Chapter 6: Normal ProbabilityDistributions

xa bµ

P a x b( )≤ ≤

Chapter Goals

• Learn about the normal, bell-shaped, orGaussian distribution.

• How probabilities are found.• How probabilities are represented.• How normal distributions are used in the

real world.

6.1: Normal ProbabilityDistributions

• The normal probability distribution is themost important distribution in all ofstatistics.

• Many continuous random variables havenormal or approximately normaldistributions.

• Need to learn how to describe a normalprobability distribution.

Normal Probability Distribution:1. A continuous random variable.2. Description involves two functions:

a. A function to determine the ordinates of the graphpicturing the distribution.

b. A function to determine probabilities.3. Normal probability distribution function:

This is the function for the normal (bell-shaped) curve.4. The probability that x lies in some interval is the area

under the curve.

f x ex

( )( )

=− −1

2

2

22σ π

µσ

The normal probability distribution:

µ µ σ+ µ σ+ 2 µ σ+ 3µ σ−µ σ− 2µ σ− 3

σ

Illustration of probabilities for a normal distribution:

P a x b f x dxa

b( ) ( )≤ ≤ = ∫

a b x

Note:1. The definite integral is a calculus topic.2. We will use a table to find probabilities for normal

distributions.3. We will learn how to compute probabilities for one special

normal distribution: the standard normal distribution.4. Transform all other normal probability questions to this

special distribution.5. Recall the empirical rule: the percentages that lie within

certain intervals about the mean come from the normalprobability distribution.

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

Percentage, proportion, and probability:1. Basically the same concepts.2. Percentage (30%) is usually used when talking about a

proportion (3/10) of a population.3. Probability is usually used when talking about the chance

that the next individual item will possess a certainproperty.

4. Area is the graphic representation of all three when wedraw a picture to illustrate the situation.

6.2: The Standard NormalDistribution

• There are infinitely many normalprobability distributions.

• They are all related to the standard normaldistribution.

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

Properties of the Standard Normal Distribution:1. The total area under the normal curve is equal to 1.2. The distribution is mounded and symmetric; it extends

indefinitely in both directions, approaching but nevertouching the horizontal axis.

3. The distribution has a mean of 0 and a standard deviationof 1.

4. The mean divides the area in half, 0.50 on each side.5. Nearly all the area is between z = −3.00 and z = 3.00.

Note:1. Table 3, Appendix B lists the probabilities associated with

the intervals from the mean (0) to a specific value of z.2. Probabilities of other intervals are found using the table

entries, addition, subtraction, and the properties above.

Table 3, Appendix B entries:

The table contains the area under the standard normal curvebetween 0 and a specific value of z.

0 z

Example: Find the area under the standard normal curvebetween z = 0 and z = 1.45.

A portion of Table 3:z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.4265

0 145.

P z( . ) .0 145 0 4265≤ ≤ =

z

M

M

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

P z( . ) . . .> = − =145 0 5000 0 4265 0 0735

0 4265.

0 145.

Area asked for

z

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

0 145.

0 5000. 0 4265.

P z( . ) . . .< = + =145 0 5000 0 4265 0 9265

z

Note:1. The addition and subtraction used in the previous examples

are correct because the “areas” represent mutuallyexclusive events.

2. The symmetry of the normal distribution is a key factor indetermining probabilities associated with values below (tothe left of) the mean. For example: the area between themean and z = −1.37 is exactly the same as the area betweenthe mean and z = +1.37.

3. When finding normal distribution probabilities, a sketch isalways helpful.

Example: Find the area between the mean (z = 0) and z = −1.26.

0−126. 1 26. z

Area asked for Area from table0.3962

P z( . ) .− < < =126 0 0 3962

Example: Find the area to the left of −.98; P(z < −.98).

0−.98 .98

Area asked for Area from table0.3365

P z( . ) . . .< − = − =98 0 5000 0 3365 01635

Example: Find the area between z = −2.3 and z = 1.8.

0 18.− 2 3.

0 4641.0 4893.

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

− < < = − < < + < <= + =

2 3 18 2 3 0 0 180 4893 0 4641 0 9534

Example: Find the area between z = −1.4 and z = −.5.

0−14. −.5 .5 14.

Area asked for

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

− < < − = < < − < <= − =

14 5 0 14 0 50 4192 01915 0 2277

Note: The normal distribution table may also be used todetermine a z-score if we are given the area (to workbackwards).

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

P85 0 z

15% 0 3500.

implies

Solution:In Table 3 Appendix B, find the “area” entry that is closest to0.3500.

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 normal distribution is 1.04.

z 0.00 0.01 0.02 0.03 0.04 0.05

1.0 0.3485 0.3500 0.3508M

M

Example: What z-scores bound the middle 90% of a normaldistribution?

0 0z z

90% 0 4500.

implies

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 = 1.645z = −1.645 and z = 1.645 bound the middle 90% of a normaldistribution.

z 0.00 0.01 0.02 0.03 0.04 0.05

1.6 0.4495 0.4500 0.4505M

M

6.3: Applications of NormalDistributions

• Apply the techniques learned for the zdistribution to all normal distributions.

• Start with a probability question in terms ofx-values.

• Convert, or transform, the question into anequivalent probability statement involvingz-values.

Standardization:Suppose x is a normal random variable with mean µ andstandard deviation σ.The random variable

has a standard normal distribution.

z x= − µσ

µ c x

0c − µσ

z

Example: A bottling machine is adjusted to fill bottles with amean of 32.0 oz of soda and standard deviation of 0.02.Assume the amount of fill is normally distributed and a bottleis selected at random.1. Find the probability the bottle contains between 32 oz and

32.025 oz.2. Find the probability the bottle contains more than 31.97 oz.

When x z= = − = − =32 025 32 32 025 3202

125. ; ..

.µσ

When x z= = − = − =32 32 32 3202

0;.

µσ

P x P x

P z

( . ). .

..

( . ) .

32 32 025 32 3202

3202

32 025 3202

0 125 0 3944

< < = − < − < −

= < < =

32 32 025. x0 125. z

Area asked forIllustration:

P x P x P z( . ).

..

( . )

. . .

> = − > −

= > −

= + =

3197 3202

3197 3202

15

0 5000 0 4332 0 9332

323197. x0−15. z

Illustration:

Note:1. The normal table may be used to answer many kinds of

questions involving a normal distribution.2. Often we need to find a cutoff point: a value of x such that

there is a certain probability in a specified interval definedby x.

Example: The waiting time x at a certain bank isapproximately normally distributed with a mean of 3.7minutes and a standard deviation of 1.4 minutes. The bankwould like to claim that 95% of all customers are waited onby a teller within c minutes. Find the value of c that makesthis statement true.

Solution:

P x c

P x c

P z c

( ) ..

..

..

..

.

≤ =−

≤−

=

≤−

=

953 7

143 7

1495

3 714

95

c

cc

− =

= + =≈

3 714

1645

1645 14 3 7 6 0036

..

.

( . )( . ) . .minutes

3 7. c x0 1645. z

0 5000. 0 4500.

0 0500.

Example: A radar unit is used to measure the speed ofautomobiles on an expressway during rush-hour traffic. Thespeeds of individual automobiles are normally distributedwith a mean of 62 mph. Find the standard deviation of allspeeds if 3% of the automobiles travel faster than 72 mph.

Illustration:

62 72 x0 188. z

0 4700.

0 0300.

Solution:

P x

P x

P z P z

( ) .

.

. ( . ) .

.

( . )( )

/ . .

> =− > −

=

> −

= > =

− =

=

= =

72 0 0362 72 62 0 03

72 62 0 03 188 0 03

72 62 188

188 10

10 188 5 32

σ σ

σ

σ

σ

σ

Notation:If x is a normal random variable with mean µ and standarddeviation σ, this is often denoted: x ~ N(µ, σ).

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

6.4: Notation

• z-score used throughout statistics in avariety of ways.

• Need convenient notation to indicate thearea under the standard normal distribution.

• z(α) is the token, or algebraic name, for thez-score (point on the z axis) such that thereis α of the area (probability) to the right ofz(α).

Illustrations:

0 z( . )010 z

010.

0z( . )0 80 z

0 80.

z(0.10) represents thevalue of z such that thearea to the right underthe standard normalcurve is 0.10

z(0.80) represents thevalue of z such that thearea to the right underthe standard normalcurve is 0.80

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

Use Table 3: look for an area as close as possible to 0.4000z(0.10) = 1.28

0 z( . )010 z

0.10 (area informationfrom notation)

Table shows this area (0.4000)

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) = −.84

0z( . )0 80 z

Look for 0.3000; rememberthat z must be negative.

Note:The values of z that will be used regularly come from one ofthe 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.

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

0z( . )0 99 z

0.01

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

Use Table 3:

0 z( . )0 005z( . )0 995or

− z( . )0 005

0 495.0 495.0 005.0 005.

z z z( . ) . ( . ) ( . ) .0 005 2 575 0 995 0 005 2 575= = − = −and

6.5: Normal Approximation ofthe Binomial• Recall: the binomial distribution is a

probability distribution of the discreterandom variable x, the number of successesobserved in n repeated independent trials.

• Binomial probabilities can be reasonablyestimated by using the normal probabilitydistribution.

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

Background: Consider the distribution of the binomialvariable x when n = 20 and p = 0.5.Histogram:

The histogram may be approximated by a normal curve.

x

P x( )

Note:1. The normal curve has mean and standard deviation from

the binomial distribution.

2. Can approximate the area of the rectangles with the areaunder the normal curve.

3. The approximation becomes more accurate as n becomeslarger.

µσ

= = == = = ≈

npnpq

( )( . )( )( . )( . ) .

20 0 5 1020 0 5 0 5 5 2 236

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 reasonableapproximation to a binomial probability distributionwhenever the values of np and n(1 − p) both equal orexceed 5.

2. The binomial distribution is discrete, and the normaldistribution is continuous.Solution: Use the continuity correction factor. Add orsubtract 0.5 to account for the width of each rectangle.

Example: Research indicates 40% of all students entering acertain university withdraw from a course during their firstyear. What is the probability that fewer than 650 of thisyear’s entering class of 1800 will withdraw from a class?

Let x be the number of students that withdraw from a courseduring their first year.x has a binomial distribution: n = 1800, p = 0.4The probability function is given by:

P xx

xx x( ) ( . ) ( . )=

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

Solution:Use the normal approximation method.

µσ

= = == = = ≈

npnpq

( )( . )( )( . )( . ) .

1800 0 4 7201800 0 4 0 6 432 20 78

P x P x xP x x

P x

P z

( ) ( )( . )

..

.( . )

. . .

is fewer than 650 (for discrete variable )(for a continuous variable )

= <= ≤

= −≤

−

= ≤ −= − =

650649 5720

20 78649 5 720

20 783 39

0 5000 0 4997 0 0003