## Statistics for Management and Economics Abbreviated 10th Edition by Gerald Keller – Test Bank

**CHAPTER 7B: RANDOM VARIABLES AND DISCRETE PROBABILITY DISTRIBUTIONS**

**TRUE/FALSE**

- The Poisson probability distribution is a continuous probability distribution.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In a Poisson distribution, the mean and variance are equal.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- The Poisson random variable is a discrete random variable with infinitely many possible values.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The mean of a Poisson distribution, where
*m*is the average number of successes occurring in a specified interval, is*m*.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The number of accidents that occur at a busy intersection in one month is an example of a Poisson random variable.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The number of customers arriving at a department store in a 5-minute period has a Poisson distribution.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The number of customers making a purchase out of 30 randomly selected customers has a Poisson distribution.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The largest value that a Poisson random variable
*X*can have is*n*.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The Poisson distribution is applied to events for which the probability of occurrence over a given span of time, space, or distance is very small.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- In a Poisson distribution, the variance and standard deviation are equal.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In a Poisson distribution, the mean and standard deviation are equal.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

**MULTIPLE CHOICE**

- Which of the following cannot have a Poisson distribution?

a.

The length of a movie.

b.

The number of telephone calls received by a switchboard in a specified time period.

c.

The number of customers arriving at a gas station in Christmas day.

d.

The number of bacteria found in a cubic yard of soil.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The Sutton police department must write, on average, 6 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean.

a.

The mean has no interpretation.

b.

The expected number of tickets written would be 6.5 per day.

c.

Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written.

d.

The number of tickets that is written most often is 6.5 tickets per day.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- The Poisson random variable is a:

a.

discrete random variable with infinitely many possible values.

b.

discrete random variable with finite number of possible values.

c.

continuous random variable with infinitely many possible values.

d.

continuous random variable with finite number of possible values.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- Given a Poisson random variable
*X*, where the average number of successes occurring in a specified interval is 1.8, then*P*(*X*= 0) is:

a.

1.8

b.

1.3416

c.

0.1653

d.

6.05

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- In a Poisson distribution, the:

a.

mean equals the standard deviation.

b.

median equals the standard deviation.

c.

mean equals the variance.

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- On the average, 1.6 customers per minute arrive at any one of the checkout counters of Sunshine food market. What type of probability distribution can be used to find out the probability that there will be no customers arriving at a checkout counter in 10 minutes?

a.

Poisson distribution

b.

Normal distribution

c.

Binomial distribution

d.

None of these choices.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- A community college has 150 word processors. The probability that any one of them will require repair on a given day is 0.025. To find the probability that exactly 25 of the word processors will require repair, one will use what type of probability distribution?

a.

Normal distribution

b.

Poisson distribution

c.

Binomial distribution

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

**COMPLETION**

- In a Poisson experiment, the number of successes that occur in any interval of time is ____________________ of the number of success that occur in any other interval.

ANS: independent

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In a(n) ____________________ experiment, the probability of a success in an interval is the same for all equal-sized intervals.

ANS: Poisson

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In a Poisson experiment, the probability of a success in an interval is ____________________ to the size of the interval.

ANS: proportional

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In Poisson experiment, the probability of more than one success in an interval approaches ____________________ as the interval becomes smaller.

ANS:

zero

0

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- A Poisson random variable is the number of successes that occur in a period of ____________________ or an interval of ____________________ in a Poisson experiment.

ANS: time; space

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- The ____________________ of a Poisson distribution is the rate at which successes occur for a given period of time or interval of space.

ANS:

mean

expected value

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- In the Poisson distribution, the mean is equal to the ____________________.

ANS: variance

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- In the Poisson distribution, the ____________________ is equal to the variance.

ANS: mean

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

- The possible values of a Poisson random variable start at ____________________.

ANS:

zero

0

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

- A Poisson random variable is a(n) ____________________ random variable.

ANS: discrete

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

**SHORT ANSWER**

- Compute the following Poisson probabilities (to 4 decimal places) using the Poisson formula:

a.

*P*(*X* = 3), if *m* = 2.5

b.

*P*(*X* £ 1), if *m* = 2.0

c.

*P*(*X* ³ 2), if *m* = 3.0

ANS:

a.

0.2138

b.

0.4060

c.

0.8009

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- Let
*X*be a Poisson random variable with*m*= 6. Use the table of Poisson probabilities to calculate:

a.

*P*(*X* £ 8)

b.

*P*(*X* = 8)

c.

*P*(*X* ³ 5)

d.

*P*(6 £ *X* £ 10)

ANS:

a.

0.8472

b.

0.1032

c.

0.7149

d.

0.5117

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- Let
*X*be a Poisson random variable with*m*= 8. Use the table of Poisson probabilities to calculate:

a.

*P*(*X* £ 6)

b.

*P*(*X* = 4)

c.

*P*(*X* ³ 3)

d.

*P*(9 £ *X* £ 14)

ANS:

a.

0.3134

b.

0.0573

c.

0.9862

d.

0.3902

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

**911 Phone Calls**

911 phone calls arrive at the rate of 30 per hour at the local call center.

- {911 Phone Calls Narrative} Find the probability of receiving two calls in a five-minute interval of time.

ANS:

*m* = 5(30/60) = 2.5; *P*(*X* = 2) = 0.2565

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {911 Phone Calls Narrative} Find the probability of receiving exactly eight calls in 15 minutes.

ANS:

*m* = 15(30/60) = 7.5; *P*(*X* = 8) = 0.1373

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {911 Phone Calls Narrative} If no calls are currently being processed, what is the probability that the desk employee can take four minutes break without being interrupted?

ANS:

*m* = 4(30/60) = 2.0; *P*(*X* = 0) = 0.1353

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

**Classified Department Phone Calls**

A classified department receives an average of 10 telephone calls each afternoon between 2 and 4 P.M. The calls occur randomly and independently of one another.

- {Classified Department Phone Calls Narrative} Find the probability that the department will receive 13 calls between 2 and 4 P.M. on a particular afternoon.

ANS:

*m* = 10; *P*(*X* = 13) = 0.0729

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Classified Department Phone Calls Narrative} Find the probability that the department will receive seven calls between 2 and 3 P.M. on a particular afternoon.

ANS:

*m * = 10; *P*(*X* = 7) = 0.0901

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Classified Department Phone Calls Narrative} Find the probability that the department will receive at least five calls between 2 and 4 P.M. on a particular afternoon.

ANS:

*m* = 10; *P*(*X* ³ 5) = 0.9707

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

**Post office**

The number of arrivals at a local post office between 3:00 and 5:00 P.M. has a Poisson distribution with a mean of 12.

- {Post Office Narrative} Find the probability that the number of arrivals between 3:00 and 5:00 P.M. is at least 10.

ANS:

*m* =12; *P*(*X* ³ 10) = 0.7576

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Post Office Narrative} Find the probability that the number of arrivals between 3:30 and 4:00 P.M. is at least 10.

ANS:

*m* = 3; *P*(*X* ³ 10) = 0.0011

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {{Post Office Narrative} Find the probability that the number of arrivals between 4:00 and 5:00 P.M. is exactly two.

ANS:

*m* = 6; *P*(*X* = 2) = 0.0446

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- Suppose that the number of buses arriving at a Depot per minute is a Poisson process. If the average number of buses arriving per minute is 3, what is the probability that exactly 6 buses arrive in the next minute?

ANS:

0.0504

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

**Unsafe Levels of Radioactivity**

The number of incidents at a nuclear power plant has a Poisson distribution with a mean of 6 incidents per year.

- {Unsafe Levels of Radioactivity Narrative} Find the probability that there will be exactly 3 incidents in a year.

ANS:

0.0892

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Unsafe Levels of Radioactivity Narrative} Find the probability that there will be at least 3 incidents in a year.

ANS:

0.9380

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Unsafe Levels of Radioactiviy Narrative} Find the probability that there will be at least 1 incident in a year.

ANS:

0.9975

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Unsafe Levels of Radioactivity Narrative} Find the probability that there will be no more than 1 incident in a year.

ANS:

0.0174

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Unsafe Levels of Radioactivity Narrative} Find the variance of the number of incidents in one year.

ANS:

6

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

- {Unsafe Levels of Radioactivity Narrative} Find the standard deviation of the number of incidents is in one year.

ANS:

2.45

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.07.05

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

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