1

ES9

xa b

P a x b( )

Chapter 6 ~ Normal Probability Distributions

2

ES9

Chapter Goals

• Learn about the normal, bell-shaped, or Gaussian distribution

• How probabilities are found

• How probabilities are represented

• How normal distributions are used in the real world

3

ES9

6.1 ~ Normal Probability Distributions

• The normal probability distribution is the most important distribution in all of statistics

• Many continuous random variables have normal or approximately normal distributions

• Need to learn how to describe a normal probability distribution

4

ES9

Normal Probability Distribution

1. A continuous random variable

2. Description involves two functions:a. A function to determine the ordinates of the graph picturing the distributionb. A function to determine probabilities

4. The probability that x lies in some interval is the area under the curve

3. Normal probability distribution function:

This is the function for the normal (bell-shaped) curvef x e

x

( )( )

12

2

2

1

5

ES9

2 3 2 3

The Normal Probability Distribution

6

ES9

• Illustration

P a x b f x dxa

b( ) ( )

a b x

Probabilities for a Normal Distribution

7

ES9

Notes The definite integral is a calculus topic We will use a table to find probabilities for normal distributions

We will learn how to compute probabilities for one special normal distribution: the standard normal distribution

Transform all other normal probability questions to this special distribution

Recall the empirical rule: the percentages that lie within certain intervals about the mean come from the normal probability distribution

We need to refine the empirical rule to be able to find the percentage that lies between any two numbers

8

ES9

Percentage, Proportion & Probability

• Basically the same concepts

• Percentage (30%) is usually used when talking about a proportion (3/10) of a population

• Probability is usually used when talking about the chance that the next individual item will possess a certain property

• Area is the graphic representation of all three when we draw a picture to illustrate the situation

9

ES96.2 ~ The Standard Normal Distribution

• There are infinitely many normal probability distributions

• They are all related to the standard normal distribution

• The standard normal distribution is thenormal distribution of the standard variable z(the z-score)

10

ES9

Standard Normal Distribution

Properties:• The total area under the normal curve is equal to 1

• The distribution is mounded and symmetric; it extends indefinitely in both directions, approaching but never touching the horizontal axis

• The distribution has a mean of 0 and a standard deviation of 1

• The mean divides the area in half, 0.50 on each side

• Nearly all the area is between z = -3.00 and z = 3.00

Notes: Table 3, Appendix B lists the probabilities associated with the intervals

from the mean (0) to a specific value of z Probabilities of other intervals are found using the table entries,

addition, subtraction, and the properties above

11

ES9

0 z

Table 3, Appendix B Entries

• The table contains the area under the standard normal curve between 0 and a specific value of z

12

ES9

0 145. z

Example

Example: Find the area under the standard normal curve between z = 0 and z = 1.45

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.4265

• A portion of Table 3:

P z( . ) .0 145 0 4265

...

...

13

ES9

P z( . ) . . . 145 0 5000 0 4265 0 0735

0 4265.

0 145.

Area asked for

z

Example

Example: Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

14

ES9

P z( . ) . . . 145 0 5000 0 4265 0 9265

0 145.

0 5000. 0 4265.

z

Example

Example: Find the area to the left of z = 1.45; P(z < 1.45)

15

ES9

Notes

The addition and subtraction used in the previous examples are correct because the “areas” represent mutually exclusive events

The symmetry of the normal distribution is a key factor in determining probabilities associated with values below (to the left of) the mean. For example: the area between the mean and z = -1.37 is exactly the same as the area between the mean and z = +1.37.

When finding normal distribution probabilities, a sketch is always helpful

16

ES9

0 126. 126. z

Area asked for Area from table 0.3962

P z( . ) . 126 0 0 3962

Example

Example: Find the area between the mean (z = 0) andz = -1.26

17

ES9

00.98 0.98

Area asked for Area from table0.3365

P z( 0. ) . . . 98 0 5000 0 3365 01635

Example

Example: Find the area to the left of -0.98; P(z < -0.98)

18

ES9

0 180. 230.

0 4641.0 4893.

P z P z P z( . . ) ( . ) ( . )

. . .

230 180 2 30 0 0 180

0 4893 0 4641 0 9534

Example

Example: Find the area between z = -2.30 and z = 1.80

19

ES9

P z P z P z( . 0. ) ( . ) ( 0.

. . .

140 50 0 140 0 50)

0 4192 01915 0 2277

Example

Example: Find the area between z = -1.40 and z = -0.50

0

Area asked for

-1.40 - 0.50 0.50 1.40

20

ES9

P85 0 z

15% 0 3500.

implies

Normal Distribution Note

The normal distribution table may also be used to determine a z-score if we are given the area (working backwards)

Example: What is the z-score associated with the 85th percentile?

21

ES9

Solution

• In Table 3 Appendix B, find the “area” entry that is closest to 0.3500:

• The area entry closest to 0.3500 is 0.3508• The z-score that corresponds to this area is 1.04• The 85th percentile in a standard normal distribution is 1.04

z 0.00 0.01 0.02 0.03 0.04 0.05

1.0 0.3485 0.3500 0.3508

...

...

22

ES9

0 0z z

90% 0 4500.

implies

Example

Example: What z-scores bound the middle 90% of a standard normal distribution?

23

ES9

Solution

• The 90% is split into two equal parts by the mean. Find the area in Table 3 closest to 0.4500:

• 0.4500 is exactly half way between 0.4495 and 0.4505

• Therefore, z = 1.645

• z = -1.645 and z = 1.645 bound the middle 90% of a normal distribution

z 0.00 0.01 0.02 0.03 0.04 0.05

1.6 0.4495 0.4500 0.4505

...

...

24

ES9

6.3 ~ Applications of Normal Distributions

• Apply the techniques learned for the z distribution to all normal distributions

• Start with a probability question in terms ofx-values

• Convert, or transform, the question into an equivalent probability statement involvingz-values

25

ES9

c x

0c

z

Standardization

• Suppose x is a normal random variable with mean and standard deviation

• The random variable has a standard normal distribution

zx

26

ES9

Example

Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally

distributed and a bottle is selected at random:

1) Find the probability the bottle contains between 32.00 oz and 32.025 oz

2) Find the probability the bottle contains more than 31.97 oz

When x z 32.0032.00 32.00 32.0

0.00;0.02

Solutions:

1)

When x z

32 02532.025 32 025 32.0

125. ;.

0.02.

27

ES9

P x Px

P z

( . )0. 0.

.

0.

( . ) .

32.0 32 02532.0 32.0

02

32.0

02

32 025 32.0

02

0 125 0 3944

32.0 32 025. x0 125. z

Area asked for

Solution Continued

28

ES9

P x Px

P z( . ).

( .

. . .

319732.0

0.023197 32.0

0.02150)

0 5000 0 4332 0 9332

32.03197. x

0 150. z

Example, Part 2

2)

29

ES9

Notes

• The normal table may be used to answer many kinds of questions involving a normal distribution

Example: The waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes. The bank would like to claim that 95% of all customers are waited on by a teller within c minutes. Find the value of c that makes this statement true.

• Often we need to find a cutoff point: a value of x such that there is a certain probability in a specified interval defined by x

30

ES9

P x c

Px c

P zc

( ) 0.

.

.

.

.0.

..

0.

95

3 7

14

3 7

1495

3 714

95

c

c

c

3 714

1645

1645 14 3 7 6 003

6

..

.

( . )( . ) . .

minutes

3 7. c x0 1645. z

0 5000. 0 4500.

0 0500.

Solution

31

ES9

Example

Example: A radar unit is used to measure the speed of automobiles on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a

mean of 62 mph. Find the standard deviation of all speeds if 3% of the automobiles travel faster than 72 mph.

62 72 x

0 188. z

0 4700.

0 0300.

32

ES9

Solution

P x( ) . 72 0 03

. 188 10

/ . . 10 188 532

P z( . ) . 188 0 03

zx

;72 621.88 =

33

ES9

Notation• If x is a normal random variable with mean and standard deviation , this is often

denoted:x ~ N(, )

Example: Suppose x is a normal random variable with = 35 and = 6. A convenient notation to identify this random variable is: x ~ N(35, 6).

34

ES9

6.4 ~ Notation

• z-score used throughout statistics in a variety of ways

• Need convenient notation to indicate the area under the standard normal distribution

• z() is the algebraic name, for the z-score (point on the z axis) such that there is of the area (probability) to the right of z()

35

ES9

Illustrations

z(0.10) represents the value of z such that the area to the right under the standard normal curve is 0.10

0 z

010.

z(0.10)

z(0.80) represents the value of z such that the area to the right under the standard normal curve is 0.80

0 z

0 80.

z(0.80)

36

ES9

Example

Example: Find the numerical value of z(0.10):

• Use Table 3: look for an area as close as possible to 0.4000

• z(0.10) = 1.28

0 z

0.10 (area information from notation)

Table shows this area (0.4000)

z(0.10)

37

ES9

Example

Example: Find the numerical value of z(0.80):

• Use Table 3: look for an area as close as possible to 0.3000

• z(0.80) = -0.84

0 z

Look for 0.3000; remember that z must be negative

z(0.80)

38

ES9

Notes

• The values of z that will be used regularly come from one of the following situations:

1. The z-score such that there is a specified area in one tail of the normal distribution

2. The z-scores that bound a specified middle proportion of the normal distribution

39

ES9

Example

Example: Find the numerical value of z(0.99):

• Because of the symmetrical nature of the normal distribution, z(0.99) = -z(0.01)

• Using Table 3: z(0.99) = -2.33

0 z

0.01

z(0.99)

40

ES9

Example

Example: Find the z-scores that bound the middle 0.99 of the normal distribution:

• Use Table 3:z(0.005) = 2.575 and z(0.995) = -z(0.005) = -2.575

0or

0 495.0 495.0 005.0 005.

z(0.995)

-z(0.005)

z(0.005)

41

ES9

6.5 ~ Normal Approximation of the Binomial

• Recall: the binomial distribution is a probability distribution of the discrete random variable x, the number of successes observed in n repeated independent trials

• Binomial probabilities can be reasonably estimated by using the normal probability distribution

42

ES9

Background & Histogram

• Background: Consider the distribution of the binomial variable x when n = 20 and p = 0.5

The histogram may be approximated by a normal curve

• Histogram:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 200.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

x

P x( )

43

ES9

np

npq

( )( . )

( )( . )( . ) .

20 0 5 10

20 0 5 0 5 5 2 236

Notes

The normal curve has mean and standard deviation from the binomial distribution:

Can approximate the area of the rectangles with the area under the normal curve

The approximation becomes more accurate as n becomes larger

44

ES9

Two Problems

1. As p moves away from 0.5, the binomial distribution is less symmetric, less normal-looking

Solution: The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 - p) both equal or exceed 5

2. The binomial distribution is discrete, and the normal distribution is continuous

Solution: Use the continuity correction factor. Add or subtract 0.5 to account for the width of each rectangle.

45

ES9

Example

Example: Research indicates 40% of all students entering a certain university withdraw from a course during their first year. What is the probability that

fewer than 650 of this year’s entering class of 1800 will withdraw from a class?

P xx

xx x( ) ( . ) ( . )

18000 4 0 6 1800 for 0, 1, 2, ... ,1800

• Let x be the number of students that withdraw from a course during their first year

• x has a binomial distribution: n = 1800, p = 0.4

• The probability function is given by:

46

ES9

np

npq

( )( . )

( )( . )( . ) .

1800 0 4 720

1800 0 4 0 6 432 20 78

P x P x x

P x x

Px

P z

( ) ( )

( . )

..

.

( . )

. . .

is fewer than 650 (for discrete variable )

(for a continuous variable )

650

649 5

72020 78

649 5 72020 78

3 39

0 5000 0 4997 0 0003

Solution

• Use the normal approximation method: