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Consider the following set of random variables:
I. Total number of points scored during a football game
II. Lifespan in hours of a halogen light bulb
III. Height in feet of the ocean's tide at a given location
IV. Number of fatalities in civilian aircraft crashes in a given year
V. Length in inches of an adult rattlesnake.
Which ones are continuous random variables?
A. II and III only
B. II, III, and V only
C. I and IV only
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Which of the following probability distributions of a discrete random variable X is not a legitimate distribution?
A.
B.
C.
3 of 24
Suppose that we have a deck of three cards, one marked with a 1, one marked with a 2, and one marked with a 5. You draw two cards at random and
without replacement from the deck. The sample space S = {(1, 2), (1, 5), (2, 5)} consists of these three equally likely outcomes. Let X be the sum of the
numbers on the two cards drawn. Which of the following is the correct probability distribution for X?
A.
B.
C.
4 of 24
In a particular game, a ball is randomly chosen from a box that contains three red balls, one green ball, and six blue balls. If a red ball is selected, you win
$2. If a green ball is selected, you win $4. If a blue ball is selected, you win nothing. Let X be the amount that you win. The expected value of X is
A. $1.
B. $2.
C. $3.
5 of 24
Let X = the number of times that a customer visits a grocery store during a one-week period. Assume that the probability distribution of x is as follows:
Use the distribution to determine the probability that a randomly chosen customer visits the grocery store at least twice during a one-week period.
A. 0.9
B. 0.5
C. 0.4
6 of 24
Let X = the number of times that a customer visits a grocery store during a one-week period. Assume that the probability distribution of X is as follows:
The standard deviation of X, σX, is approximately
A. 1.50.
B. 0.65.
C. 0.81.
7 of 24
The time in minutes X that you must wait before a train arrives at your local subway station is a uniformly distributed random variable between 5 minutes and
15 minutes. That is, the density curve of the distribution of x has constant height between 5 and 15 and height 0 outside this interval. Determine P (6 < X <
8).
A. 0.1
B. 0.5
C. 0.2
8 of 24
The weight of a medium-sized orange selected at random from a large bin of oranges at a local supermarket is a random variable with mean μ = 12 ounces
and standard deviation σ = 1.2 ounces. Suppose we independently select two oranges at random from the bin. The difference in the weights of the two
oranges (the weight of the first orange minus the weight of the second orange) is a random variable with a standard deviation equal to
A. 0 ounces.
B. 1.70 ounces.
C. 2.88 ounces.
9 of 24
A widget manufacturer estimates that the total weekly cost in dollars, C, to produce x widgets is given by the linear function C(x) = 500 + 10x, where the
intercept 500 represents a "fixed" cost of manufacture and the slope 10 represents the "variable" cost of producing a certain number of widgets. Analysis of
weekly widget production reveals that the number of widgets X produced in a week is a random variable with mean μX = 200 and standard deviation σX = 20.
What are the mean and the standard deviation of C?
A. mean of C = $2500, standard deviation of C = $200
B. mean of C = $2500, standard deviation of C = $700
C. mean of C = $100,010, standard deviation of C = $10,000
10 of 24
Suppose that we are given random variables X, Y for which we know the means μ X, μ Y and the variances σ2X, σ2Y. Which of the following quantities could
we not compute without knowing some additional information about X, Y?
A. μ X - Y
B. μ 3X - 2Y
C. σ X+Y
11 of 24
The daily total sales (except for Saturday) at a small restaurant have a probability distribution that is approximately Normal with a mean of μ = $530 and a
standard deviation of σ = $120. The probability the sales will exceed $700 on a given day is approximately
A. 0.9222.
B. 0.5778.
C. 0.0778.
12 of 24
Suppose we have a "loaded" (unfair) die that gives the outcomes 1 through 6 according to the probability distribution
If this die is rolled 6000 times, then the sample mean number of spots per roll for the 6000 rolls should be about
A. 3.00.
B. 3.30.
C. 3.50.
13 of 24
A set of 10 playing cards consists of five red cards and five black cards. The cards are shuffled thoroughly, and we draw four cards one at a time and without
replacement. Let X = the number of red cards drawn. The random variable X has which of the following probability distributions?
A. binomial distribution with parameters n = 10 and p = 0.5
B. binomial distribution with parameters n = 4 and p = 0.5
C. neither (A) nor (B)
14 of 24
There are 20 multiple-choice questions on an exam, each having four possible responses, of which only one is correct. Each question is worth 5 points if
answered correctly. Suppose that a student guesses the answer to each question, with her guesses from question to question being independent. If the
student needs at least 40 points to pass the exam, the probability that she passes is closest to
A. 0.0609.
B. 0.1018.
C. 0.9591.
15 of 24
There are 20 multiple-choice questions on an exam, each having four possible responses, of which only one is correct. Each question is worth 5 points if
answered correctly. Suppose that a student guesses the answer to each question, with her guesses from question to question being independent. The
probability that the student scores lower than a 60 on the exam is
A. 0.0009.
B. 0.9998.
C. 0.9991.
16 of 24
There are 20 multiple-choice questions on an exam, each having four possible responses, of which only one is correct. Each question is worth 5 points if
answered correctly. Suppose that a student guesses the answer to each question, with her guesses from question to question being independent. The
student's expected (mean) score on this exam is
A. 25.
B. 5.
C. 50.
17 of 24
There are 20 multiple-choice questions on an exam, each having four possible responses, of which only one is correct. Each question is worth 5 points if
answered correctly. Suppose that a student guesses the answer to each question, with her guesses from question to question being independent. The
standard deviation of the student's score on the exam is
A. 9.68.
B. 1.94.
C. 93.75.
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For which of the following choices of n, p can we not use the Normal approximation to the binomial distribution?
A. n = 25, p = 0.6
B. n = 40, p = 0.4
C. n = 60, p = 0.9
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In the gambling game of chuck-a-luck, three dice are rolled using a rotating, hourglass-shaped cage. The player chooses one of the six possible sides (1, 2,
3, 4, 5, or 6) and receives a payoff the amount of which depends on how many dice turn up on that particular side. Let X = the number of times the dice have
to be rolled until we see "three of a kind" (of any type). Which of the following probability distributions does X have?
A. geometric with p = 1/216
B. geometric with p = 6/216
C. binomial with n = 3 and p = 6/216
20 of 24
In the old children's game of "rock-scissors-paper," two players simultaneously use their hands to show one of three objects: a rock (a closed fist), a pair of
scissors (two fingers extended in a V-shape), or a piece of paper (an open palm). The winner is chosen in the following ways:
Rock beats scissors (rock can crush scissors).
Scissors beats paper (scissors can cut paper).
Paper beats rock (paper can wrap around rock).
If the players each show the same object, then the game is played again, with additional repeats as needed until one player wins. Assume that each player
selects an object independently and that each player is equally likely to choose any of the three objects. Let X = the number of games that must be played to
decide a winner. (We assume that the identity of the winner is unimportant.) Then X is a geometric random variable with probability of success p equal to
A. 1/3.
B. 2/3.
C. 1/9.
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In a certain large population, 70% are right-handed. You need a left-handed pitcher for your softball team and decide to find one by asking people chosen
from the population at random. (We assume that once you do find a left-hander, he or she will be happy to join your team and will not say no.) The
probability that the first left-hander you find is the fourth person you ask is approximately
A. 0.1029.
B. 0.019.
C. 0.072.
22 of 24
In a certain large population, 70% are right-handed. You need a left-handed pitcher for your softball team and decide to find one by asking people chosen
from the population at random. (We assume that once you do find a left-hander, he or she will be happy to join your team and will not say no.) The
probability that you will have to ask more than three people before finding your first left-hander is approximately
A. 0.027.
B. 0.240.
C. 0.343.
23 of 24
In a certain large population, 70% are right-handed. You need a left-handed pitcher for your softball team and decide to find one by asking people chosen
from the population at random. (We assume that once you do find a left-hander, he or she will be happy to join your team and will not say no.) The
probability that you will have to ask at most two people to find your first left-hander is approximately
A. 0.51.
B. 0.49.
C. 0.91.
24 of 24
In a certain large population, 70% are right-handed. You need a left-handed pitcher for your softball team and decide to find one by asking people chosen
from the population at random. (We assume that once you do find a left-hander, he or she will be happy to join your team and will not say no.) The mean and
variance of the number of people you will have to ask to find your first left-hander are
A. mean = 3.33, variance = 0.612.
B. mean = 3.33, variance = 7.78.
C. mean = 0.3, variance = 7.78.
1 B All these variables take values on an interval of real numbers, whereas I and IV take values on only a discrete set of numbers (the counting numbers).
2 B The probabilities for the three values -1, 0, and 1 do not add to 1.
3 A If (1, 2) is drawn, then the sum is a 3. If (1, 5) is drawn, then the sum is a 6. If (2, 5) is drawn, then the sum is a 7. Each of the three possible pairs is
equally likely, so the three sums are equally likely as well, and each must have probability 1/3.
4 A The amount X takes the value $0 with probability 6/10, the value $2 with probability 3/10, and the value $4 with probability 1/10. The expected value
(mean) of X is then given by μX = Σ xp (x) = ($0)(6/10) + ($2)(3/10) + ($4)(1/10) = 0 + 6/10 + 4/10 = $1.
5 B The event "a randomly chosen customer visits the grocery store at least twice during a one-week period" can be written in terms of X as X ≥ 2. The
probability of this event is p (2) + p (3) = 0.4 + 0.1 = 0.5 by the simple addition rule for disjoint events.
6 C Computing the mean of X yields μX = Σ xp (x) = (0)(0.1) + (1)(0.4) + (2)(0.4) + (3)(0.1) = 0 + 0.4 + 0.8 + 0.3 = 1.5. The variance of X, σX2, is Σ (x - μX)2p
(x) = (0 - 1.5)2(0.1) + (1 - 1.5)2(0.4) + (2 - 1.5)2(0.4) + (3 - 1.5)2(0.1) = 0.225 + 0.1 + 0.1 + 0.225 = 0.65. The standard deviation σX is therefore √0.65 = 0.81.
7 C Since the total area under any density curve is 1, and the interval over which X is defined has length 10, the height of the curve must be 1/10 = 0.1. P(6
< X < 8) is the area under the density curve of X between 6 and 8, or the area of a rectangle with height 0.1 and width 8 - 6 = 2; that is, (0.1)(2) = 0.2.
8 B Because we have drawn a simple random sample of size 2 from a large population, the rule for the variance of the difference of two independent random
variables applies. Letting X, Y = the weight of the first orange and the second orange, respectively, we have σ2X - Y = σ2X + σ2Y. Since σX = σY = 1.2 ounces,
we have σ2X - Y = (1.2)2 + (1.2)2 = 1.44 + 1.44 = 2.88. The standard deviation σX - Y is therefore √2.88 = 1.70.
9 A The mean of a linear function of X, a + bX, is given by μa+bX = a + bμX, and the variance is given by σ2a+bX = b2σ2X. In this case, a = 500 and b = 10, so
we have μC = 500 + 10(200) = 2500, σ2C = (10)2(20)2 = 40,000, and σC = √40,000 = 200.
10 C We know that the variance of the sum of two independent random variables is given by σ2X+Y = σ2X + σ2Y and that the standard deviation of the sum is
equal to the square root of this quantity. However, we are given no information about the independence or dependence of X, Y. Without this additional
information, we cannot compute σ X+Y.
11 C Letting X = daily total sales, and recalling that probability is equivalent to area for continuous distributions like the Normal, we have P(X > 700) = P((X μ)/σ > (700 - 530)/120) = P(Z > 1.42) = area to the right of 1.42 under the standard Normal curve = 1 - 0.9222 = 0.0778.
12 B The law of large numbers says that for a large number of repetitions, the sample mean should be very close to the population mean. In this case, the
latter is μ X = (1)(0.1) + (2)(0.2) + (3)(0.3) + (4)(0.2) + (5)(0.1) + (6)(0.1) = 3.30.
13 C This distribution is not binomial because the conditions for an experiment to be a binomial experiment are not satisfied. In particular, the results of the
four trials (draws) are not independent, and the probability of a success (getting a red card) varies from trial to trial.
14 B The random variable X = "number correct out of 20" has a binomial distribution with n = 20 and p = 0.25. To get at least 40 points, the student must get
at least eight of the 20 questions correct. The desired probability, expressed in terms of X, is therefore P(X ≥ 8). Adding binomial probabilities for all values
of X from 8 through 20 yields the desired answer.
15 C The random variable X = "number correct out of 20" has a binomial distribution with n = 20 and p = 0.25. Getting a score of 60 corresponds to getting
12 of the 20 questions right, so in terms of X, the event whose probability we seek is X < 12, that is, X ≤ 11. Adding binomial probabilities for all values of X
between 0 and 11 yields the desired answer.
16 A The random variable X = "number correct out of 20" has a binomial distribution with n = 20 and p = 0.25. The mean of X is μ X = np = (20)(0.25) = 5.
The student's score on the exam can be written as the linear function 5X, so the student's mean score is μ 5X = 5μ X = 5(5) = 25 by the rule for the mean of a
linear function of a random variable.
17 A The random variable X = "number correct out of 20" has a binomial distribution with n = 20 and p = 0.25. The variance of X is σ2X = np(1 - p) =
(20)(0.25)(0.75) = 3.75. The student's score on the exam can be written as the linear function 5X, so the variance of the student's score is σ25X = 25σ2X =
25(3.75) = 93.75 by the rule for the variance of a linear function of a random variable. The standard deviation of the score is therefore √93.75 = 9.68.
18 C In order for the approximation to be valid, np and n(1 - p) must both be 10 or more. Here, np = (60)(0.9) = 54 > 10, but n(1 - p) = (60)(0.1) = 6 < 10, so
the approximation can't be used.
19 B This is an example of a geometric setting because the results of individual rolls of the dice are independent and the variable of interest is the number of
trials required to obtain the first success ("three of a kind"). Since there are (6)(6)(6) = 216 possible outcomes, and six of them represent "three of a kind"
(that is, (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), and (6, 6, 6)), the probability of success is p = 6/216.
20 B There are 3 x 3 = 9 possible pairs of objects that can be displayed: PP, PR, PS, RP, RR, RS, SP, SR, SS. (Note that the order is unimportant when all
we want to do is declare a winner and we don't care which person wins.) Six of the nine pairs (PR, PS, RP, RS, SP, SR) will result in a winner. By the stated
conditions, this is a geometric setting with p = probability of success = probability of getting PR, PS, RP, RS, SP, or SR = 6/9 = 2/3.
21 A We want P(X = 4), where X = "number of people you ask to get your first left-hander" is geometric with p = 0.3. By the geometric probability formula,
this probability is (1 - 0.3)4-1(0.3) = (0.7)3(0.3) = 0.1029.
22 C We want P(X > 3), where X = "number of people you ask to get your first left-hander" is geometric with p = 0.3. The probability P(X > n) for a geometric
random variable X with probability of success p is equal to (1 - p)n. Here, n = 3, and P(X > 3) = (1 - 0.3)3 = (0.7)3 = 0.343.
23 A We want P(X ≤ 2), where X = "number of people you ask to get your first left-hander" is geometric with p = 0.3. By the geometric probability formula
and the addition rule for probabilities, P(X ≤ 2) = p(1) + p(2) = 0.3 + (0.7)(0.3) = 0.51.
24 B The expected value (mean) and variance of a geometric random variable with probability of success p are 1/p and (1 - p)/p2, respectively. Here, p = 0.3,
so μ X = 1/0.3 = 3.33 and σ2X = (1 - 0.3)/(0.3)2 = 0.7/0.09 = 7.78.