Chapter 6: Normal Probability Distributions

xa b

P a x b( )

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.

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.

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 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

22

The normal probability distribution:

2 3 2 3

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 normal probability distribution.

6. We need to refine the empirical rule to be able to find the percentage 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 certain property.

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

6.2: 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 the normal distribution of the standard variable z (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 never touching the horizontal axis.3. The distribution has a mean of 0 and a standard deviation

of 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 curve between 0 and a specific value of z.

0 z

Example: Find the area under the standard normal curve between 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

Example: Find the area under the normal curve to the right of z = 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 mutually exclusive events.2. 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 = +1.37.3. When finding normal distribution probabilities, a sketch is

always helpful.

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

0 126. 126. z

Area asked for Area from table 0.3962

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 to determine a z-score if we are given the area (to work backwards).

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

P85 0 z

15% 0 3500.

implies

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 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

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

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 normal distribution.

z 0.00 0.01 0.02 0.03 0.04 0.05

1.6 0.4495 0.4500 0.4505

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 of x-values.

• Convert, or transform, the question into an equivalent probability statement involving z-values.

Standardization:Suppose x is a normal random variable with mean and standard deviation .The random variable

has a standard normal distribution.

z x

c x

0c

z

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 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. ; ..

.

When x z 32 32 32 3202

0;.

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 defined by x.

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.

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 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.

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

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).

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

Illustrations:

0 z( . )010 z

010.

0z( . )0 80 z

0 80.

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

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 information from 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; remember that z must be negative.

Note: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.

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 the normal 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 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.

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 binomial variable 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 area under the normal curve.

3. The approximation becomes more accurate as n becomes larger.

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 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.

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?

Let x be the number of students that withdraw from a course during their first year.x has a binomial distribution: n = 1800, p = 0.4The 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.

np

npq

( )( . )

( )( . )( . ) .

1800 0 4 720

1800 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