graph curves downward between μ - σ and μ + σ and curves upward to the left of μ - σ and to the right of μ + σ. The points at which the curvature changes

are called inflection points.

μ - σ μ + σ

3

Means and Standard Deviations

Curves with different means, same standard deviation

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

4

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

+2σ-2σ +1σ +3σ-1σ-3σ μ

68%

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

5

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

μ = 4.2 hrsσ = 0.3 hrs

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

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

3.6 and 4.8 hrs.

Larson/Farber Ch 56

The Standard Score

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

xz

deviation standardmean - 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)

Larson/Farber Ch 57

From z-Scores to Raw Scores

To transform a standard z-score to a data value, x use the formula

x = μ + z σ

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

Larson/Farber Ch 58

The Standard Normal Distribution

The 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

9

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 3-1-2-3z

10

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

Then move across that row to the column under 0.05.

0 1 2 3-1-2-3z

Cumulative Areas

0.1026

Read down the z column on the left to z = -1.2.

The probability that z is at most -1.25 is 0.1056. P ( z -1.25) = 0.1026

The value in the cell is 0.1056, the cumulative area.

11

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 that corresponds to a cumulative area of 0.9803.

z = 2.06 is roughly the 98th percentile.

4 3 2 1 0 1 2 3 4

0.9803

12

Finding Probabilities

To 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 3-1-2-3z

Read down the z-column to -1.2 and across to .04. The cumulative area is 0.1075.

Find P( z < -1.24)

P ( z < 1.24) = 0.1075

Larson/Farber Ch 513


To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1.

0 1 2 3-1-2-3z

P( z > -1.24) = 0.8925

Required area

Find P( z > -1.24)

The cumulative area (area to the left) is 0.1075.

0.1075

So the area to the right is 1 - 0.1075 = 0.8925.

0.8925

14


To 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 3-1-2-3z

150 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 z

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

16

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.


B 4 3.99 1

0 1

Probabilities and Normal Distributions

μ = 0σ = 1

Find P(z < 1)

B 4 3.99 1

115100

Standard Normal Distribution

Find P(x < 115)

μ = 100σ = 15

Normal Distribution

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

SAM

E SAM

E

18

Other Normal Distributions

If 0 or 1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution.

19

Other Normal Distributions

If 0 or 1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution.

Formula 5-2

x - µz =

20

Converting to Standard Normal Distribution

x(a)

Figure 5-13

P

21

Converting to Standard Normal Distribution

x 0

Figure 5-13

z

x -

z =

(a) (b)

P P

22

Probability of Weight between 143 pounds and 201 pounds

143 201

z

0 2.00

z = 201 - 143

29 = 2.00

s =29x = 143

Figure 5-14

Weight

Example

23


143 201

z

0 2.00

x = 143

Figure 5-14

Value foundin Table

Weight

s =29

Example

24


143 201

z

0 2.00

x = 143

Figure 5-14

0.4772

Weight

s =29

Example

25


143 201

z

0 2.00

x = 143

Figure 5-14

There is a 0.4772 probability of randomly selecting a woman with a weight between 143 and 201 lbs.

Weight

s =29

26


143 201

z

0 2.00

x = 143

Figure 5-14

OR - 47.72% of women have weights between 143 lb and 201 lb.

Weight

s =29

Example

27


• Probability = 0.5 – Φ(2.00) = 0.5 – 0.228 = 0.4772 note : Φ is the Greek letter phi

x - µz =

= 143

s =29

z = 201 - 143

= 2.0029

228.0)00.2(

x

28

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)μ = 100σ = 12

Normal Distribution

67.112

10080

z 25.1

12100115

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

29

Finding Percentiles

Monthly 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.

To find the corresponding x-value, use x = μ + z σx = 100 + 1.28(12) = 115.36.$115.36 is the smallest value for the top 10%.

z

30

ISL

• Reading Handout [1];‘The Normal distribution

pg 33 – 43’

31

Tutorial

• Exercise 2A pg 45 – 48 from Handout [1] ; The Normal distribution.

• Discuss the solutions within groups.

• Present the solutions during tutorial session.

32

Good Luck

• Meet again in Elementary Sampling (Sampling Distribution)

33

Sampling Distributions

A 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

34

μ x

x

x

xx

x

xx

x

xx

xxx

xxxxxx

xxx

x

xx

the sample means will have a normal distribution

The Central Limit Theorem

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

μ

x

with a mean

nx

and standard deviation


x

xx

x

xx

x

xx

xxx

xxxxxx

x

xx

xx

If a sample of any size is taken from a population with a normal distribution and mean = and standard deviation = ,

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

standard deviation

nx

x

The Central Limit Theorem

μ

μ x

36

Application

The mean height of American men (ages 20-29) is = 69.2 inches and σ = 2.9 inches. Random samples of 60 men in this age group are selected. Find the mean and standard deviation (standard error) of the sampling distribution.

69.2

x

xx

x

xx

x

xx

xxx

xxxxxx

x

xx

xx

3744.0609.2

x

Distribution of means of sample size 60 , will be normal with a meanstandard deviation (standard error)

2.69x

μ = 69.2σ = 2.9

37

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 σ = 2.9”.

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

14.23744.0

2.6970

x

xz

3744.0609.2

xstandard deviation

2.69xmean

since n > 30The sampling distribution of will be normalx

38

t 1.87 1.88 4

2.14

Interpreting the Central Limit Theorem

P ( > 70)x

z

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

= P (z > 2.14)

= 1 - 0.9838

= 0.0162

39

Application Central Limit Theorem

During 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 σ = $0.049.

63.00079.0

164.1169.1

z 90.1

0079.0164.1179.1

z

0079.038049.0

nx

standard deviation

mean164.1 xx

The sampling distribution of will be normalx

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

40

.63 1.90

z

Application Central Limit Theorem

P( 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.

41

Normal Approximations to the Binomial

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

repeated under identical conditions• Each trial has 2 outcomes,

S = Success or F = Failure.• The probability of success on a single

trial is p and the probability of failure is q. P(S) = p P(F) =q p + q = 1

• The central problem is to find the probability of x successes out of n trials. Where x = 0 or 1 or 2 … n.

Characteristics of a Binomial Experiment

x is a count of the number of successes in n trials.

42

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

mean μ = np and npq

Application

34% 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.

43

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


Binomial Probabilities

The 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 and x = 16.

14 15 16

0.1110.089

0.065

0.111 + 0.089 + 0.065 = 0.265

P (14 x 16) = 0.265

45

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.


Normal Approximation to the Binomial

Use 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)

33.00618.3

5.125.13

z 31.1

0618.35.125.16

z

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

Convert each endpoint to a standard score

5.12)25(.50 np

0618.3)75)(.25(.50 npq

Find the mean and standard deviation using binomial distribution formulas.

47

Application

A 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 you can use the normal approximation to the binomial.

Use the correction for continuity P(x < 139.5)

71.11237.6

1505.139

z

P(z < -1.71) = 0.0436The probability that fewer than 140 are infavor of government regulation is 0.0436

1237.6)25)(.75(.200 npq150)75(.200 np

Documents

2 Normal Distribution