**INSTANT DOWNLOAD WITH ANSWERS**

#### BSTAT 1st Edition by Keller – Test Bank

**CHAPTER 6: RANDOM VARIABLES AND DISCRETE PROBABILITY DISTRIBUTIONS** **TRUE/FALSE**

- The time required to drive from New York to New Mexico is a discrete random variable.

ANS: F NAT: Analytic; Probability Distributions

- A random variable is a function or rule that assigns a number to each outcome of an experiment.

ANS: T NAT: Analytic; Probability Distributions

- The number of home insurance policy holders is an example of a discrete random variable

ANS: T NAT: Analytic; Probability Distributions

- The mean of a discrete probability distribution for
*X*is the sum of all possible values of*X*, divided by the number of possible values of*X*.

ANS: F NAT: Analytic; Probability Distributions

- The length of time for which an apartment in a large complex remains vacant is a discrete random variable.

ANS: F NAT: Analytic; Probability Distributions

- The number of homeless people in Boston is an example of a discrete random variable.

ANS: T NAT: Analytic; Probability Distributions

- A continuous variable may take on any value within its relevant range even though the measurement device may not be precise enough to record it.

ANS: T NAT: Analytic; Probability Distributions

- Given that
*X*is a discrete random variable, then the laws of expected value and variance can be applied to show that*E*(*X*+ 5) =*E*(*X*) + 5, and*V*(*X*+ 5) =*V*(*X*) + 25.

ANS: F NAT: Analytic; Probability Distributions

- A table, formula, or graph that shows all possible values a random variable can assume, together with their associated probabilities, is referred to as discrete probability distribution.

ANS: T NAT: Analytic; Probability Distributions

- Faculty rank (professor, associate professor, assistant professor, and lecturer) is an example of a discrete random variable.

ANS: F NAT: Analytic; Probability Distributions

- For a random variable
*X*, if*V*(*cX*) = 4*V*(*X*), where*V*refers to the variance, then*c*must be 2.

ANS: T NAT: Analytic; Probability Distributions

- The amount of milk consumed by a baby in a day is an example of a discrete random variable.

ANS: F NAT: Analytic; Probability Distributions

- Another name for the mean of a probability distribution is its expected value.

ANS: T NAT: Analytic; Probability Distributions

- For a random variable
*X*,*E*(*X*+ 2) – 5 =*E*(*X*) – 3, where*E*refers to the expected value.

ANS: T NAT: Analytic; Probability Distributions

- For a random variable
*X*,*V*(*X*+ 3) =*V*(*X*+ 6), where*V*refers to the variance.

ANS: T NAT: Analytic; Probability Distributions

- The binomial random variable is the number of successes that occur in a fixed period of time.

ANS: F NAT: Analytic; Probability Distributions

- The binomial probability distribution is a discrete probability distribution.

ANS: T NAT: Analytic; Probability Distributions

- The binomial distribution deals with consecutive trials, each of which has two possible outcomes.

ANS: T NAT: Analytic; Probability Distributions

- The binomial distribution deals with consecutive trials, each of which has two possible outcomes.

ANS: T NAT: Analytic; Probability Distributions

- The variance of a binomial distribution for which
*n*= 50 and*p*= 0.20 is 8.0.

ANS: T NAT: Analytic; Probability Distributions

- The expected number of heads in 250 tosses of an unbiased coin is 125.

ANS: T NAT: Analytic; Probability Distributions

- If
*X*is a binomial random variable with*n*= 25, and*p*= 0.25, then*P*(*X*= 25) = 1.0.

ANS: F NAT: Analytic; Probability Distributions

- The standard deviation of a binomial random variable
*X*is given by the formula*s*^{2}=*np*(1 –*p*), where*n*is the number of trials, and*p*is the probability of success.

ANS: F NAT: Analytic; Probability Distributions

- The number of female customers out of a random sample of 100 customers arriving at a department store has a binomial distribution.

ANS: T NAT: Analytic; Probability Distributions

- If the probability of success
*p*remains constant in a binomial distribution, an increase in*n*will increase the variance.

ANS: T NAT: Analytic; Probability Distributions

- If the probability of success
*p*remains constant in a binomial distribution, an increase in*n*will not change the mean.

ANS: F NAT: Analytic; Probability Distributions

- The Poisson probability distribution is a continuous probability distribution.

ANS: F NAT: Analytic; Probability Distributions

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

ANS: T NAT: Analytic; Probability Distributions

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

ANS: T NAT: Analytic; Probability Distributions

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

ANS: T NAT: Analytic; Probability Distributions

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

ANS: T NAT: Analytic; Probability Distributions

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

ANS: T NAT: Analytic; Probability Distributions

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

ANS: F NAT: Analytic; Probability Distributions

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

ANS: F NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

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

ANS: F NAT: Analytic; Probability Distributions

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

ANS: F NAT: Analytic; Probability Distributions **MULTIPLE CHOICE**

- A table, formula, or graph that shows all possible values a random variable can assume, together with their associated probabilities, is called a(n):

a. | discrete probability distribution. |

b. | discrete random variable. |

c. | expected value of a discrete random variable. |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- A function or rule that assigns a numerical value to each simple event of an experiment is called:

a. | a sample space. |

b. | a probability distribution. |

c. | a random variable. |

d. |
None of these choices. |

ANS: C NAT: Analytic; Probability Distributions

- The weighted average of the possible values that a random variable
*X*can assume, where the weights are the probabilities of occurrence of those values, is referred to as the:

a. | variance. |

b. | standard deviation. |

c. | expected value. |

d. |
None of these choices. |

ANS: C NAT: Analytic; Probability Distributions

- The number of accidents that occur annually on a busy stretch of highway is an example of:

a. | a discrete random variable. |

b. | a continuous random variable. |

c. | expected value of a discrete random variable. |

d. | expected value of a continuous random variable. |

ANS: A NAT: Analytic; Probability Distributions

- Which of the following are required conditions for the distribution of a discrete random variable
*X*that can assume values*x*?_{i}

a. | 0 £ p(x) £ 1 for all _{i}x_{i} |

b. | |

c. |
Both a and b are required conditions. |

d. |
Neither a nor b are required conditions. |

ANS: C NAT: Analytic; Probability Distributions

- Which of the following is not a required condition for the distribution of a discrete random variable
*X*that can assume values*x*?_{i}

a. | 0 £ p(x) £ 1 for all _{i}x_{i} |

b. | |

c. | p(x) > 1 for all _{i}x_{i} |

d. |
All of these choices are true. |

ANS: C NAT: Analytic; Probability Distributions

- A lab at the DeBakey Institute orders 150 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Suppose the mean cost of rats used in lab experiments turned out to be $20.00 per week. Interpret this value.

a. | Most of the weeks resulted in rat costs of $20.00 |

b. | The median cost for the distribution of rat costs is $20.00 |

c. | The expected or average costs for all weekly rat purchases is $20.00 |

d. | The rat cost that occurs more often than any other is $20.00 |

ANS: C NAT: Analytic; Probability Distributions

- In the notation below,
*X*is the random variable,*c*is a constant, and*V*refers to the variance. Which of the following laws of variance is not true?

a. | V(c) = 0 |

b. | V(X + c) = V(X) + c |

c. | V(cX) = c^{2} V(X) |

d. |
None of these choices. |

ANS: B NAT: Analytic; Probability Distributions

- Which of the following is a discrete random variable?

a. | The Dow Jones Industrial average. |

b. | The volume of water in Michigan Lakes. |

c. | The time it takes you to drive to school. |

d. | The number of employees of a soft drink company. |

ANS: D NAT: Analytic; Probability Distributions

- Which of the following is a continuous random variable?

a. | The number of employees of an automobile company. |

b. | The amount of milk produced by a cow in one 24-hour period. |

c. | The number of gallons of milk sold at Albertson’s grocery store last week. |

d. |
None of these choices. |

ANS: B NAT: Analytic; Probability Distributions

- In the notation below,
*X*is the random variable,*E*and*V*refer to the expected value and variance, respectively. Which of the following is false?

a. | E(3X) = 3E(X) |

b. | V(2) = 0 |

c. | E(X + 1) = E(X) + 1 |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- Which of the following about the binomial distribution is not a true statement?

a. | The probability of success must be constant from trial to trial. |

b. | The random variable of interest is continuous. |

c. | Each outcome may be classified as either “success” or “failure”. |

d. | Each outcome is independent of the other. |

ANS: B NAT: Analytic; Probability Distributions

- The expected number of heads in 100 tosses of an unbiased coin is

a. | 25 |

b. | 50 |

c. | 75 |

d. | 100 |

ANS: C NAT: Analytic; Probability Distributions

- Which of the following is not a characteristic of a binomial experiment?

a. | Each trial results in two or more outcomes. |

b. | There is a sequence of identical trials. |

c. | The trials are independent of each other. |

d. | The probability of success p is the same from one trial to another. |

ANS: A NAT: Analytic; Probability Distributions

- The variance of a binomial distribution for which
*n*= 100 and*p*= 0.20 is:

a. | 100 |

b. | 80 |

c. | 20 |

d. | 16 |

ANS: D NAT: Analytic; Probability Distributions

- If
*n*= 10 and*p*= 0.60, then the mean of the binomial distribution is

a. | 0.06 |

b. | 2.65 |

c. | 6.00 |

d. | 5.76 |

ANS: C NAT: Analytic; Probability Distributions

- If
*n*= 20 and*p*= 0.70, then the standard deviation of the binomial distribution is

a. | 0.14 |

b. | 2.05 |

c. | 14.0 |

d. | 14.7 |

ANS: B NAT: Analytic; Probability Distributions

- The expected value,
*E*(*X*), of a binomial probability distribution with*n*trials and probability*p*of success is:

a. | n + p |

b. | np(1 – p) |

c. | np |

d. | n + p – 1 |

ANS: C NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions **COMPLETION**

- A motorcycle insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for motorcycle insurance. How long a person has been a licensed rider is an example of a(n) ____________________ random variable.

ANS: continuous NAT: Analytic; Probability Distributions

- An auto insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of claims a person has made in the last 3 years is an example of a(n) ____________________ random variable.

ANS: discrete NAT: Analytic; Probability Distributions

- An auto insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A person’s age is an example of a(n) ____________________ random variable.

ANS: continuous NAT: Analytic; Probability Distributions

- A motorcycle insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for motorcycle insurance. The number of tickets a person has received in the last 3 years is an example of a(n) ____________________ random variable.

ANS: discrete NAT: Analytic; Probability Distributions

- A motorcycle insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for motorcycle insurance. The distance a person rides in a year is an example of a(n) ____________________ random variable.

ANS: continuous NAT: Analytic; Probability Distributions

- The dean of students conducted a survey on campus. Grade point average (GPA) is an example of a(n) ____________________ random variable.

ANS: continuous NAT: Analytic; Probability Distributions

- The amount of time that a microcomputer is used per week is an example of a(n) ____________________ random variable.

ANS: continuous NAT: Analytic; Probability Distributions

- The number of days that a microcomputer goes without a breakdown is an example of a(n) ____________________ random variable.

ANS: discrete NAT: Analytic; Probability Distributions

- A(n) ____________________ random variable is one whose values are uncountable.

ANS: continuous NAT: Analytic; Probability Distributions

- A(n) ____________________ random variable is one whose values are countable.

ANS: discrete NAT: Analytic; Probability Distributions

- A binomial experiment consists of a(n) ____________________ number of trials,
*n*.

ANS: fixed NAT: Analytic; Probability Distributions

- In each trial of a binomial experiment, there are ____________________ possible outcomes.

ANS:two2 NAT: Analytic; Probability Distributions

- The probability of a success in a binomial experiment is denoted by ____________________.

ANS: *p* NAT: Analytic; Probability Distributions

- The probability of a failure in a binomial experiment is denoted by ____________________.

ANS: *1* – *p* NAT: Analytic; Probability Distributions

- The trials in a binomial experiment are ____________________, meaning the outcome of one trial does not affect the outcomes of any other trials.

ANS: independent NAT: Analytic; Probability Distributions

- The mean of a binomial distribution is equal to ____________________.

ANS: *np* NAT: Analytic; Probability Distributions

- The variance of a binomial distribution is equal to ____________________.

ANS: *np(1* – *p)* NAT: Analytic; Probability Distributions

- The probability
*P*(*X*£*x*) is called a(n) ____________________ probability. The binomial table reports these probabilities.

ANS: cumulative NAT: Analytic; Probability Distributions

- To find the probability that
*X*is at least 10, you should find the probability that*X*is 10 or ____________________.

ANS: more NAT: Analytic; Probability Distributions

- To find the probability that
*X*is at most 10, you should find the probability that*X*is 10 or ____________________.

ANS: less NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

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

ANS: Poisson NAT: Analytic; Probability Distributions

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

ANS: proportional NAT: Analytic; Probability Distributions

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

ANS:zero0 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions

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

ANS:meanexpected value NAT: Analytic; Probability Distributions

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

ANS: variance NAT: Analytic; Probability Distributions

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

ANS: mean NAT: Analytic; Probability Distributions

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

ANS:zero0 NAT: Analytic; Probability Distributions

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

ANS: discrete NAT: Analytic; Probability Distributions **SHORT ANSWER**

- For each of the following random variables, indicate whether the variable is discrete or continuous, and specify the possible values that it can assume.

a. | X = the number of traffic accidents in Albuquerque on a given day. |

b. | X = the amount of weight lost in a month by a randomly selected dieter. |

c. | X = the average number of children per family in a random sample of 175 families. |

d. | X = the number of households out of 10 surveyed that own a convection oven. |

e. | X = the time in minutes required to obtain service in a restaurant. |

ANS:

a. | discrete; x = 0, 1, 2, 3, . . . |

b. | continuous; -¥ < x < ¥ |

c. | continuous; x ³ 0 |

d. | discrete; x = 0, 1, 2, . . . , 10 |

e. | continuous; x > 0 |

NAT: Analytic; Probability Distributions NARRBEGIN: Number of Motorcycles**Number of Motorcycles** The probability distribution of a discrete random variable *X* is shown below, where *X* represents the number of motorcycles owned by a family.

x |
0 | 1 | 2 | 3 |

p(x) |
0.25 | 0.40 | 0.20 | 0.15 |

NARREND

- {Number of Motorcycles Narrative} Find the following probabilities:

a. | P(X > 1) |

b. | P(X £ 2) |

c. | P(1 £ X £ 2) |

d. | P(0 < X < 1) |

e. | P(1 £ X < 3) |

ANS:

a. | 0.35 |

b. | 0.85 |

c. | 0.60 |

d. | 0.00 |

e. | 0.60 |

NAT: Analytic; Probability Distributions

- {Number of Motorcycles Narrative} Find the expected value of
*X*.

ANS:*E*(*X*) = 1.25 cars NAT: Analytic; Probability Distributions

- {Number of Motorcycles Narrative} Find the standard deviation of
*X*.

ANS:*s* = 0.9937 cars NAT: Analytic; Probability Distributions

- {Number of Motorcycles Narrative} Apply the laws of expected value to find the following:

a. | E(X^{2}) |

b. | E(2X^{2} + 5) |

c. | E(X – 2)^{2} |

ANS:

a. | 2.55 |

b. | 10.1 |

c. | 1.55 |

NAT: Analytic; Probability Distributions

- {Number of Motorcycles Narrative} Apply the laws of expected value and variance to find the following:

a. | V(3X) |

b. | V(3X – 2) |

c. | V(3) |

d. | V(3X) – 2 |

ANS:

a. | 8.89 |

b. | 8.89 |

c. | 0 |

d. | 8.89 |

NAT: Analytic; Probability Distributions NARRBEGIN: Number of Horses**Number of Horses** The random variable *X* represents the number of horses per family in a rural area in Iowa, with the probability distribution: *p*(*x*) = 0.05*x*, *x* = 2, 3, 4, 5, or 6.NARREND

- {Number of Horses Narrative} Express the probability distribution in tabular form.

ANS:

x |
2 | 3 | 4 | 5 | 6 |

p(x) |
0.10 | 0.15 | 0.20 | 0.25 | 0.30 |

NAT: Analytic; Probability Distributions

- {Number of Horses Narrative} Find the expected number of horses per family.

ANS:*E*(*X*) = 4.5 NAT: Analytic; Probability Distributions

- {Number of Horses Narrative} Find the variance and standard deviation of
*X*.

ANS:*s*^{2} = 1.75, and *s* = 1.323 NAT: Analytic; Probability Distributions

- {Number of Horses Narrative} Find the following probabilities:

a. | P(X ³ 4) |

b. | P(X > 4) |

c. | P(3 £ X £ 5) |

d. | P(2 < X < 4) |

e. | P(X = 4.5) |

ANS:

a. | 0.75 |

b. | 0.55 |

c. | 0.60 |

d. | 0.15 |

e. | 0.00 |

NAT: Analytic; Probability Distributions

- Determine which of the following are not valid probability distributions, and explain why not.

a. | x |
0 | 1 | 2 | 3 | |

p(x) |
0.15 | 0.25 | 0.35 | 0.45 | ||

b. | x |
2 | 3 | 4 | 5 | |

p(x) |
-0.10 | 0.40 | 0.50 | 0.25 | ||

c. | x |
-2 | -1 | 0 | 1 | 2 |

p(x) |
0.10 | 0.20 | 0.40 | 0.20 | 0.10 |

ANS:

a. | This is not a valid probability distribution because the probabilities don’t sum to one. |

b. | This is not a valid probability distribution because it contains a negative probability. |

c. | This is a valid probability distribution. |

NAT: Analytic; Probability Distributions NARRBEGIN: Blackjack**Blackjack** The probability distribution of a random variable *X* is shown below, where *X* represents the amount of money (in $1,000s) gained or lost in a particular game of Blackjack.

x |
-4 | 0 | 4 | 8 |

p(x) |
0.15 | 0.25 | 0.20 | 0.40 |

NARREND

- {Blackjack Narrative} Find the following probabilities:

a. | P(X £ 0) |

b. | P(X > 3) |

c. | P(0 £ X £ 4) |

d. | P(X = 5) |

ANS:

a. | 0.40 |

b. | 0.60 |

c. | 0.45 |

d. | 0.00 |

NAT: Analytic; Probability Distributions

- {Blackjack Narrative} Find the following values, and indicate their units.

a. | E(X) |

b. | V(X) |

c. | Standard deviation of X |

ANS:

a. | $3.40 |

b. | 19.64 (dollars squared) |

c. | $4.43 |

NAT: Analytic; Probability Distributions NARRBEGIN: Gym Visits**Gym Visits** Let *X* represent the number of times a student visits a gym in a one month period. Assume that the probability distribution of *X* is as follows:

x |
0 | 1 | 2 | 3 |

p(x) |
0.05 | 0.25 | 0.50 | 0.20 |

NARREND

- {Gym Visits Narrative} Find the mean
*m*and the standard deviation*s*of this distribution.

ANS:*m** _{x}* =1.85, and

*s*

*= 0.792 NAT: Analytic; Probability Distributions*

_{x}- {Gym Visits Narrative} Find the mean and the standard deviation of
*Y*= 2*X*– 1.

ANS:*m** _{y}* = 2.70, and

*s*

*= 1.584 NAT: Analytic; Probability Distributions*

_{y}- {Gym Visits Narrative} What is the probability that the student visits the gym at least once in a month?

ANS:*P*(1) + *P*(2) + *P*(3) = 0.95 NAT: Analytic; Probability Distributions

- {Gym Visits Narrative} What is the probability that the student visits the gym at most twice in a month?

ANS:*P*(0) + *P*(1) + *P*(2) = 0.80 NAT: Analytic; Probability Distributions

- The monthly sales at a Gas Station have a mean of $50,000 and a standard deviation of $6,000. Profits are calculated by multiplying sales by 40% and subtracting fixed costs of $12,000. Find the mean and standard deviation of monthly profits.

ANS:Let *P* = profit, and *X* = sales. Then *P* = 0.40*X* – 12,000.*E*(*P*) = *E*(0.40*X* – 12,000) = 0.40 *E*(*X*) – 12,000 = 0.40($50,000) – $12,000 = $8,000*V*(*P*) = *V*(0.40*X* – 12,000) = (0.40)^{2} *V*(*X*) = (0.40)^{2} (6,000)^{2} = 5,760,000.Thus, the mean and standard deviation of monthly profits are $8,000 and $2,400, respectively. NAT: Analytic; Probability Distributions NARRBEGIN: Shopping Outlet**Shopping Outlet** A shopping outlet estimates the probability distribution of the number of stores shoppers actually enter as shown in the table below.

x |
0 | 1 | 2 | 3 | 4 |

p(x) |
0.05 | 0.35 | 0.25 | 0.20 | 0.15 |

NARREND

- {Shopping Outlet Narrative} Find the expected value of the number of stores entered.

ANS:*E*(*X*) = 2.05 NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Find the variance and standard deviation of the number of stores entered.

ANS: NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Suppose
*Y*= 2*X*+ 1 for each value of*X*. What is the probability distribution of*Y*?

ANS:

y |
1 | 3 | 5 | 7 | 9 |

P(y) |
0.05 | 0.35 | 0.25 | 0.20 | 0.15 |

NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Calculate the expected value of
*Y*directly from the probability distribution of*Y*.

ANS:*E*(*Y*) = 5.10 NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Use the laws of expected value to calculate the mean of
*Y*from the probability distribution of*X*.

ANS:*E*(*Y*) = *E*(2*X* + 1) = 2*E*(*X*) + 1 = 2(2.05) + 1 = 5.10 NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Calculate the variance and standard deviation of
*Y*directly from the probability distribution of*Y*.

ANS: NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} Use the laws of variance to calculate the variance and standard deviation of
*Y*from the probability distribution of*X*.

ANS:*V*(*Y*) = *V*(2*X* + 1) = 4*V*(*X*) = 4(1.3475) = 5.39; *SD*(*Y*) = Ö*V*(*X*) = Ö5.39 = 2.32 NAT: Analytic; Probability Distributions

- {Shopping Outlet Narrative} What did you notice about the mean, variance, and standard deviation of
*Y*= 2*X*+ 1 in terms of the mean, variance, and standard deviation of*X*?

ANS:*E*(*Y*) = 2*E*(*X*) + 1, *V*(*Y*) = 4*V*(*X*), and *SD(Y)* = 2*SD(X).* NAT: Analytic; Probability Distributions NARRBEGIN: Retries**Retries** The following table contains the probability distribution for *X* = the number of retries necessary to successfully transmit a 1024K data package through a double satellite media.

x |
0 | 1 | 2 | 3 |

p(x) |
0.35 | 0.35 | 0.25 | 0.05 |

NARREND

- {Retries Narrative} What is the probability of no retries?

ANS:*p*(0) = 0.35 NAT: Analytic; Probability Distributions

- {Retries Narrative} What is the probability of a least one retry?

ANS:*p*(1) + *p*(2) + *p*(3) = 0.65 NAT: Analytic; Probability Distributions

- {Retries Narrative} What is the mean or expected value for the number of retries?

ANS:*E*(*X*) = 1.0 NAT: Analytic; Probability Distributions

- {Retries Narrative} What is the variance for the number of retries?

ANS:*V*(*X*) = 0.80 NAT: Analytic; Probability Distributions

- {Retries Narrative} What is the standard deviation of the number of retries?

ANS:*s* = 0.894 NAT: Analytic; Probability Distributions

- Evaluate the following binomial coefficients.

a. | |

b. | |

c. | |

d. |

ANS:

a. | 15 |

b. | 20 |

c. | 35 |

d. | 35 |

NAT: Analytic; Probability Distributions NARRBEGIN: Stress**Stress** Consider a binomial random variable *X* with *n* = 5 and *p* = *0*. 40, where *X* represents the number of times in the final exam week a student with 18 credit hours may feel stressed.NARREND

- {Stress Narrative} Find the probability distribution of
*X*.

ANS:

x |
0 | 1 | 2 | 3 | 4 | 5 |

p(x) |
.0778 | .2592 | .3456 | .2304 | .0768 | .0102 |

NAT: Analytic; Probability Distributions

- {Stress Narrative} Find
*P*(*X*< 3).

ANS:0.6826 NAT: Analytic; Probability Distributions

- {Stress Narrative} Find
*P*(2 £*X*£ 4).

ANS:0.6528 NAT: Analytic; Probability Distributions

- {Stress Narrative} Find the expected number of times a student may feel stressed during the final exam week.

ANS:*E*(*X*) = 2 NAT: Analytic; Probability Distributions

- {Stress Narrative} Find the variance and standard deviation.

ANS:*s*^{2} = 1.2 and *s* = 1.095 NAT: Analytic; Probability Distributions

- Given a binomial random variable with
*n*= 20 and*p*= 0.60, find the following probabilities using the binomial table.

a. | P(X £ 13) |

b. | P(X ³ 15) |

c. | P(X = 17) |

d. | P(11 £ X £ 14) |

e. | P(11 < X < 14) |

ANS:

a. | 0.75 |

b. | 0.126 |

c. | 0.012 |

d. | 0.629 |

e. | 0.346 |

NAT: Analytic; Probability Distributions NARRBEGIN: Montana Highways**Montana Highways** A recent survey in Montana revealed that 60% of the vehicles traveling on highways, where speed limits are posted at 70 miles per hour, were exceeding the limit. Suppose you randomly record the speeds of ten vehicles traveling on US 131 where the speed limit is 70 miles per hour. Let *X *denote the number of vehicles that were exceeding the limit.NARREND

- {Montana Highways Narrative} What is the distribution of
*X*?

ANS:*X* is a binomial random variable with *n* = 10 and *p* = 0.60. NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find
*P*(*X*= 10).

ANS:0.006 NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find
*P*(4 <*X*< 9).

ANS:0.788 NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find
*P*(*X*= 2).

ANS:0.01 NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find
*P*(3 £*X*£ 6).

ANS:0.606 NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find the expected number of vehicles that are traveling on Montana highways and exceeding the speed limit.

ANS:*E*(*X*) = 6 NAT: Analytic; Probability Distributions

- {Montana Highways Narrative} Find the standard deviation of number of vehicles that are traveling on Montana highways and exceeding the speed limit.

ANS:*s* = 1.549 NAT: Analytic; Probability Distributions NARRBEGIN: Online Bankers**Online Bankers** An official from the securities commission estimates that 75% of all online bankers have profited from the use of insider information. Assume that 15 online bankers are selected at random from the commission’s registry.NARREND

- {Online Bankers Narrative} Find the probability that at most 10 have profited from insider information.

ANS:0.314 NAT: Analytic; Probability Distributions

- {Online Bankers Narrative} Find the probability that at least 6 have profited from insider information.

ANS:0.999 NAT: Analytic; Probability Distributions

- {Online Bankers Narrative} Find the probability that all 15 have profited from insider information.

ANS:0.013 NAT: Analytic; Probability Distributions

- {Online Bankers Narrative} What is the expected number of Online bankers who have profited from the use of insider information?

ANS:*E*(*X*) = 11.25 NAT: Analytic; Probability Distributions

- {Online Bankers Narrative} Find the variance and standard deviation of the number of Online bankers who have profited from the use of insider information.

ANS:*s*^{2} = 2.8125, and *s* = 1.677 NAT: Analytic; Probability Distributions

- Let
*X*be a binomial random variable with*n*= 25 and*p*= 0.01.

a. | Use the binomial table to find P(X = 0), P(X = 1), and P(X = 2). |

b. | Find the variance and standard deviation of X. |

ANS:

a. | P(X = 0) = 0.778, P(X = 1) = 0.196, and P(X = 2) = 0.024 |

b. | s^{2} = np(1 – p) = 25(0.01)(0.99) = 0.2475, and s = 0.4975 |

NAT: Analytic; Probability Distributions

- A remedial program evenly enrolls tradition and non-traditional students. If a random sample of 4 students is selected from the program to be interviewed about the introduction of a new on-line class, what is the probability that all 4 students selected are traditional students?

ANS:.0625 NAT: Analytic; Probability Distributions

- If
*X*has a binomial distribution with*n*= 4 and*p**=*0.3, find*P*(*X*= 0).

ANS:0.4116 NAT: Analytic; Probability Distributions

- If
*X*has a binomial distribution with*n*= 4 and*p*= 0.3, find*P*(*X*> 1).

ANS:0.3483 NAT: Analytic; Probability Distributions

- If
*X*has a binomial distribution with*n*= 4 and*p*= 0.3, find the probability that*X*is at most one.

ANS:.6517 NAT: Analytic; Probability Distributions

- If
*X*has a binomial distribution with*n*= 4 and*p*= 0.3, find the probability that*X*is at least one.

ANS:.7599 NAT: Analytic; Probability Distributions NARRBEGIN: Sports Fans**Sports Fans** Suppose that past history shows that 5% of college students are sports fans. A sample of 10 students is to be selected.NARREND

- {Sports Fans Narrative} Find the probability that exactly 1 student is a sports fan.

ANS:0.315 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} Find the probability that at least 1 student is a sports fan.

ANS:0.401 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} Find the probability that less than 1 student is a sports fan.

ANS:0.599 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} Find the probability that at most 1 student is a sports fan.

ANS:0.914 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} Find the probability that more than 1 student is a sports fan.

ANS:0.086 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} A sample of 100 students is to be selected. What is the average number that you would expect to sports fan?

ANS:5 NAT: Analytic; Probability Distributions

- {Sports Fans Narrative} A sample of 100 students is to be selected. What is the standard deviation of the number of sports fans you expect?

ANS:2.18 NAT: Analytic; Probability Distributions

- 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 |

NAT: Analytic; Probability Distributions

- 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. | 0.847 |

b. | 0.103 |

c. | 0.715 |

d. | 0.511 |

NAT: Analytic; Probability Distributions

- 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. | 0.313 |

b. | 0.058 |

c. | 0.986 |

d. | 0.390 |

NAT: Analytic; Probability Distributions NARRBEGIN: 911 Phone Calls**911 Phone Calls** 911 phone calls arrive at the rate of 30 per hour at the local call center.NARREND

- {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 NAT: Analytic; Probability Distributions

- {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 NAT: Analytic; Probability Distributions

- {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 NAT: Analytic; Probability Distributions NARRBEGIN: Classified Department Pho**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.NARREND

- {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) = 0.072 NAT: Analytic; Probability Distributions

- {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 * = 5; *P*(*X* = 7) = 0.105 NAT: Analytic; Probability Distributions

- {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) = 0.971 NAT: Analytic; Probability Distributions NARRBEGIN: Post Office**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.NARREND

- {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) = 0.758 NAT: Analytic; Probability Distributions

- {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) = 0.001 NAT: Analytic; Probability Distributions

- {{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) = 0.045 NAT: Analytic; Probability Distributions

- 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 NAT: Analytic; Probability Distributions NARRBEGIN: Unsafe Levels of Radioact** ****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.NARREND

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

ANS:0.0892 NAT: Analytic; Probability Distributions

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

ANS:0.9380 NAT: Analytic; Probability Distributions

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

ANS:0.9975 NAT: Analytic; Probability Distributions

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

ANS:0.0174 NAT: Analytic; Probability Distributions

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

ANS:6 NAT: Analytic; Probability Distributions

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

ANS:2.45 NAT: Analytic; Probability Distributions**CHAPTER 7: CONTINUOUS PROBABILITY DISTRIBUTIONS** **TRUE/FALSE**

- Since there is an infinite number of values a continuous random variable can assume, the probability of each individual value is virtually 0.

ANS: T NAT: Analytic; Probability Distributions

- A continuous probability distribution represents a random variable having an infinite number of outcomes which may assume any number of values within an interval.

ANS: T NAT: Analytic; Probability Distributions

- Continuous probability distributions describe probabilities associated with random variables that are able to assume any finite number of values along an interval.

ANS: F NAT: Analytic; Probability Distributions

- A continuous random variable
*X*has a uniform distribution between 10 and 20 (inclusive), then the probability that*X*falls between 12 and 15 is 0.30.

ANS: T NAT: Analytic; Probability Distributions

- A continuous random variable is one that can assume an uncountable number of values.

ANS: T NAT: Analytic; Probability Distributions

- A continuous random variable
*X*has a uniform distribution between 5 and 15 (inclusive), then the probability that*X*falls between 10 and 20 is 1.0.

ANS: F NAT: Analytic; Probability Distributions

- A continuous random variable
*X*has a uniform distribution between 5 and 25 (inclusive), then*P*(*X*= 15) = 0.05.

ANS: F NAT: Analytic; Probability Distributions

- We distinguish between discrete and continuous random variables by noting whether the number of possible values is countable or uncountable.

ANS: T NAT: Analytic; Probability Distributions

- In practice, we frequently use a continuous distribution to approximate a discrete one when the number of values the variable can assume is countable but very large.

ANS: T NAT: Analytic; Probability Distributions

- Let
*X*represent weekly income expressed in dollars. Since there is no set upper limit, we cannot identify (and thus cannot count) all the possible values. Consequently, weekly income is regarded as a continuous random variable.

ANS: T NAT: Analytic; Probability Distributions

- To be a legitimate probability density function, all possible values of
*f(x)*must be non-negative.

ANS: T NAT: Analytic; Probability Distributions

- To be a legitimate probability density function, all possible values of
*f(x)*must lie between 0 and 1 (inclusive).

ANS: F NAT: Analytic; Probability Distributions

- The sum of all values of
*f(x)*over the range of [*a*,*b*] must equal one.

ANS: F NAT: Analytic; Probability Distributions

- A probability density function shows the probability for each value of
*X*.

ANS: F NAT: Analytic; Probability Distributions

- If
*X*is a continuous random variable on the interval [0, 10], then*P*(*X*> 5) =*P*(*X*³ 5).

ANS: T NAT: Analytic; Probability Distributions

- If
*X*is a continuous random variable on the interval [0, 10], then*P*(*X*= 5) =*f*(5) = 1/10.

ANS: F NAT: Analytic; Probability Distributions

- If a point
*y*lies outside the range of the possible values of a random variable*X*, then*f(y)*must equal zero.

ANS: T NAT: Analytic; Probability Distributions

- A national standardized testing company can tell you your relative standing on an exam without divulging the mean or the standard deviation of the exam scores.

ANS: T NAT: Analytic; Probability Distributions

- If your golf score is 3 standard deviations below the mean, its corresponding value on the
*Z*distribution is -3.

ANS: T NAT: Analytic; Probability Distributions

- If we standardize the normal curve, we express the original
*X*values in terms of their number of standard deviations away from the mean.

ANS: T NAT: Analytic; Probability Distributions

- A normal distribution is symmetric; therefore the probability of being below the mean is 0.50 and the probability of being above the mean is 0.50.

ANS: T NAT: Analytic; Probability Distributions

- A random variable
*X*is standardized by subtracting the mean and dividing by the variance.

ANS: F NAT: Analytic; Probability Distributions

- A random variable
*X*has a normal distribution with mean 132 and variance 36. If*x*= 120, its corresponding value of*Z*is 2.0.

ANS: F NAT: Analytic; Probability Distributions

- A random variable
*X*has a normal distribution with a mean of 250 and a standard deviation of 50. Given that*X*= 175, its corresponding value of*Z*is -1.50.

ANS: T NAT: Analytic; Probability Distributions

- Given that
*Z*is a standard normal random variable, a negative value of*Z*indicates that the standard deviation of*Z*is negative.

ANS: F NAT: Analytic; Probability Distributions

- In the standard normal distribution,
*z*_{0.05}= 1.645 means that 5% of all values of*z*are below 1.645 and 95% are above it.

ANS: F NAT: Analytic; Probability Distributions

- The probability that a standard normal random variable
*Z*is less than -3.5 is approximately 0.

ANS: T NAT: Analytic; Probability Distributions

- If the value of
*Z*is*z*= 99, that means you are at the 99^{th}percentile on the*Z*distribution.

ANS: F NAT: Analytic; Probability Distributions

- The 10
^{th}percentile of a*Z*distribution has 10% of the*Z*-values lying above it.

ANS: F NAT: Analytic; Probability Distributions

- The probability that
*Z*is less than -2 is the same as one minus the probability that*Z*is greater than +2.

ANS: F NAT: Analytic; Probability Distributions

- Suppose
*X*has a normal distribution with mean 70 and standard deviation 5. The 50^{th}percentile of*X*is 70.

ANS: T NAT: Analytic; Probability Distributions

- Like that of the Student
*t*distribution, the shape of the chi-squared distribution depends on its number of degrees of freedom.

ANS: T NAT: Analytic; Probability Distributions

- The value of
*c*^{2}with*v*degrees of freedom such that the area to its right under the chi-squared curve is equal to*A*is denoted by , while denotes the value such that the area to its left is*A*.

ANS: T NAT: Analytic; Probability Distributions

- The variance of a Student
*t*random variable with*v*degrees of freedom (*v*> 2) is always greater than 1.

ANS: T NAT: Analytic; Probability Distributions

- We define as the value of the
*F*with*v*_{1}and*v*_{2}degrees of freedom such that the area to its right under the*F*curve is*A*, while is defined as the value such that the area to its left is*A*.

ANS: T NAT: Analytic; Probability Distributions

- The value of
*A*such that*P*(-*A*£*t*£*A*) = 0.95, where the degrees of freedom are 20, is 2.086.

ANS: T NAT: Analytic; Probability Distributions

- The variance of a
*c*^{2}distribution is twice the value of its mean.

ANS: T NAT: Analytic; Probability Distributions

- As the degrees of freedom approach infinity, the values of a Student
*t*distribution approach those of a standard normal distribution.

ANS: T NAT: Analytic; Probability Distributions

- The value of an
*F*distribution with*v*_{1}= 5 and*v*_{2}= 10 degrees of freedom such that the area to its left is 0.95 is 3.33.

ANS: T NAT: Analytic; Probability Distributions

- The expected value of the Student
*t*distribution is zero.

ANS: T NAT: Analytic; Probability Distributions

- The variance of a Student
*t*distribution approaches zero as the degrees of freedom approaches infinity.

ANS: F NAT: Analytic; Probability Distributions

- The value of an
*F*distribution with*v*_{1}= 6 and*v*_{2}= 9 degrees of freedom such that the area to its right is 0.05 is 3.37.

ANS: T NAT: Analytic; Probability Distributions

- The value of a chi-squared distribution with 5 degrees of freedom such that the area to its left is 0.10 is 1.61.

ANS: T NAT: Analytic; Probability Distributions

- The Student
*t*distribution looks similar in shape to a standard normal distribution, except it is not as widely spread.

ANS: F NAT: Analytic; Probability Distributions

- The value of a chi-squared distribution with 8 degrees of freedom such that the area to its left is 0.95 is 2.73.

ANS: F NAT: Analytic; Probability Distributions **MULTIPLE CHOICE**

- Which of the following is always true for all probability density functions of continuous random variables?

a. | The probability at any single point is zero. |

b. | They contain an uncountable number of possible values. |

c. | The total area under the density function f(x) equals 1. |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- The probability density function,
*f(x)*, for any continuous random variable*X*, represents:

a. | all possible values that X will assume within some interval a £ x £ b. |

b. | the probability that X takes on a specific value x. |

c. | the height of the density function at x. |

d. |
None of these choices. |

ANS: C NAT: Analytic; Probability Distributions

- Which of the following represents a difference between continuous and discrete random variables?

a. | Continuous random variables assume an uncountable number of values, and discrete random variables do not. |

b. | The probability for any individual value of a continuous random variable is zero, but for discrete random variables it is not. |

c. | Probability for continuous random variables means finding the area under a curve, while for discrete random variables it means summing individual probabilities. |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- Suppose
*f(x)*= 0.25. What range of possible values can*X*take on and still have the density function be legitimate?

a. | [0, 4] |

b. | [4, 8] |

c. | [-2, +2] |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- What is the shape of the probability density function for a uniform random variable on the interval [
*a*,*b*]?

a. | A rectangle whose X values go from a to b. |

b. | A straight line whose height is 1/(b – a) over the range [a, b]. |

c. | A continuous probability density function with the same value of f(x) from a to b. |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- Which of the following is true about
*f(x)*when*X*has a uniform distribution over the interval [*a*,*b*]?

a. | The values of f(x) are different for various values of the random variable X. |

b. | f(x) equals one for each possible value of X. |

c. | f(x) equals one divided by the length of the interval from a to b. |

d. |
None of these choices. |

ANS: C NAT: Analytic; Probability Distributions

- Suppose
*f(x)*= 1/4 over the range*a*£*x*£*b*, and suppose*P*(*X*> 4) = 1/2. What are the values for*a*and*b*?

a. | 0 and 4 |

b. | 2 and 6 |

c. | Can be any range of x values whose length (b – a) equals 4. |

d. |
Cannot answer with the information given. |

ANS: B NAT: Analytic; Probability Distributions

- The probability density function
*f(x)*for a uniform random variable*X*defined over the interval [2, 10] is

a. | 0.20 |

b. | 8 |

c. | 4 |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- If the random variable
*X*has a uniform distribution between 40 and 50, then*P*(35 £*X*£ 45) is:

a. | 1.0 |

b. | 0.5 |

c. | 0.1 |

d. | undefined. |

ANS: B NAT: Analytic; Probability Distributions

- The probability density function
*f(x)*of a random variable*X*that has a uniform distribution between*a*and*b*is

a. | (b + a)/2 |

b. | 1/b – 1/a |

c. | (a – b)/2 |

d. |
None of these choices. |

ANS: D NAT: Analytic; Probability Distributions

- Which of the following does not represent a continuous uniform random variable?

a. | f(x) = 1/2 for x between -1 and 1, inclusive. |

b. | f(x) = 10 for x between 0 and 1/10, inclusive. |

c. | f(x) = 1/3 for x = 4, 5, 6. |

d. |
None of these choices represents a continuous uniform random variable. |

ANS: C NAT: Analytic; Probability Distributions

- Which of the following is not a characteristic for a normal distribution?

a. | It is symmetrical. |

b. | The mean is always zero. |

c. | The mean, median, and mode are all equal. |

d. | It is a bell-shaped distribution. |

ANS: B NAT: Analytic; Probability Distributions

- If
*X*has a normal distribution with mean 60 and standard deviation 6, which value of*X*corresponds with the value*z*= 1.96?

a. | x = 71.76 |

b. | x = 67.96 |

c. | x = 61.96 |

d. | x = 48.24 |

ANS: A NAT: Analytic; Probability Distributions

- A standard normal distribution is a normal distribution with:

a. | a mean of zero and a standard deviation of one. |

b. | a mean of one and a standard deviation of zero. |

c. | a mean always larger than the standard deviation. |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- What proportion of the data from a normal distribution is within two standard deviations from the mean?

a. | 0.3413 |

b. | 0.4772 |

c. | 0.6826 |

d. | 0.9544 |

ANS: D NAT: Analytic; Probability Distributions

- Given that
*Z*is a standard normal random variable, the area to the left of a value*z*is expressed as

a. | P(Z ³ z) |

b. | P(Z £ z) |

c. | P(0 £ Z £ z) |

d. | P(Z ³ –z) |

ANS: B NAT: Analytic; Probability Distributions

- Given that
*Z*is a standard normal variable, the variance of*Z*:

a. | is always greater than 2.0. |

b. | is always greater than 1.0. |

c. | is always equal to 1.0. |

d. | cannot assume a specific value. |

ANS: C NAT: Analytic; Probability Distributions

- Given that
*Z*is a standard normal random variable, a negative value (*z*) on its distribution would indicate:

a. | z is to the left of the mean. |

b. | the standard deviation of this Z distribution is negative. |

c. | the area between zero and the value z is negative. |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- A larger standard deviation of a normal distribution indicates that the distribution becomes:

a. | narrower and more peaked. |

b. | flatter and wider. |

c. | more skewed to the right. |

d. | more skewed to the left. |

ANS: B NAT: Analytic; Probability Distributions

- In its standardized form, the normal distribution:

a. | has a mean of 0 and a standard deviation of 1. |

b. | has a mean of 1 and a variance of 0. |

c. | has an area equal to 0.5. |

d. | cannot be used to approximate discrete probability distributions. |

ANS: A NAT: Analytic; Probability Distributions

- Most values of a standard normal distribution lie between:

a. | 0 and 1 |

b. | -3 and 3 |

c. | 0 and 3 |

d. | minus infinity and plus infinity |

ANS: B NAT: Analytic; Probability Distributions

- Stacy took a math test whose mean was 70 and standard deviation was 5. The total points possible was 100. Stacey’s results were reported to be at the 95th percentile. What was Stacey’s actual exam score, rounded to the nearest whole number?

a. | 95 |

b. | 78 |

c. | 75 |

d. | 62 |

ANS: B NAT: Analytic; Probability Distributions

- Tanner took a statistics test whose mean was 80 and standard deviation was 5. The total points possible was 100. Tanner’s score was 2 standard deviations below the mean. What was Tanner’s score, rounded to the nearest whole number?

a. | 78 |

b. | 70 |

c. | 90 |

d. |
None of these choices. |

ANS: B NAT: Analytic; Probability Distributions

- Lamont took a psychology exam whose mean was 70 with standard deviation 5. He also took a calculus exam whose mean was 80 with standard deviation 10. He scored 85 on both exams. On which exam did he do better compared to the other students who took the exam?

a. | He did better on the psychology exam, comparatively speaking. |

b. | He did better on the calculus exam, comparatively speaking. |

c. | He did the same on both exams, relatively speaking. |

d. |
Cannot tell without more information. |

ANS: A NAT: Analytic; Probability Distributions

- Suppose Lamont’s exam score was at the 80
^{th}percentile on an exam whose mean was 90. What was Lamont’s exam score?

a. | 76.81 |

b. | 72.00 |

c. | 80.00 |

d. |
Cannot tell without more information. |

ANS: D NAT: Analytic; Probability Distributions

- The Student
*t*distribution:

a. | is symmetrical. |

b. | approaches the normal distribution as the degrees of freedom increase. |

c. | has more area in the tails than the standard normal distribution does. |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- The Student
*t*distribution with parameter*v*= 2 has a mean*E*(*t*) equal to:

a. | 0 |

b. | 1 |

c. | 2 |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- The Student
*t*distribution with parameter*v*= 4 has a variance*V*(*t*) equal to:

a. | 4 |

b. | 0 |

c. | 2 |

d. | 1 |

ANS: C NAT: Analytic; Probability Distributions

- Which of the following statements is correct regarding the percentile points of the
*F*distribution?

a. | F_{0.10,10,20} = 1/F_{0.90,10,20} |

b. | F_{0.90,10,20} = 1/F_{0.10,20,10} |

c. | F_{0.90,10,20} = 1/F_{0.90,20,10} |

d. | F_{0.10,10,20} = 1/F_{0.10,20,10} |

ANS: B NAT: Analytic; Probability Distributions

- Which of the following statements is false?

a. | The chi-squared distribution is positively skewed. |

b. | All the values of the chi-squared distribution are non-negative. |

c. | The shape of the chi-squared distribution depends on its degrees of freedom. |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions

- What number corresponds to
*t*_{0.05,10}?

a. | 1.812 |

b. | 1.372 |

c. | 2.228 |

d. | 1.833 |

ANS: A NAT: Analytic; Probability Distributions

- If
*P*(*t*>*t*_{.01,v}) = 2.50, then the number of degrees of freedom*v*is:

a. | 20 |

b. | 21 |

c. | 22 |

d. | 23 |

ANS: D NAT: Analytic; Probability Distributions

- What number corresponds to ?

a. | 28.30 |

b. | 26.22 |

c. | 21.00 |

d. | 5.23 |

ANS: C NAT: Analytic; Probability Distributions

- If , then the number of degrees of freedom
*v*is:

a. | 5 |

b. | 6 |

c. | 7 |

d. | 8 |

ANS: B NAT: Analytic; Probability Distributions

- What number corresponds to
*F*_{0.95,4,8}?

a. | 6.040 |

b. | 3.840 |

c. | 0.260 |

d. | 0.166 |

ANS: D NAT: Analytic; Probability Distributions

- What number corresponds to
*F*_{.025,3,5}?

a. | 14.88 |

b. | 7.76 |

c. | 12.06 |

d. | 5.41 |

ANS: B NAT: Analytic; Probability Distributions

- Suppose
*X*has a chi-squared distribution with 10 degrees of freedom. The mean of*X*is:

a. | 10 |

b. | 9 |

c. | 20 |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- Suppose
*X*has a chi-squared distribution with 10 degrees of freedom. The variance of*X*is:

a. | 20 |

b. | 10 |

c. | 9 |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- The number of parameters for an
*F*distribution is:

a. | 1 |

b. | 2 |

c. | 0 |

d. |
None of these choices. |

ANS: B NAT: Analytic; Probability Distributions

- Suppose
*X*has an*F*distribution. Which of the following is true?

a. | f(x) is symmetrical. |

b. | All the values of X are non-negative. |

c. | The mean of X is zero. |

d. |
All of these choices. |

ANS: B NAT: Analytic; Probability Distributions

- Which of the following distributions can take on negative values?

a. | Student t |

b. | c^{2} |

c. | F |

d. |
None of these choices. |

ANS: A NAT: Analytic; Probability Distributions

- Which of the following distributions is not skewed?

a. | Student t |

b. | c^{2} |

c. | F |

d. |
All of these distributions are skewed. |

ANS: A NAT: Analytic; Probability Distributions

- Which of the following has a mean and variance that depend on degrees of freedom?

a. | Student t |

b. | c^{2} |

c. | F |

d. |
All of these choices are true. |

ANS: D NAT: Analytic; Probability Distributions **COMPLETION**

- A(n) ____________________ random variable is one that assumes an uncountable number of possible values.

ANS: continuous NAT: Analytic; Probability Distributions

- For a continuous random variable, the probability for each individual value of
*X*is ____________________.

ANS:zero0 NAT: Analytic; Probability Distributions

- Probability for continuous random variables is found by finding the ____________________ under a curve.

ANS: area NAT: Analytic; Probability Distributions

- A(n) ____________________ random variable has a density function that looks like a rectangle and you can use areas of a rectangle to find probabilities for it.

ANS: uniform NAT: Analytic; Probability Distributions

- Suppose
*X*is a continuous random variable for*X*between*a*and*b*. Then its probability ____________________ function must non-negative for all values of*X*between*a*and*b*.

ANS: density NAT: Analytic; Probability Distributions

- The total area under
*f(x)*for a continuous random variable must equal ____________________.

ANS:1one NAT: Analytic; Probability Distributions

- The probability density function of a uniform random variable on the interval [0, 5] must be ____________________ for 0 £
*x*£ 5.

ANS:1/50.20 NAT: Analytic; Probability Distributions

- To find the probability for a uniform random variable you take the ____________________ times the ____________________ of its corresponding rectangle.

ANS:base; heightheight; baselength; widthwidth; length NAT: Analytic; Probability Distributions

- You can use a continuous random variable to ____________________ a discrete random variable that takes on a countable, but very large, number of possible values.

ANS: approximate NAT: Analytic; Probability Distributions

- Suppose
*X*has a normal distribution with mean 40 and standard deviation 2. Shifting all the*X*values to the right 10 units results in a normal distribution with mean ____________________ and standard deviation ____________________.

ANS:50; 2fifty; two NAT: Analytic; Probability Distributions

- ____________________ the value of
*s*in a normal distribution will make it wider.

ANS: Increasing NAT: Analytic; Probability Distributions

- We standardize a random variable by subtracting its ____________________ and dividing by its ____________________.

ANS: mean; standard deviation NAT: Analytic; Probability Distributions

- Suppose
*X*has a normal distribution with mean 10 and standard deviation 2. The probability that*X*is less than 8 is equal to the probability that*Z*is less than ____________________.

ANS: -1 NAT: Analytic; Probability Distributions

*P*(*Z*> 1.9) = ____________________*P*(*Z*< 1.9).

ANS:1 -1- NAT: Analytic; Probability Distributions

*P*(1 <*Z*< 2) =*P*(*Z*< 2) – ____________________.

ANS:*P*(*Z* < 1)*P*(*Z*<1) NAT: Analytic; Probability Distributions

- The mean of the standard normal distribution is ____________________ and the standard deviation is ____________________.

ANS:0; 1zero; one NAT: Analytic; Probability Distributions

*P*(*Z*> 3.00) is approximately ____________________.

ANS:0zero NAT: Analytic; Probability Distributions

*P*(*Z*< 3.00) is approximately ____________________.

ANS:1one NAT: Analytic; Probability Distributions

- Suppose
*X*is a normal random variable with mean 70 and standard deviation 3. Then*P*(*X*= 3) = ____________________.

ANS:0zero NAT: Analytic; Probability Distributions

*Z*_{.025}is the value of*Z*such that the area to the ____________________ of*Z*is .9750.

ANS: left NAT: Analytic; Probability Distributions

- The shape of the ____________________ distribution is similar to a normal distribution, except it has more area in the tails.

ANS:Student *t**t* NAT: Analytic; Probability Distributions

- The mean of a Student
*t*distribution is ____________________.

ANS:zero0 NAT: Analytic; Probability Distributions

- The variance of a Student
*t*distribution approaches ____________________ as the degrees of freedom increase to infinity.

ANS:one1 NAT: Analytic; Probability Distributions

- A
*X*^{2}distribution with 5 degrees of freedom has a mean of ____________________ and a variance of ____________________.

ANS:5; 10five; ten NAT: Analytic; Probability Distributions

- The mean and variance of a
*X*^{2}distribution approach ____________________ as the degrees of freedom increase.

ANS: infinity NAT: Analytic; Probability Distributions

- For values of degrees of freedom greater than 100, the
*X*^{2}distribution can be approximated by a(n) ____________________ distribution.

ANS: normal NAT: Analytic; Probability Distributions

- The shape of a
*X*^{2}distribution is ____________________.

ANS:positively skewedskewed NAT: Analytic; Probability Distributions

- The
*F*distribution has two parameters called degrees of freedom,*n*_{1}and*n*_{2}. We call*n*_{1}the ____________________ degrees of freedom, and we call*n*_{2}the ____________________ degrees of freedom.

ANS: numerator; denominator NAT: Analytic; Probability Distributions

- As the ____________________ degrees of freedom increase, the mean of the
*F*distribution approaches 1.

ANS: denominator NAT: Analytic; Probability Distributions

- The shape of an
*F*distribution is ____________________.

ANS:positively skewedskewed NAT: Analytic; Probability Distributions **SHORT ANSWER**

- A continuous random variable
*X*has the following probability density function:

*f(x)* = 1/4, 0 £ *x* £ 4Find the following probabilities:

a. | P(X £ 1) |

b. | P(X ³ 2) |

c. | P(1 £ X £ 2) |

d. | P(X = 3) |

ANS:

a. | 0.25 |

b. | 0.50 |

c. | 0.25 |

d. | 0 |

NAT: Analytic; Probability Distributions NARRBEGIN: Waiting Time**Waiting Time** The length of time patients must wait to see a doctor at an emergency room in a large hospital has a uniform distribution between 40 minutes and 3 hours.NARREND

- {Waiting Time Narrative} What is the probability density function for this uniform distribution?

ANS:*f(x)* = 1/140, 40 £ *x* £ 180 (minutes) NAT: Analytic; Probability Distributions

- {Waiting Time Narrative} What is the probability that a patient would have to wait between one and two hours?

ANS:0.43 NAT: Analytic; Probability Distributions

- {Waiting Time Narrative} What is the probability that a patient would have to wait exactly one hour?

ANS:0 NAT: Analytic; Probability Distributions

- {Waiting Time Narrative} What is the probability that a patient would have to wait no more than one hour?

ANS:0.143 NAT: Analytic; Probability Distributions

- The time required to complete a particular assembly operation has a uniform distribution between 25 and 50 minutes.

a. | What is the probability density function for this uniform distribution? |

b. | What is the probability that the assembly operation will require more than 40 minutes to complete? |

c. | Suppose more time was allowed to complete the operation, and the values of X were extended to the range from 25 to 60 minutes. What would f(x) be in this case? |

ANS:

a. | f(x) = 1/25, 25 £ x £ 50 |

b. | 0.40 |

c. | f(x) = 1/35, 25 £ x £ 60 |

NAT: Analytic; Probability Distributions

- Suppose
*f(x)*equals 1/50 on the interval [0, 50].

a. | What is the distribution of X? |

b. | What does the graph of f(x) look like? |

c. | Find P(X £ 25) |

d. | Find P(X ³ 25) |

e. | Find P(X = 25) |

f. | Find P(0 < X < 3) |

g. | Find P(-3 < X < 0) |

h. | Find P(0 < X < 50) |

ANS:

a. | X has a uniform distribution on the interval [0, 50]. |

b. | f(x) forms a rectangle of height 1/50 from x = 0 to x = 50. |

c. | 0.50 |

d. | 0.50 |

e. | 0 |

f. | 0.06 |

g. | 0.06 |

h. | 1.00 |

NAT: Analytic; Probability Distributions NARRBEGIN: Electronics Test**Electronics Test** The time it takes a student to finish a electronics test has a uniform distribution between 50 and 70 minutes.NARREND

- {Electronics Test Narrative} What is the probability density function for this uniform distribution?

ANS:*f(x)* = 1/20, 50 £ *x* £ 70 NAT: Analytic; Probability Distributions

- {Electronics Test Narrative} Find the probability that a student will take more than 60 minutes to finish the test.

ANS:0.50 NAT: Analytic; Probability Distributions

- {Electronics Test Narrative} Find the probability that a student will take no less than 55 minutes to finish the test.

ANS:0.75 NAT: Analytic; Probability Distributions

- {Electronics Test Narrative} Find the probability that a student will take exactly one hour to finish the test.

ANS:0 NAT: Analytic; Probability Distributions

- {Electronics Test Narrative} What is the median amount of time it takes a student to finish the test?

ANS:60 minutes NAT: Analytic; Probability Distributions

- {Electronics Test Narrative} What is the mean amount of time it takes a student to finish the test?

ANS:60 minutes NAT: Analytic; Probability Distributions NARRBEGIN: Subway Waiting Time**Subway Waiting Time** At a subway station the waiting time for a subway is found to be uniformly distributed between 1 and 5 minutes.NARREND

- {Subway Waiting Time Narrative} What is the probability density function for this uniform distribution?

ANS:*f(x)* = 1/4, 1 £ *x* £ 5 NAT: Analytic; Probability Distributions

- {Subway Waiting Time Narrative} What is the probability of waiting no more than 3 minutes?

ANS:0.50 NAT: Analytic; Probability Distributions

- {Subway Waiting Time Narrative} What is the probability that the subway arrives in the first minute and a half?

ANS:0.125 NAT: Analytic; Probability Distributions

- {Subway Waiting Time Narrative} What is the median waiting time for this subway?

ANS:3 minutes NAT: Analytic; Probability Distributions NARRBEGIN: Battery Life**Battery Life** A certain brand of batteries has a lifetime that *has a normal distribution* with a mean of 3,750 hours and a standard deviation of 300 hours.NARREND

- {Battery Life Narrative} What proportion of these batteries will last for more than 4,000 hours?

ANS:0.2033 NAT: Analytic; Probability Distributions

- {Battery Life Narrative} What proportion of these batteries will last less than 3,600 hours?

ANS:0.3085 NAT: Analytic; Probability Distributions

- {Battery Life Narrative} What proportion of these batteries will last between 3,800 and 4,100 hours?

ANS:0.3115 NAT: Analytic; Probability Distributions

- {Battery Life Narrative} What lifetime should the manufacturer advertise for these batteries in order that only 2% of the lamps will wear out before the advertised lifetime?

ANS:3135 NAT: Analytic; Probability Distributions NARRBEGIN: Diet**Diet** Researchers studying the effects of a new diet found that the weight loss over a one-month period by those on the diet was normally distributed with a mean of 10 pounds and a standard deviation of 5 pounds.NARREND

- {Diet Narrative} What proportion of the dieters lost more than 12 pounds?

ANS:0.3446 NAT: Analytic; Probability Distributions

- {Diet Narrative} What proportion of the dieters
__gained__weight?

ANS:0.0028 NAT: Analytic; Probability Distributions

- {Diet Narrative} If a dieter is selected at random, what is the probability that the dieter lost more than 7.5 pounds?

ANS:0.6915 NAT: Analytic; Probability Distributions

- Let
*X*be a normally distributed random variable with a mean of 12 and a standard deviation of 1.5. What proportions of the values of*X*are:

a. | less than 14 |

b. | more than 8 |

c. | between 10 and 13 |

ANS:

a. | 0.9082 |

b. | 0.9962 |

c. | 0.6568 |

NAT: Analytic; Probability Distributions

- If
*Z*is a standard normal random variable, find the value*z*for which:

a. | the area between 0 and z is 0.3729 |

b. | the area to the right of z is 0.7123 |

c. | the area to the left of z is 0.1736 |

d. | the area between –z and z is 0.6630 |

ANS:

a. | 1.14 |

b. | -.56 |

c. | -.94 |

d. | 0.96 |

NAT: Analytic; Probability Distributions

- If
*Z*is a standard normal random variable, find the following probabilities:

a. | P(Z £ -1.77) |

b. | P(Z ³ -1.96) |

c. | P(0.35 £ Z £ 0.85) |

d. | P(-2.88 £ Z £ -2.15) |

e. | P(Z £ 1.45) |

ANS:

a. | 0.0384 |

b. | 0.9750 |

c. | 0.1655 |

d. | 0.0138 |

e. | 0.9265 |

NAT: Analytic; Probability Distributions NARRBEGIN: Calculus Scores**Calculus Scores** Scores of high school students on a national calculus exam were normally distributed with a mean of 86 and a standard deviation of 4. (Total possible points = 100.)NARREND

- {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score of 80 or higher?

ANS:0.9332 NAT: Analytic; Probability Distributions

- {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score between 80 and 90?

ANS:0.7745 NAT: Analytic; Probability Distributions

- {Calculus Scores Narrative} What is the probability that a randomly selected student will have a score of 94 or lower?

ANS:0.9772 NAT: Analytic; Probability Distributions NARRBEGIN: Checking Accounts**Checking Accounts** A bank has determined that the monthly balances of the checking accounts of its customers are normally distributed with an average balance of $1,200 and a standard deviation of $250.NARREND

- {Checking Accounts Narrative} What proportion of customers have monthly balances less than $1,000?

ANS:0.2119 NAT: Analytic; Probability Distributions

- {Checking Accounts Narrative} What proportion of customers have monthly balances more than $1,125?

ANS:0.6179 NAT: Analytic; Probability Distributions

- {Checking Accounts Narrative} What proportion of customers have monthly balances between $950 and $1,075?

ANS:0.1498 NAT: Analytic; Probability Distributions NARRBEGIN: IT Graduates Salary**IT Graduates Salary** The recent average starting salary for new college graduates in IT systems is $47,500. Assume salaries are normally distributed with a standard deviation of $4,500.NARREND

**{**IT Graduates Salary Narrative} What is the probability of a new graduate receiving a salary between $45,000 and $50,000?

ANS:0.4246 NAT: Analytic; Probability Distributions

- {IT Graduates Salary Narrative} What is the probability of a new graduate getting a starting salary in excess of $55,000?

ANS:0.0475 NAT: Analytic; Probability Distributions

- {IT Graduates Salary Narrative} What percent of starting salaries are no more than $42,250?

ANS:12.10% NAT: Analytic; Probability Distributions

- {IT Graduates Salary Narrative} What is the cutoff for the bottom 5% of the salaries?

ANS:$40,097.50 NAT: Analytic; Probability Distributions

- {IT Graduates Salary Narrative} What is the cutoff for the top 3% of the salaries?

ANS:$55,960 NAT: Analytic; Probability Distributions

- A worker earns $16 per hour at a plant and is told that only 5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, find the average wage for the plant workers per hour.

ANS:*P*(*X* > 16) = .05 Þ (16 – *m*) / 5 = 1.645 Þ *m* = $7.78 NAT: Analytic; Probability Distributions

- Use the
*t*-table to find the following values of*t*.

a. | t_{.10,9} |

b. | t_{.10,20} |

c. | t_{.025,82} |

d. | t_{.05,196} |

ANS:

a. | 1.383 |

b. | 1.325 |

c. | 1.990 |

d. | 1.653 |

NAT: Analytic; Probability Distributions

- Use the
*t*-table to find the following probabilities.

a. | P(t_{8} > 2.306) |

b. | P(t_{80} > 2.639) |

c. | P(t_{24} > 1.711) |

d. | P(t_{35} > 1.306) |

ANS:

a. | 0.025 |

b. | 0.005 |

c. | 0.050 |

d. | 0.100 |

NAT: Analytic; Probability Distributions

- Use the
*X*^{2}table to find the following values of*X*^{2}.

a. | |

b. | |

c. | |

d. |

ANS:

a. | 14.61 |

b. | 37.6 |

c. | 4.61 |

d. | 45.44 |

NAT: Analytic; Probability Distributions

- Use the
*X*^{2}table to find the following probabilities.

a. | |

b. | |

c. | |

d. |

ANS:

a. | 0.900 |

b. | 0.050 |

c. | 0.025 |

d. | 0.010 |

NAT: Analytic; Probability Distributions

- Use the
*F*table to find the following values of*F*.

a. | F_{.01,12,20} |

b. | F_{.05,20,40} |

c. | F_{.025,5,15} |

d. | F_{.01,8,30} |

ANS:

a. | 3.23 |

b. | 1.84 |

c. | 3.58 |

d. | 3.17 |

NAT: Analytic; Probability Distributions

- Use the
*F*table to find the following values of*F*.

a. | F_{.99,12,20} |

b. | F_{.95,20,40} |

c. | F_{.975,5,15} |

d. | F_{.99,8,30} |

ANS:

a. | 1 / 3.86 = 0.2591 |

b. | 1 / 1.99 = 0.5025 |

c. | 1 / 6.43 = 0.1555 |

d. | 1 / 5.20 = 0.1923 |

NAT: Analytic; Probability Distributions

- Use the
*F*table to find the following probabilities.

a. | P(F_{6,14} > 2.85) |

b. | P(F_{20,60} > 2.20) |

c. | P(F_{12,25} > 2.51) |

d. | P(F_{15,30} > 2.01) |

ANS:

a. | 0.05 |

b. | 0.01 |

c. | 0.025 |

d. | 0.05 |

NAT: Analytic; Probability Distributions

- Suppose you have a Student
*t*distribution with 20 degrees of freedom.

a. | Find the mean. |

b. | Find the variance. |

c. | Find the standard deviation. |

ANS:

a. | 0 |

b. | 20/18 = 1.11 |

c. | 1.05 |

NAT: Analytic; Probability Distributions

- Suppose you have a
*X*^{2}distribution with 20 degrees of freedom.

a. | Find the mean. |

b. | Find the variance. |

c. | Find the standard deviation. |

ANS:

a. | 20 |

b. | 40 |

c. | 6.32 |

NAT: Analytic; Probability Distributions

- Suppose you have an
*F*distribution with degrees of freedom 10, 20.

a. | Find the mean |

b. | Find the variance. |

c. | Find the standard deviation. |

ANS:

a. | 20/18 = 1.11 |

b. | [2(20)^{2} (10 + 20 – 2)] / [(10)(18)^{2} (20 – 4)] = 0.4321 |

c. | Ö0.4321 = 0.6573 |

NAT: Analytic; Probability Distributions

- What happens to the shape, mean, and variance of a Student
*t*distribution as the degrees of freedom increase?

ANS:As the degrees of freedom of the Student *t* distribution increase, the shape approaches a standard normal distribution; the mean remains 0; and the variance approaches 1. NAT: Analytic; Probability Distributions

- What happens to the shape, mean, and variance of a
*X*^{2}distribution as the degrees of freedom increase?

ANS:The shape becomes less positively skewed; the mean increases to infinity, as does the variance.