Continuous Probability Distributions

Chapter 7

McGraw-Hill/Irwin Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved.

Learning Objectives

LO1 List the characteristics of the uniform distribution.LO2 Compute probabilities by using the uniform distribution.LO3 List the characteristics of the normal probability distribution.LO4 Convert a normal distribution to the standard normal distribution.LO5 Find the probability that an observation on a normally distributed random variable is between two values.LO6 Find probabilities using the Empirical Rule.LO7 Approximate the binomial distribution using the normal distribution.LO8 Describe the characteristics and compute probabilities using the exponential distribution.

7-2

The Uniform DistributionThe uniform probability distribution is perhaps the simplest distribution for a continuous random variable.

This distribution is rectangular in shape and is defined by minimum and maximum values.

LO1 List the characteristics of the uniform distribution.

7-3

The Uniform Distribution – Mean and Standard Deviation

LO1

7-4

The Uniform Distribution - Example

Southwest Arizona State University provides bus service to students while they are on campus. A bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 A.M. and 11 P.M. during weekdays. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed from 0 to 30 minutes.

1. Draw a graph of this distribution.2. Show that the area of this uniform distribution is 1.00.3. How long will a student “typically” have to wait for a bus? In other

words what is the mean waiting time? What is the standard deviation of the waiting times?

4. What is the probability a student will wait more than 25 minutes5. What is the probability a student will wait between 10 and 20

minutes?

LO2 Compute probabilities by using the uniform distribution.

7-5

The Uniform Distribution - Example

1. Graph of this distribution.

LO2

7-6

The Uniform Distribution - Example

2. Show that the area of this distribution is 1.00

LO2

7-7

The Uniform Distribution - Example

3. How long will a student “typically” have to wait for a bus? In other words what is the mean waiting time?

What is the standard deviation of the waiting times?

LO2

7-8

The Uniform Distribution - Example

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LO2

7-9

The Uniform Distribution - Example

5. What is the probability a student will wait between 10 and 20 minutes? 0.3333

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LO2

7-10

Characteristics of a Normal Probability Distribution

1. It is bell-shaped and has a single peak at the center of the distribution.

2. It is symmetrical about the mean

3. It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions.

4. The location of a normal distribution is determined by the mean,, the dispersion or spread of the distribution is determined by the standard deviation,σ .

5. The arithmetic mean, median, and mode are equal

6. The total area under the curve is 1.00; half the area under the normal curve is to the right of this center point, the mean, and the other half to the left of it.

LO3 List the characteristics of the normal probability distribution.

7-11

The Normal Distribution - Graphically

LO3

7-12

The Family of Normal Distribution

Different Means and Standard Deviations

Equal Means and Different Standard Deviations

Different Means and Equal Standard Deviations

LO3

7-13

The Standard Normal Probability Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

It is also called the z distribution. A z-value is the signed distance between a

selected value, designated X, and the population mean , divided by the population standard deviation, σ.

The formula is:

LO4 Convert a normal distribution to the standard normal distribution.

7-14

Areas Under the Normal Curve

LO4

7-15

The Normal Distribution – ExampleThe weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a mean of $1,000 and a standard deviation of $100.

What is the z value for the income, let’s call it X, of a foreman who earns $1,100 per week? For a foreman who earns $900 per week?

LO5 Find the probability that an observation on a normally distributed random variable is between two values.

7-16

Normal Distribution – Finding Probabilities

In an earlier example we reported that the mean weekly income of a shift foreman in the glass industry is normally distributed with a mean of $1,000 and a standard deviation of $100.

What is the likelihood of selecting a foreman whose weekly income is between $1,000 and $1,100?

LO5

7-17

Normal Distribution – Finding Probabilities

LO5

7-18

Finding Areas for Z Using Excel

The Excel function=NORMDIST(x,Mean,Standard_dev,Cumu)=NORMDIST(1100,1000,100,true)generates area (probability) fromZ=1 and below

LO5

7-19

Normal Distribution – Finding Probabilities (Example 2)

Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is:Between $790 and $1,000?

LO5

7-20

Normal Distribution – Finding Probabilities (Example 3)

Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is:Less than $790?

LO5

7-21

Normal Distribution – Finding Probabilities (Example 4)

Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is:Between $840 and $1,200?

LO5

7-22

Normal Distribution – Finding Probabilities (Example 5)

Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a shift foreman in the glass industry whose income is:Between $1,150 and $1,250

LO5

7-23

Using Z in Finding X Given Area - Example

Layton Tire and Rubber Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 with a standard deviation of 2,050 miles and that the distribution of miles follows the normal probability distribution. Layton wants to set the minimum guaranteed mileage so that no more than 4 percent of the tires will have to be replaced. What minimum guaranteed mileage should Layton announce?

24

LO5

7-24

Using Z in Finding X Given Area - Example

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LO5

7-25

Using Z in Finding X Given Area - Excel

LO5

7-26

The Empirical Rule About 68 percent of

the area under the normal curve is within one standard deviation of the mean.

About 95 percent is within two standard deviations of the mean.

Practically all is within three standard deviations of the mean.

LO6 Find probabilities using the Empirical Rule.

7-27

The Empirical Rule - ExampleAs part of its quality assurance program, the Autolite Battery Company conducts tests on battery life. For a particular D-cell alkaline battery, the mean life is 19 hours. The useful life of the battery follows a normal distribution with a standard deviation of 1.2 hours.

Answer the following questions.1. About 68 percent of the

batteries failed between what two values?

2. About 95 percent of the batteries failed between what two values?

3. Virtually all of the batteries failed between what two values?

LO6

7-28

Normal Approximation to the Binomial

The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.

The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

LO7 Approximate the binomial distribution using the normal distribution.

7-29

Normal Approximation to the Binomial

Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems reasonable because, as n increases, a binomial distribution gets closer and closer to a normal distribution.

LO7

7-30

Continuity Correction Factor

The value .5 subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).

LO7

7-31

How to Apply the Correction Factor

Only one of four cases may arise:

1. For the probability at least X occurs, use the area above (X -.5).

2. For the probability that more than X occurs, use the area above (X+.5).

3. For the probability that X or fewer occurs, use the area below (X -.5).

4. For the probability that fewer than X occurs, use the area below (X+.5).

LO7

7-32

Suppose the management of the Santoni Pizza Restaurant found that 70 percent of its new customers return for another meal. For a week in which 80 new (first-time) customers dined at Santoni’s, what is the probability that 60 or more will return for another meal?

Normal Approximation to the Binomial - Example

LO7

7-33

Normal Approximation to the Binomial - Example

P(X ≥ 60) = 0.063+0.048+ … + 0.001) = 0.197

Binomial distribution solution:

LO7

7-34

Step 1. Find the mean and the variance of a binomial distribution and find the z corresponding to an X of 59.5 (x-.5, the correction factor)

Step 2: Determine the area from 59.5 and beyond

Normal Approximation to the Binomial - Example

LO7

7-35

The Family of Exponential Distributions

Characteristics and Uses:

1. Positively skewed, similar to the Poisson distribution (for discrete variables).

2. Not symmetric like the uniform and normal distributions.

3. Described by only one parameter, which we identify as λ, often referred to as the “rate” of occurrence parameter.

4. As λ decreases, the shape of the distribution becomes “less skewed.”

LO8 Describe the characteristics and compute probabilities using the exponential distribution.

The exponential distribution usually describes inter-arrival situations such as:• The service times in a system.• The time between “hits” on a web site.• The lifetime of an electrical component.• The time until the next phone call arrives in a customer service center

7-36

Exponential Distribution - Example

Orders for prescriptions arrive at a pharmacy management website according to an exponential probability distribution at a mean of one every twenty seconds.

Find the probability the next order arrives in:

1) in less than 5 seconds,

2) in more than 40 seconds,

3) or between 5 and 40 seconds.

LO8

7-37

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7-38

Exponential Distribution - Example

Compton Computers wishes to set a minimum lifetime guarantee on it new power supply unit. Quality testing shows the time to failure follows an exponential distribution with a mean of 4000 hours. Note that 4000 hours is a mean and not a rate. Therefore, we must compute λ as 1/4000 or 0.00025 failures per hour.

Compton wants a warranty period such that only five percent of the power supply units fail during that period. What value should they set for the warranty period?

LO8

7-39

Use formula (7–7) . In this case, the rate parameter is 4,000 hours and we want the area, as shown in the diagram, to be .05.

Now, we need to solve this equation for x.

Obtain the natural log of both sides of the equation:

X = 205.17. Hence, Compton can set the warranty period at 205 hours and expect about 5 percent of the power supply units to be returned.

LO8

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7-40