This chapter continues the study of probability distributions by examining continuous probability distributions. We learn about three continuous probability distributions: the uniform probability distribution, the normal probability distribution, and the exponential probability distribution.

Learning Objectives

LO7-1 Describe the uniform probability distribution and use it to calculate probabilities

LO7-2 Describe the characteristics of a normal probability distribution

LO7-3 Describe the standard normal probability distribution and use it to calculate

probabilities

LO7-4 Describe the exponential probability distribution and use it to calculate probabilities

7-2

Uniform Distribution

The uniform distribution characteristics

It is rectangular in shape

The mean and the median are equal

It is completely described by its minimum value a and its maximum value b

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The uniform distribution is a continuous distribution. If you know the minimum and maximum values, designated a and b, you can calculate the mean, the standard deviation, and any probability you wish to find.

Uniform Distribution Formulas

The mean and standard deviation of a uniform distribution are computed as follows

The following equation describes the region from a to b

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These are the formulas to use when working with uniform distributions.

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.

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Here is a graph of the distribution of student wait times for a bus. The minimum wait time is 0 minutes and the maximum wait time is 30 minutes, so the range of the distribution is 30 minutes. The height is 1/(b-a)= 1/30 = .0333

Uniform Distribution Example (2 of 4)

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The area of the uniform distribution is found by multiplying height *base

Area = (30-0) = 1.00

The mean is = = 15

The standard deviation is = = 8.66

With continuous distributions, area represents probability and the uniform distribution’s rectangular shape allows us to use the area formula for a rectangle. Here, with a minimum wait time of 0 minutes and a maximum wait time of 30 minutes, the entire area of the distribution is 1.00. The mean is 15, so the typical wait time for bus service is 15 minutes. The standard deviation is 8.66; this measures the variation in student wait times.

Uniform Distribution Example (3 of 4)

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To find the probability that a student will wait more than 25 minutes, find the area between 25 and 30 minutes.

P(25 < wait time < 30) = (height)(base) = (5) = .1667

To find the probability of various wait times, use formula 7-3. In this example, the probability a student will wait more than 25 minutes is .1667.

Uniform Distribution Example (4 of 4)

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To find the probability that a student will wait between 10 and 20 minutes, find the area between 10 and 20 minutes.

P(10 < wait time < 20) = (height)(base) = (10) = .3333

Using formula 7-3, we find the probability that a student waits between 10 and 20 minutes is .3333.

Normal Probability Distribution

The normal probability distribution is a continuous distribution with the following characteristics:

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

The distribution is symmetric

It is asymptotic, meaning the curve approaches but never touches the X-axis

It is completely described by its mean and standard deviation

There is a family of normal probability distributions

Another normal probability distribution is created when either the mean or the standard deviation changes

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The normal probability distribution has a complex formula (see Formula 7-4) to find probabilities, but we will not need to use it. Instead, we use the table given in Appendix B.3. You can also use Excel to find these probabilities.

The Normal Curve

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Here is a graphical display of the characteristics of the normal distribution. The location of a normal distribution is determined by the mean, which is in the center of the bell-shaped curve. The dispersion or spread of the distribution is determined by the standard deviation. The area below the curve defines probabilities and the total area under the curve is 1.0. Therefore, the area to the left of the mean is .5 and the area to the right of the mean is .5.

Family of Normal Probability Distributions

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There is not just one normal probability distribution, but a family of them. In the top left chart of length of employee service at three different plants, the mean is the same but the standard deviations are different. In the bottom chart of box weights of three different cereals, the means are different but the standard deviations are identical. And, in the top right chart of tensile strength of three different cables, we see they have different means and different standard deviations.

Standard Normal Probability Distribution

The standard normal probability distribution is a particular normal distribution

It has a mean of 0 and a standard deviation of 1

Any normal probability distribution can be converted to the standard normal probability distribution with the following formula:

z VALUE The signed distance between a selected value, designated x, and the mean, , divided by the standard deviation, .

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The number of normal distributions is unlimited, but they can all be converted to a standard normal probability distribution. By standardizing a normal probability distribution, the distance of a value from the mean can be expressed in units of the standard deviation. The results are called z values or z scores.

Areas Under the Normal Curve

Here is a portion of the “z” Table

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For example, if you have a z=1.50, this reflects an area (or probability) of .4332. The entire table can be found in Appendix B.3.

Standard Normal Probability Example

Rideshare services are available internationally. A customer uses a smartphone app to request a ride. Then, a driver receives the request, picks up the customer, and takes the customer to the desired location. No cash is involved; the payment for the transaction is handled digitally.

Suppose the weekly income of rideshare drivers follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. What is the z value of income for a driver who earns $1,100 per week? For a driver who earns $900 per week?

What is the z-value of income for a driver who earns $1,100?

Z = = = 1.00

What is the z-value of income for a driver who earns $900?

Z = = = -1.00

Regardless of whether z is +1or -1, the area under the curve is .3413

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This idea was first developed by Uber Technologies, which is headquartered in San Francisco, California. A z of 1.00 indicates that a weekly income of $1,100 is one standard deviation above the mean and a z of -1.00 shows that a $900 weekly income is one standard deviation below the mean. Both incomes are the same distance from the mean.

The Empirical Rule

To verify the Empirical Rule:

z of 1.00 = .3413 so .3413 × 2 = .6826 or about 68%

z of 2.00 = .4772 so .4772 × 2 = .9544 or about 95%

z of 3.00 = .4987 so .4987 × 2 = .9974 or about 99.7%

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The Empirical Rule was introduced in chapter 3. To verify it, use the table found in Appendix B.3 to find a z of 1.00. This corresponds to .3413, so plus and minus 1 standard deviation is .3413 × 2=.6826, which is approximately 68% of the observations. A z of 2.00 corresponds to .4772, so plus and minus 2 standard deviations equals .4772 × 2= .9544, or about 95%. And a z of 3.00 corresponds to .4987, so plus and minus 3 standard deviations equals .4987 × 2= .9974, or practically all at about 99.7%.

The Empirical Rule Example

As 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.

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1. About 68% of the batteries failed between what two values?

19 ± 1(1.2) hours;

About 68% of batteries will fail between 17.8 and 20.2 hours.

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

19 ± 2(1.2) hours;

About 95% of batteries will fail between 16.6 and 21.4 hours.

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

19 ± 3(1.2) hours;

Practically all of the batteries will fail between 15.4 and 22.6 hours.

If the distribution is normal and you know its mean and standard deviation, then you can use the Empirical Rule to find probabilities of plus and/or minus 1, 2, or 3 standard deviations.

Finding Areas under the Normal Curve

Using the weekly incomes of Uber drivers:

P($1,000 < weekly income < $1,100) = .3413

P(weekly income < $1,100) = .3413 + .5000 =.8413

What is the z-value of income for a driver who earns $1,100?

Z = = = 1.00

7-17

To find the area between 0 and z or (-z), look up the probability directly in the table. In this example, the probability of finding a driver that earns between $1000 and $1100 is .3413 and the probability of finding a driver that earns less than $1,100 is .8413. Therefore, 34.13% of drivers earn between $1000 and $1100 and 84.13% of drivers earn less than $1,100. You can also use Excel to find probabilities; the software commands are in Appendix C.

Finding Areas under the Normal Curve (2 of 4)

Using the weekly incomes of Uber drivers:

P($790

P(weekly income < $790) = .5000 − .4821 = .0179

What is the z-value of income for a driver who earns $790?

Z = = = -2.10

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To find the area beyond z or (-z), locate the probability of z in the table and subtract that probability from .5000. The probability of finding a driver earning between $790 and $1000 is .4821 and the probability of finding a driver earning less than $790 is .0179. Therefore, 48.21% of drivers earn between $790 and $1000 and 1.79% of drivers earn less than $790.

Finding Areas Under the Normal Curve (3 of 4)

Using the weekly incomes of Uber drivers:

P($840

P($1,000

P($840 < weekly income < $1,200) = .4452 + .4772 = .9224

What is the z-value of income for a driver who earns $840?

Z = = = -1.60

What is the z-value of income for a driver who earns $1,200?

Z = = = 2.00

7-19

In this example, we combine two areas or probabilities under the curve; one area to the right of the mean and one area to the left of the mean. To find the area between two points on different sides of the mean, determine the z values and add the corresponding probabilities. Here, 92.24% of the drivers have weekly incomes between $840 and $1,200.

Finding Areas Under the Normal Curve (4 of 4)

What is the z-value of income for a driver who earns $1,250?

Z = = = 2.50

What is the z-value of income for a driver who earns $1,150?

Z = = = 1.50

Using the weekly incomes of Uber drivers:

P($1,000

P($1,150 < weekly income < $1,250) = .4938 − .4332 = .0606

7-20

In this example, we find the area of two values on the same side of the mean. To find the area between two points on the same side of the mean, determine the z values and subtract the smaller probability from the larger. In this example, about 6% of the drivers earn between $1,150 and $1,250.

Finding a Value for x Using z

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 follows the normal distribution. Let x represent the minimum guaranteed mileage and use the formula for z to solve so that no more than 4% of tires need to be replaced.

7-21

z = = and from the table we find z = -1.75

so -1.75 = = therefore, x = 64,312 miles

The minimum guaranteed mileage should be set at 64,312 miles, so that only 4% of the tires will be replaced under this plan.

Approximate a Binomial Distribution

The normal probability distribution can approximate a binomial distribution under certain conditions

n and n(1-) must both be at least 5

n is the number of observations

is the probability of a success

The four conditions for a binomial probability distribution are

There are only two possible outcomes

(pi) remains the same from trial to trial

The trials are independent

The distribution results from a count of the number of successes in a fixed number of trials

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We can use the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n because, as n increases, a binomial gets closer and closer to a normal distribution. But first, you must make sure that that the distribution of interest is really a binomial distribution.

Exponential Probability Distribution

The exponential probability distribution describes times between events in a sequence

The actions occur independently at a constant rate per unit of time or length

It is non-negative, is positively skewed, declines steadily to the right, and is asymptotic

Examples of situations using the exponential distributions

The service time for customers at the information desk at Dallas Public Library

The time until the next phone call arrives in a customer service center

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Since time is never negative, the exponential variable is always positive.

The Family of Exponential Distributions

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There is not just one exponential distribution but a family of them

lambda, is the rate parameter

The lower the rate parameter, the “less skewed” the shape of the distribution

This chart shows three exponential distributions. The graphs start at the value of when the random variable’s x value is 0 and then declines steadily as we move to the right with increasing values of x. Compare the shapes of the different exponential distributions as changes from 1/3 to 1 to 2.

Exponential Distribution Formulas

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This formula describes the exponential distribution

The area under the curve is given by this formula

Both the mean and the standard deviation are

With continuous distributions, we do not find the probability that a particular value of x occurs but rather, the probability that a random variable x occurs in a specified interval. The area of that interval under the curve is the probability and rather than use a table, we use a formula. In the formula, e represents a constant 2.71828. The mean and the standard deviation are equal in an exponential distribution.

Exponential Distribution Example

Orders for prescriptions arrive at a pharmacy website according to an exponential probability distribution at a mean of one every 20 seconds.

Find the probability the next order arrives in less than 5 seconds.

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P(Arrival time < 5) = 1 -

= 1 -

= 1 - .7788 = .2212

Use formula 7-7 to find this probability with a We conclude there is a 22% chance the next order will arrive in less than 5 seconds and we see this is in the left-tail area of the exponential distribution.

Exponential Distribution Example (2 of 2)

Orders for prescriptions arrive at a pharmacy website according to an exponential probability distribution at a mean of one every 20 seconds.

Find the probability the next order arrives in more than 40 seconds.

7-27

P(Arrival time > 40) = = .1353

When you wish to find a probability in the right-tail area, you can first use the formula 7-7 to find the probability an order arrives in less than 40 seconds and then use the complement rule. But 1-(1-e-) = e-; so we can just use this. Here, we find the likelihood that it will be 40 seconds or more before the next order is received at the pharmacy is 13.5%.

Chapter 7 Practice Problems

7-28

Question 1

7-29

A uniform distribution is defined over the interval from 6 to 10.

What are the values for a and b?

What is the mean of this uniform distribution?

What is the standard deviation?

Show that the probability of any value between 6 and 10 is equal to 1.0.

What is the probability that the random variable is more than 7?

What is the probability that the random variable is between 7 and 9?

What is the probability that the random variable is equal to 7.91?

LO-1

Question 9

7-30

The mean of a normal probability distribution is 500; the standard deviation is 10.

About 68% of the observations lie between what two values?

About 95% of the observations lie between what two values?

Practically all of the observations lie between what two values?

LO7-3

Question 13

7-31

A normal population has a mean of 20.0 and a standard deviation of 4.0.

Compute the z value associated with 25.0.

What proportion of the population is between 20.0 and 25.0?

What proportion of the population is less than 18.0?

LO7-3

Question 19

7-32

The Internal Revenue Service reported the average refund in 2017 was $2,878 with a standard deviation of $520. Assume the amount refunded is normally distributed.

What percent of the refunds are more than $3,500?

What percent of the refunds are more than $3,500 but less than $4,000?

What percent of the refunds are more than $2,400 but less than $4,000?

LO7-3

Question 27

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According to media research, the typical American listened to 195 hours of music in the last year. This is down from 290 hours 4 years earlier. Dick Trythall is a big country and western music fan. He listens to music while working around the house, reading, and riding in his truck. Assume the number of hours spent listening to music follows a normal probability distribution with a standard deviation of 8.5 hours.

If Dick is in the top 1% in terms of listening time, how many hours did he listen last year?

Assume that the distribution of times 4 years earlier also follows the normal probability distribution with a standard deviation of 8.5 hours. How many hours did the 1% who listen to the least music actually listen?

LO7-3

Question 31

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Waiting times to receive food after placing an order at the local Subway sandwich shop follow an exponential distribution with a mean of 60 seconds. Calculate the probability a customer waits:

Less than 30 seconds.

More than 120 seconds.

Between 45 and 75 seconds.

Fifty percent of the patrons wait less than how many seconds? What is the median?

LO7-4

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