6-1

Business Statistics: A Decision-Making Approach

8th Edition

Chapter 6Introduction to Continuous

Probability Distributions

6-2

Chapter Goals

After completing this chapter, you should be able to:

Convert values from any normal distribution to a standardized z-score

Find probabilities using a normal distribution table Apply the normal distribution to business problems Recognize when to apply the uniform and

exponential distributions (not covered)

6-3

Continuous Probability Distributions

A continuous probability distribution differs from a discrete probability distribution in several ways. A continuous random variable can take on an infinite

number of values. That is, the exact probability for a specific variable is always zero.

As a result, a continuous probability distribution cannot be expressed in tabular form.

Instead, an equation or formula is used to describe a continuous probability distribution.

Probability Density Function

The equation for describing a continuous probability distribution is called a probability density function and noted as “f(x).” Sometimes refer to as Density function pdf or PDF

6-4

f x e x( ) ( ) / 1

2

2 22

Probability Density Function Example

The shaded area in the graph represents the probability that the random variable X is less than or equal to a. Cumulative probability Always to the left

The probability that the random variable X  is exactly equal to a would be zero.

6-5

6-6

Types of Continuous Distributions

Three types Normal Uniform (so easy…by yourself) Exponential (not cover)

6-7

The Normal Distribution

‘Bell Shaped’ Symmetrical Mean=Median=Mode

Location is determined by the mean, μSpread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + to

Mean Median Mode

x

f(x)

μ

σ

Position and Shape of the Normal Curve

6-8

9

68.26% of the area under the curve falls within +/- 1 Stdev. 95.44% of the area under the curve falls within +/- 2 Stdev. 99.73% of the area under the curve falls within +/- 3 Stdev.

Basis for the empirical rules

6-10

f(x)

xμ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)xP(

0.5)xP(μ 0.5μ)xP(

Normal Probability Distribution

The normal distribution refers to a family of continuous probability distributions: same equation.

6-11

f x e x( ) ( ) / 1

2

2 22

Where: = mean

= standard deviation

= 3.14159

e = 2.71828

6-12

The Standard Normal Distribution

Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values Values below the mean have negative z-values

6-13

Translation to the Standard Normal Distribution

Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation:

z is the number of standard deviations units that

x is away from the population mean

σ

μxz

6-14

Example 1

If x is distributed normally with mean of 100 and standard deviation of 50, the z value for x = 250 is

This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

3.050

100250

σ

μxz

6-15

Comparing x and z units

z100

3.00250 x

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)

μ = 100

σ = 50

Why standardizing?

To measure variable with different means and/or standard deviations on a single scale.

Statistically (based standard normal distribution), are they really or fundamentally different matters? SAT mean 1020 with standard deviation of 160. You scored

1260. What is your percentile?

Average adult male height is 70 inches with standard deviation of 2 inches. You are 73 inches tall. What is your height percentile?

6-16

Z-scores for Two Problems

6-17

1.5160

10201260

σ

μxz

1.52

7073

σ

μxz

Both are 1.5 Std Dev above the mean….

Download the Normal Dist. using Excel file from the class website and then select the first tab “Example 1”….

Four Excel Functions

NORMDIST(x, mean, standard_dev, cumulative) Gives the probability that a number falls at or below a given

value of a normal distribution

NORMSDIST(z) Translates the number of standard deviations (z) into

cumulative probabilities

6-18

Four Excel Functions

The distribution of heights of American women aged 18 to 24 is approximately normally distributed with a mean of 65.5 inches (166.37 cm) and a standard deviation of 2.5 inches (6.35 cm).

What percentage of these women is taller than 5' 8", that is, 68 inches (172.72 cm)? NORMDIST(68, 65.5, 2.5, TRUE)  =  84.13% STANDARDIZE(68, 65.5, 2.5) = 1, NORMSDIST(1) = 84.13% Cumulative probability (to the left on the normal curve) Thus, 1- 84.13% = 15.87% or NORMSDIST(-1) = 15.87%

6-19

Four Excel Functions

NORMINV(probability, mean, standard_dev) Calculates the x variable given a probability

NORMSINV(probability) Given the probability that a variable is within a certain distance

of the mean, it finds the z value

6-20

Four Excel Functions

How tall would a woman need to be if she wanted to be among the tallest 75% of American women (find 25% shorter instead as the right figure shows)? NORMINV(0.25, 65.5, 2.5)  = 63.81 inches

Find the z values that mark the boundary that is 25% less than the mean and 25% (75%) more than the mean. NORMSINV(0.25) = - 0.674 NORMSINV(0.75) = 0.674

6-21

Formula Table

6-22

We have

X Z p

We want

X X=μ+zσ Norminv(p,μ,σ)

Z Normsinv(p)

p Normdist(x,μ,σ,1)

Normsdist(z)

),,(estandardiz

xz

xz

Example 1

An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days?

6-23

Example 2

Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110?

6-24

Example 3

Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)

6-25

6-26

The Standard Normal Table

The Standard Normal table in the textbook (Appendix D)

Gives the probability from the mean (zero)

up to a desired value for z

z0 2.00

0.4772Example:

P(0 < z < 2.00) = 0.4772

6-27

The Standard Normal Table gives the probability between the mean and a certain z value

The z value ALWAYS refers to the area between some value (-z or +z) and the mean

Since the distribution is symmetrical, the Standard Normal Table only displays probabilities for ½ of the full distribution

The Standard Normal Table(continued)

6-28

The Standard Normal Table

The value within the table gives the probability from z = 0 up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = 0.4772

The row shows the value of z to the first decimal point

The column gives the value of z to the second decimal point

2.0

.

.

.

(continued)

6-29

General Procedure for Finding Probabilities

1. Determine m and s

2. Define the event of interest

e.g., P(x > x1)

3. Convert to standard normal

4. Use the table to find the probability

σ

μxz

6-30

z Table Example

Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.12 0

x8.6 8

05

88

σ

μxz

0.125

88.6

σ

μxz

Calculate z-values:

6-31

z Table Example

Suppose x is normal with mean 8.0 and standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.12 0x8.6 8

P(8 < x < 8.6)

m = 8 = 5

m = 0 = 1

(continued)

6-32

Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

0.0478.02

0.1 .0478

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.12)P(8 < x < 8.6)

6-33

Finding Normal Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(x < 8.6) The probability of obtaining a value less than 8.6

Z

8.6

8.0

P = 0.5

6-34

Finding Normal Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(x < 8.6)

(continued)

Z

0.12

0.0478

0.00

0.5000 P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= 0.5000 + 0.0478 = 0.5478

6-35

Upper Tail Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(x > 8.6)

Z

8.6

8.0

6-36

P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= 0.5000 - 0.0478 = 0.4522

Now Find P(x > 8.6)…(continued)

Z

0.12

0Z

0.12

0.0478

0

0.5000 0.4522

Upper Tail Probabilities

6-37

Lower Tail Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(7.4 < x < 8)

Z

7.48.0

6-38

Lower Tail Probabilities

Now Find P(7.4 < x < 8)…the probability between 7.4 and the mean of 8

Z

7.48.0

The Normal distribution is symmetric, so we use the same table even if z-values are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= 0.0478

(continued)

0.0478

6-39

Empirical Rules

μ ± 1σ covers about 68% of x’s

f(x)

xμ μ+1σμ-1σ

What can we say about the distribution of values around the mean if the distribution is normal?

σσ

68.26%

Recall Tchebyshev from Chpt. 3

6-40

The Empirical Rule

μ ± 2σ covers about 95% of x’s

μ ± 3σ covers about 99.7% of x’s

xμ

2σ 2σ

xμ

3σ 3σ

95.44% 99.72%

(continued)

6-41

Importance of the Rule

If a value is about 2 or more standard deviations away from the mean in a normal distribution, then it is far from the mean

The chance that a value that far or farther away from the mean is highly unlikely, given that particular mean and standard deviation

6-42

Normal Probabilities in PHStat

We can use Excel and PHStat to quickly

generate probabilities for any normal

distribution

We will find P(8 < x < 8.6) when x is

normally distributed with mean 8 and

standard deviation 5

6-43

PHStat Dialogue Box

Select desired options and enter values

6-44

PHStat Output

6-45

The Uniform Distribution

The uniform distribution is a probability distribution that has equal probabilities for all possible

outcomes of the random variable Referred to as the distribution of “little information” Probability is the same for ANY interval of the same width Useful when you have limited information about how the

data “behaves” (e.g., is it skewed left?)

6-46

The Continuous Uniform Distribution:

otherwise 0

bxaifab

1

where

f(x) = value of the density function at any x value

a = lower limit of the interval of interest

b = upper limit of the interval of interest

The Uniform Distribution(continued)

f(x) =

6-47

The mean (expected value) is:

2

baμE(x)

where

a = lower limit of the interval from a to b

b = upper limit of the interval from a to b

The Mean and Standard Deviation for the Uniform Distribution

The standard deviation is

12

a)(bσ

2

6-48

Steps for Using the Uniform Distribution

1. Define the density function

2. Define the event of interest

3. Calculate the required probability

x

f(x)

6-49

Uniform Distribution

Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

x

f(x)

6-50

Uniform Distribution

Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6:

42

62μE(x)

1.154712

2)(6

12

a)(bσ

22

6-51

The Exponential Distribution

Used to measure the time that elapses between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at a dock Time between transactions at an ATM Machine Time between phone calls to the main operator

Recall l = mean for Poisson (see Chpt. 5)

6-52

The Exponential Distribution

aλe1a)xP(0

The probability that an arrival time is equal to or less than some specified time a is

where 1/ is the mean time between events and e = 2.7183

NOTE: If the number of occurrences per time period is Poisson with mean , then the time between occurrences is exponential with mean time 1/ and the standard deviation also is 1/l

6-53

Exponential Distribution

Shape of the exponential distribution

(continued)

f(x)

x

= 1.0(mean = 1.0)

l = 0.5 (mean = 2.0)

= 3.0(mean = .333)

6-54

Time between arrivals is exponentially distributed with mean time between arrivals of 4 minutes (15 per 60 minutes, on average)

1/ = 4.0, so = .25

P(x < 5) = 1 - e-a = 1 – e-(.25)(5) = 0.7135

There is a 71.35% chance that the arrival time between consecutive customers is less than 5 minutes

Example

Example: Customers arrive at the claims counter at the rate of 15 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes?

6-55

Using PHStat

6-56

Chapter Summary

Reviewed key continuous distributions normal uniform exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems

6-57

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying,

recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.