Essentials of Business Analytics 2nd Edition By Camm – Test Bank

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Sample Questions Posted Below

 

 

 

 

 

1. Probability is​ the

  a. number of successes divided by the number of failures.
  b. numerical measure likelihood that an outcome occurs.
  c. chance that an event will not happen.
  d. number of successes divided by the standard deviation of the distribution.

 

ANSWER:   b
RATIONALE:   Probability is the numerical measure likelihood that an outcome occurs.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

2. A(n) ________________ is a description of how probabilities are distributed over the values of a random variable.​

  a. ​probability distribution
  b. ​mass function of an event
  c. ​density function
  d. ​expected value

 

ANSWER:   a
RATIONALE:   A probability distribution is a description of how probabilities are distributed over the values of a random variable.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 186
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

3. An initial estimate of the probabilities of events is a ______________ probability.

  a. ​posterior
  b. ​conditional
  c. empirical
  d. ​prior

 

ANSWER:   d
RATIONALE:   Prior probability is an initial estimate of the probabilities of events.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

4. Bayes’ theorem is a method used to compute ___________________ probabilities.

  a. ​posterior
  b. ​conditional
  c. empirical
  d. ​prior

 

ANSWER:   a
RATIONALE:   Bayes’ theorem is a method used to compute posterior probabilities.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 182
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

5. All the events in the sample space that are not part of the specified event are called

  a. joint events.
  b. the complement of the event.
  c. simple events.
  d.  independent events.

 

ANSWER:   b
RATIONALE:   The complement of an event is all the events in the sample space that are not part of the specified event.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 173
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

6. Sample space is

  a. ​a process that results in some outcome.
  b. ​the collection of all possible outcomes.
  c. the collection of events
  d. ​a subgroup of a population/the likelihood of an outcome.

 

ANSWER:   b
RATIONALE:   By specifying all possible outcomes, we identify the sample space of a random experiment.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

7. The event containing the outcomes belonging to A or B or both is the ________________ of A and B.

  a. union
  b. Venn diagram
  c. intersection
  d. subset

 

ANSWER:   a
RATIONALE:   The event containing the outcomes belonging to A or B or both is the Union of A and B.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   SOME BASIC RELATIONSHIPS OF PROBABILITY, Page 174
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

8. Two events are independent if

  a. the two events occur at the same time.
  b. the probability of one or both events is greater than 1.
  c. the probability of each event is not affected by the other.
  d. none of these.

 

ANSWER:   c
RATIONALE:   Two events are independent if the probability of each event is not affected by the other.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

9. Which statement is true about mutually exclusive events?

  a. If events A and B cannot occur at the same time, they are called mutually exclusive.
  b. If either event A or event B must occur, they are called mutually exclusive.
  c. P(A) + P(B) = 1 for any events A and B that are mutually exclusive.
  d. None of these.

 

ANSWER:   a
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   CONDITIONAL PROBABILITY, Page 175
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

10. A joint probability is the

  a. sum of the probabilities of two events.
  b. probability of the intersection of two events.
  c. probability of the union of two events.
  d. sum of the probabilities of two independent events.

 

ANSWER:   b
RATIONALE:   The probability of the intersection of two events is called a joint probability.​
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   CONDITIONAL PROBABILITY, Page 179
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

11. In the probability table below, which value is a marginal probability?

​​

   Completed
Obstacle Course Level  No Yes Total
Challenging  0.4 0.3 0.7
Easy  0.1 0.2 0.3
Total  0.5 0.5 1

 

  a. 0.1
  b. 1
  c. 0.5
  d. 0.4

 

ANSWER:   c
RATIONALE:   The values in the Total column and Total row (the margins) provide the probabilities of each event separately. These probabilities are referred to as marginal probabilities because of their location in the joint probability table.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   CONDITIONAL PROBABILITY, Page 179
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

12. A variable that can only take on specific numeric values is called a

​​

  a. categorical variable.
  b. discrete random variable.
  c. continuous random variable.
  d. categorical variable.

 

ANSWER:   b
RATIONALE:   A random variable that can take on only specified discrete values is referred to as a discrete random variable.
POINTS:   1
DIFFICULTY:   Easy
REFERENCES:   RANDOM VARIABLES, Page 184
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

13. An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable in this experiment is a

  a. discrete random variable.
  b. continuous random variable.
  c. complex random variable.
  d. categorical random variable.

 

ANSWER:   b
RATIONALE:   Probability is the numerical measure likelihood that an outcome occurs.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   RANDOM VARIABLES, Page 184
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

14. Which of the following statements is correct?

  a. The binomial and normal distributions are both discrete probability distributions.
  b. The binomial and normal distributions are both continuous probability distributions.
  c. The binomial distribution is a continuous probability distribution and the normal distribution is a discrete probability distribution.
  d. The binomial distribution is a discrete probability distribution and the normal distribution is a continuous probability distribution.

 

ANSWER:   d
RATIONALE:   The binomial distribution is a discrete probability distribution and the normal distribution is a continuous probability distribution.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 192 and CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 202
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

15. Which of the following is a discrete random variable?

  a. The number of times a student guesses the answers to questions on a certain test.
  b. The amount of gasoline purchased by a customer.
  c. ​The amount of mercury found in fish caught in the Gulf of Mexico.
  d. ​The height of water-oak trees.

 

ANSWER:   a
RATIONALE:   A discrete random variable can only take on specified discrete values.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   RANDOM VARIABLES, Page 184
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

16. All of the following are examples of discrete random variables except

​​

  a. number of tickets sold.
  b. marital status.
  c. time.
  d. population of a city.

 

ANSWER:   c
RATIONALE:   Discrete random variables take on only specific discrete values. Time is an example of a continuous random variable, which takes on any numerical value in an interval or collection of intervals.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   RANDOM VARIABLES, Page 184
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

17. The _______________ probability distribution can be used to estimate the number of vehicles that go through an intersection during the lunch hour.

​​

  a. binomial
  b. normal
  c. triangular
  d. Poisson

 

ANSWER:   d
RATIONALE:   The Poisson probability distribution is often useful in estimating the number occurrences of an event over a specified interval of time or space.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 194
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

18. The random variable X is known to be uniformly distributed between 2 and 12. Compute E(X), the expected value of the distribution.

  a. 4
  b. 5
  c. 6
  d. 7

 

ANSWER:   d
RATIONALE:   The expected value is the average of the endpoints of a uniform distribution. Therefore E(X) = (a+b)/2 = (2+12)/2 = 7.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 200
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

19. The random variable X is known to be uniformly distributed between 2 and 12. Compute the standard deviation of X.

  a. ​2.887
  b. ​3.464
  c. 8.333
  d. ​12

 

ANSWER:   a
RATIONALE:   The variance of a random variable that is uniformly distributed is Var(X) = (b – a)2/12 = (12 – 2)2 = 12 = 8.333.  The standard deviation is the square root of the variance, so the standard deviation is 2.887.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 200
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

20. If a​ z-score is​ zero, then the corresponding x-value must be equal to the

​​

  a. mean.
  b. median.
  c. mode.
  d. standard deviation.

 

ANSWER:   a
RATIONALE:   You can transform an x-value to a z-score using the following formula. z-score = (x-value – mean)/standard deviation. If the z-score is zero, then the x-value is equal to the mean.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 204
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

21. In a normal​ distribution, which is​ greater, the mean or the​ median?

​​

  a. Mean
  b. Median
  c. Neither (they are equal)
  d. Cannot be determined with the information provided

 

ANSWER:   c
RATIONALE:   One of the properties of a normal distribution is that the​ mean, median, and mode are equal.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 203
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

22. The center of a normal curve is

  a. always equal to zero.
  b. the mean of the distribution.
  c. always a positive number.
  d. equal to the standard deviation.

 

ANSWER:   b
RATIONALE:   The center of a normal curve (the highest point on the normal curve) is at the mean.​
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 203
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

23. Which of the following is not a characteristic of the normal probability distribution?

  a. The mean, median, and the mode are equal.
  b. The mean of the distribution can be negative, zero, or positive.
  c. The distribution is symmetrical.
  d. The standard deviation must be 1.

 

ANSWER:   d
RATIONALE:   The standard deviation of a normal probability distribution does not have to be 1.  It can be any positive number.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 204
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

24. A health conscious student faithfully wears a device that tracks his steps.  Suppose that the distribution of the number of steps he takes is normally distributed with a mean of 10,000 and a standard deviation of 1,500 steps.  What percent of the time does he exceed 13,000 steps?

  a. 2.28%
  b. 5.0%
  c. 95%
  d. 97.72%

 

ANSWER:   a
RATIONALE:   We are asked to calculate P(X > 13,000) given a normal distribution with a mean of 10.000 and a standard deviation of 1,500 steps.  This probability can be calculated using the Excel function 1 – NORM.DIST(13000, 10000, 1500, TRUE) = 0.0228.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 205
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

25. The newest model of smart car is supposed to get excellent gas mileage.  A thorough study showed that gas mileage (measured in miles per gallon) is normally distributed with a mean of 75 miles per gallon and a standard deviation of 10 miles per gallon. What is the probability that, if driven normally, the car will get 100 miles per gallon or better?

​​

  a. 0.6%
  b. 2.5%
  c. 6%
  d. 25%

 

ANSWER:   a
RATIONALE:   We are asked to calculate P(X > 100) given a normal distribution with a mean of 75 miles per gallon and a standard deviation of 10 miles per gallon.  This probability can be calculated using the Excel function 1 – NORM.DIST(100, 75, 10, TRUE) = 0.006.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 205
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

26. A health conscious student faithfully wears a device that tracks his steps.  Suppose that the distribution of the number of steps he takes is normally distributed with a mean of 10,000 and a standard deviation of 1,500 steps.  One day he took 15,000 steps.  What was his percentile on that day?

​​

  a. 95%
  b. 97.7%
  c. 99.7%
  d. 100%

 

ANSWER:   c
RATIONALE:   We are asked to calculate P(X < 15,000) given a normal distribution with a mean of 10.000 and a standard deviation of 1,500 steps.  This probability can be calculated using the Excel function  NORM.DIST(15000, 10000, 1500, TRUE) = 0.977.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 205
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

27. The newest model of smart car is supposed to get excellent gas mileage.  A thorough study showed that gas mileage (measured in miles per gallon) is normally distributed with a mean of 75 miles per gallon and a standard deviation of 10 miles per gallon. What value represents the 50th percentile of this distribution?

​​

  a. 75
  b. 85
  c. 95
  d. 105

 

ANSWER:   a
RATIONALE:   The 50th percentile of a normal distribution is located at its mean. In this case, the mean is 75 mpg, so the 50th percentile is 75.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 205
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

28. A health conscious student faithfully wears a device that tracks his steps.  Suppose that the distribution of the number of steps he takes is normally distributed with a mean of 10,000 and a standard deviation of 1,500 steps.  How many steps would he have to take to make the cut for the top 5% for his distribution?

​​

  a. 7,533
  b. 8,078
  c. 10,000
  d. 12,467

 

ANSWER:   d
RATIONALE:   We are asked to calculate the number of steps needed to put the student at the top 5% of days.  This is equivalent to finding the 95th percentile of the distribution. Given a normal distribution with a mean of 10.000 and a standard deviation of 1,500 steps this value can be calculated using the Excel function NORM.INV(0.95, 10000, 1500) = 12,467.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 207
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

29. What is the mean of x, given the exponential probability function

  a. 0.05
  b. 20
  c. 100
  d. 2,000

 

ANSWER:   b
RATIONALE:   The exponential probability function is

so the mean, m,  is 20.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 208
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Comprehension

 

30. Fast food restaurants pride themselves in being able to fill orders quickly. A study was done at a local fast food restaurant to determine how long it took customers to receive their order at the drive thru.   It was discovered that the time it takes for orders to be filled is exponentially distributed with a mean of 1.5 minutes. What is the probability density function for the time it takes to fill an order?

​​

  a.
  b.
  c.
  d. None of the above answers are correct.

 

ANSWER:   c
RATIONALE:   The mean is 1.5, which is equivalent to 3/2.  The exponential probability distribution function is    Substituting 3/2 for m gives us , which simplifies to

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 208
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

31. What is the total area under the normal distribution curve?

​​

  a. It depends upon the mean and standard deviation
  b. It must be calculated
  c. 1
  d. 100

 

ANSWER:   c
RATIONALE:   The total area under the normal distribution curve is 1.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 204
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

32. The triangular distribution is a good model for____________ distributions.

​​

  a. uniform
  b. skewed
  c. normal
  d. normal

 

ANSWER:   b
RATIONALE:   The triangular distribution is a good model for skewed distributions.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 201
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

33. Fast food restaurants pride themselves in being able to fill orders quickly. A study was done at a local fast food restaurant to determine how long it took customers to receive their order at the drive thru.   It was discovered that The time it takes for orders to be filled is exponentially distributed with a mean of 1.5 minutes. What is the probability that it takes less than one minute to fill an order?

  a. 0.1813
  b. 0.4866
  c. 0.6321
  d. 0.7769

 

ANSWER:   b
RATIONALE:   The cumulative probability P(X < 1) =
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 209
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

34. A nickel and a dime are tossed. How many possible outcomes are in this event?​

ANSWER:   There are 4 possible outcomes.​
RATIONALE:  
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

35. A nickel and a dime are tossed. We are interested only in the event that includes at least one head appears on a single toss of both coins. What are the possible outcomes?

ANSWER:   There are 3 possible outcomes.​
RATIONALE:  
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

36. Consider a random experiment of rolling 2 dice. The sample space for rolling two dice is shown. Let S be the set of all ordered pairs listed in the figure. What are the possible outcomes for the event of rolling a 7?

ANSWER:  

{(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)}

RATIONALE:   The probability of rolling a seven corresponds to the event {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)}​
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

37. Consider a random experiment of rolling 2 dice. The sample space for rolling two dice is shown. Let S be the set of all ordered pairs listed in the figure. What is probability of rolling a 7?

ANSWER:   0.1667 or 16.7% or 1/6
RATIONALE:   The probability of an event is equal to the sum of the probabilities of outcomes for the event. The probability of rolling a seven corresponds to the event {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)} which is 6 out of 36 = 1/6.​
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

38. Consider a random experiment of rolling 2 dice. The sample space for rolling two dice is shown. Let S be the set of all ordered pairs listed in the figure. What is probability of rolling a sum larger than 10?

ANSWER:   1/12 or 0.0833 or 8.3%
RATIONALE:   The probability of rolling a sum of 11 corresponds to the event {(6, 5), (5, 6)} which is 2 out of 36 = 2/36.

The probability of rolling a sum of 12 corresponds to the event {(6, 6)} which is 1 out of 36 = 1/36.

Using the addition rule we add the two we have 2/36 + 1/36 = 3/36 = 1/12​.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Pages 171-173
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

39. James has two fair coins. When he flips them, what is the sample space?

ANSWER:   heads-heads, heads-tails, tails-heads, and tails-tails​
RATIONALE:  
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 171
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

40. A nickel and a dime are tossed. If an event is defined as a single toss of both coins where at least one head appears, what is the complement of that event?

ANSWER:   ​tails-tails
RATIONALE:   The event has 3 possible outcomes–heads-heads, tails-heads, and heads-tails. Given an event A, the complement of A is defined to be the event consisting of all outcomes that are not in A. The only outcome not in the event is tails-tails.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   EVENTS AND PROBABILITIES, Page 173
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

41. Given that A and B are independent with P(AB) = 0.8 and P(Bc) = 0.3, find P(A).

ANSWER:   0.33
RATIONALE:   First note that P(B) = 1 − 0.3 = 0.7

The general addition law states that

Since A and B are independent

​Therefore, in this situation

​Substituting the given values yields 0.8 = P(A) + 0.7 – P(A) × 0.7.

Subtract 0.7 from both sides 0.1 = P(A) – P(A) × 0.7.

Combine like terms on the right side 0.1 = 0.3P(A).

Divide

POINTS:   1
DIFFICULTY:   Challenging
REFERENCES:   CONDITIONAL PROBABILITY, Independent Events, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

42. A bucket contains 2 red balls, 4 yellow balls, and 5 purple balls.  One ball is taken from the bucket and then replaced.  Another ball is taken from the bucket.  Are the events of pulling first ball is red then a purple one independent or dependent?

ANSWER:   Independent
RATIONALE:   The sample space of 11 balls does not change from the first event to the second event.  The events are independent.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

43. A bucket contains 2 red balls, 4 yellow balls, and 5 purple balls.  One ball is taken from the bucket and then replaced.  Another ball is taken from the bucket.  What is the probability that the first ball is red and the second ball is purple?

ANSWER:   10/121
RATIONALE:   The events are independent, so the multiplication law for independent events can be used to calculate the joint probability.

P(red then purple) = P(red) × P(purple) = 2/11 × 5/11 = 10/121

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

44. A bucket contains 3 red balls, 4 yellow balls, and 5 purple balls.  One ball is taken from the bucket and is not replaced.  Another ball is taken from the bucket.  Are the events of pulling first ball is red then a purple one independent or dependent?

ANSWER:   dependent
RATIONALE:   The sample space of the first event is 12 balls, but the sample space of the second event is now 11 balls (because the first ball is not replaced).  Therefore, the events are dependent.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

45. A bucket contains 3 red balls, 4 yellow balls, and 5 purple balls.  One ball is taken from the bucket and is not replaced.  Another ball is taken from the bucket.  What is the probability that the first ball is red and the second ball is purple?

ANSWER:   5/44
RATIONALE:   The events are dependent therefore the probability is calculated as P(red then purple) = P(red) • P(purple) = 3/12 • 5/11 = 15/132 = 5/44

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

46. Given that P(A) = 0.3, P(A|B) = 0.4, and P(B) = 0.5, compute P(A ∩ B)

ANSWER:   0.2
RATIONALE:   P(AB) = P(A|B) × P(B) = 0.4 × 0.5 = 0.2
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

47. The cross tabulation below classifies employees of a communications company by age and field of expertise. Use the given information to create a joint probability table.

   Under 35  35-44 45+ Total
 Engineering  8,399  8,663  7,072  24,134
 Business  14,515  14,988  26,683  56,186
 Education  6,738  5,657  8,669  21,064
 Liberal Arts  11,415  11,484  12,111  35,010
 Total  41,067  40,792  54,535  136,394

ANSWER:  
   Under 35 35-44 45+ Total
 Engineering  0.06  0.06  0.05  0.18
 Business  0.11  0.11  0.20  0.41
 Education  0.05  0.04  0.06  0.15
 Liberal Arts  0.08  0.08  0.09  0.26
 Total  0.30  0.30  0.40  1.00

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 179
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

48. The contingency table below represents employees of a communications company classified by age and field of expertise. Fill in the missing entries.

   Under 35  35-44 45+ Total
 Engineering  7,635  7,875  6,429  21,939
 Business  13,195      38,814
 Education  4,802  5,143    
 Liberal Arts  10,377  10,440  11,010 31,827
 Total  36,009    37,313  110,405

ANSWER:  
   Under 35 35-44 45+ Total
 Engineering 7,635  7,875  6,429 21,939
 Business 13,195 13,625  24,257 38,814
 Education 4,802 5,143 7,881 17,825
 Liberal Arts 10,377 10,440 11,010  31,827
 Total 36,009 37,083 37,313 110,405

110,405 – (36,009 + 37,313) = 37,083

110,405 – (21,939 + 38,814 + 31,827) = 17,825

38,814 – (13,195 + 1,362) = 24,257

37,083 – (7,875 + 5,143 + 10,440) = 13,625

17,825 – (4,802 + 5,142) = 7,881

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 177
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

49. The contingency table below represents employees of a communications company classified by age and field of expertise. What is the probability that a randomly selected employee age 35-45 years old has business expertise?

   Under 35  35-44 45+ Total
 Engineering  8,399  8,663  7,072  24,134
 Business  14,515  14,988  26,683  56,186
 Education  6,738  5,657  8,669  21,064
 Liberal Arts  11,415  11,484  12,111  35,010
 Total  41,067  40,792  54,535  136,394

ANSWER:   14,988/40,791 = 0.37
RATIONALE:   There are 40,792 employees that are 35-44 years old.  Of those, 14,988 have business expertise. Therefore, the probability that a randomly selected employee that is age 35-44 years old has business expertise is 14,988 / 40,792 = 0.37.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 179
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

50. The cross tabulation shown below shows employees of a communications company classified by age and field of expertise. What is the probability that a randomly selected engineer is under the age of 35?

   Under 35  35-44 45+ Total
 Engineering  8,399  8,663  7,072  24,134
 Business  14,515  14,988  26,683  56,186
 Education  6,738  5,657  8,669  21,064
 Liberal Arts  11,415  11,484  12,111  35,010
 Total  41,067  40,792  54,535  136,394

ANSWER:   8,399 / 24,134 = 0.35
RATIONALE:   There are 24,134 engineers.  Of those, 8,300 of them are under 35 years old.  Therefore, the probability that a randomly selected engineer is under the age of 35 is 8,399 / 24,134 = 0.35.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONDITIONAL PROBABILITY, Page 179
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

51. The random variable X is known to be uniformly distributed between 2 and 12. Compute P(X = 3).

ANSWER:   0

RATIONALE:   A single point is an interval of zero width.  This implies that the probability of a continuous random variable assuming any particular value exactly is zero.

POINTS:   1
DIFFICULTY:   Challenging
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 199
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

52. The random variable X is known to be uniformly distributed between 2 and 12. Compute P(X > 10).

ANSWER:   0.2

RATIONALE:   P(X > 6) = (base)(height) = (2)(1/10) = 0.2. The probability of a continuous random variable assuming a value in any interval is the same whether or not the endpoints are included.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 199
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

53. For the standard normal probability distribution, what percent of the curve lies to the left of the mean?

ANSWER:   50%

RATIONALE:   In a standard normal probability distribution, the mean is located at the center of the curve which means that half of the area is below the mean and half is above the mean.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 203
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

54. Participants at the state fair were given 8 rings to toss. The number x of rings tossed onto a stick can be approximated by the probability distribution in the table.  Use the probability distribution to find the mean and variance of the probability distribution.

 x  f(x)
 0  0.010
 1  0.030
 2  0.070
 3  0.070
 4  0.100
 5  0.210
 6  0.320
 7  0.130
 8  0.060

ANSWER:  
 x f(x) xf(x)
 0  0.010  0
 1  0.030  0.03
 2  0.070  0.14
 3  0.070  0.21
 4  0.100  0.4
 5  0.210  1.05
 6  0.320  1.92
 7  0.130  0.91
 8  0.060  0.48
RATIONALE:   The mean of a discrete random variable is given by the formula, μ = ∑ x × f(x). To calculate the mean, multiply each value of x by its corresponding probability f(x) and then add the products.

μ = ∑ x × f(x)  = 0 + 0.03 + 0.16 + 0.255 + 0.56 + 1.05 + 1.92 + 0.91 + 0.48 = 5.14

The variance of a discrete random variable is given by the formula, σ2 =∑(x−μ)2 f(x)

 x  f(x)  u
 0  0.01  5.14  0.264
 1  0.03  5.14  0.514
 2  0.07  5.14  0.690
 3  0.07  5.14  0.321
 4  0.1  5.14  0.130
 5  0.21  5.14  0.004
 6  0.32  5.14  0.237
 7  0.13  5.14  0.450
 8  0.06  5.14  0.491
       3.100

= 0.264 + 0.514 + 0.690 + 0.321 + 0.130 + 0.004 + 0.237 + 0.450 + 0.491 = 3.1

 

POINTS:   1
DIFFICULTY:   Challenging
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 190
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

55. In a binomial​ experiment, what does it mean to say that each trial is independent of the other​ trials?

ANSWER:   Each trial is independent of the other trials if the outcome of one trial does not affect the outcome of any of the other trials.
RATIONALE:   This is the definition of independence: the events do not influence one another. The probability of one event is not changed by the existence of another event. Each trial is independent of the other trials if the outcome of one trial does not affect the outcome of any of the other trials.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 181
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

56. What type of distribution models the number of occurrences of an event over a specified interval of time or space?

ANSWER:   A Poisson Distribution

RATIONALE:   A random variable follows a Poisson probability distribution if (1) the probability of an occurrence is the same for any two intervals (of time or space) of equal length; and (2) the occurrence or nonoccurrence in any interval (of time or space) is independent of the occurrence or nonoccurrence in any other interval.

 

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 195
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Knowledge

 

57. Let X be a random variable with a Uniform distribution between 8 and 20. Find the probability that X is less than 10?

ANSWER:   2/12 = 0.1667

RATIONALE:   A continuous random variable with a uniform distribution takes the shape of a rectangle.  The base of the rectangle runs from 8 to 20.  The height of the rectangle is

To calculate the probability that X is less than 10, calculate the area of the rectangle with a base that runs from 8 to 10 (base = 2) and a height of 1/12.  P(X < 10) = (2)(1/12) = 2/12 = 0.1667.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 199
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

58. Could this curve represent a normal distribution?

ANSWER:   No

RATIONALE:   The normal distribution is symmetric, whereas this distribution is heavily right skewed.
POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 203
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

59. You recently took a standardized test in which scores follow a normal distribution with a mean of 18 and a standard deviation of 3. You were told that your score is at the 75th percentile of this distribution. What is your score?

ANSWER:   Your score  is 20.

RATIONALE:   If your score is at the 75th percentile, then you have a higher score than 75% of the test takers.  Your score can be calculated using the Excel function NORM.INV(0.75,18,3) = 20.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 207
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

60. The time in seconds that it takes a production worker to inspect an item has an exponential distribution with mean 15 seconds.  What proportion of inspection times is less than 10 seconds?

ANSWER:   0.4866

RATIONALE:   For exponential distributions the cumulative probability can be calculated using the formula

​​

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 209
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

61. The random variable X is normally distributed with mean of 80 and standard deviation of 10. What is the probability that a value of X chosen at random will be between 70 and 90?

ANSWER:   P(70 < x < 90) =0.683

RATIONALE:   In a normal distribution, 68.3% of the values are within plus or minus one standard deviation of the mean.

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   CONTINUOUS PROBABILITY DISTRIBUTIONS, Page 205
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

62. Reviews of call center representatives over the last 3 years showed that 10% of all call center representatives were rated as outstanding, 75% were rated as excellent/good, 10% percent were rated as satisfactory, and 5% were considered unsatisfactory.  For a sample of 10 reps selected at random, what is the probability that two will be rated as unsatisfactory?

ANSWER:   0.0746

RATIONALE:   For n = 10 and p = 0.05 and x = 2

POINTS:   1
DIFFICULTY:   Moderate
REFERENCES:   DISCRETE PROBABILITY DISTRIBUTIONS, Page 193
NATIONAL STANDARDS:   United States – BUSPROG: Analytic skills – and DISC: Descriptive Statistics
KEYWORDS:   Blooms: Application

 

 

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