Probability and Random Variables - Review the Knowledge You Need to Score High - 5 Steps to a 5 AP Statistics 2017 (2016)

5 Steps to a 5 AP Statistics 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 9

Probability and Random Variables

IN THIS CHAPTER

Summary: We”ve completed the basics of data analysis, and we now begin the transition to inference. In order to do inference, we need to use the language of probability. In order to use the language of probability, we need an understanding of random variables and probabilities. The next two chapters lay the probability foundation for inference. In this chapter, we”ll learn about the basic rules of probability, what it means for events to be independent, and about discrete and continuous random variables, simulation, and rules for combining random variables.

Key Ideas

Probability

Random Variables

Discrete Random Variables

Continuous Random Variables

Probability Distributions

Normal Probability

Simulation

Transforming and Combining Random Variables

Probability

The second major part of a course in statistics involves making inferences about populations based on sample data (the first was exploratory data analysis ). The ability to do this is based on being able to make statements such as, “If we assume the null hypothesis to be true then the probability of getting results such as ours (or more extreme) by chance alone is 0.06.” To make sense of this statement, you need to have a understanding of what is meant by the term “probability” as well as an understanding of some of the basics of probability theory.

A chance experiment (random phenomenon): An activity whose outcome we can observe or measure but we do not know how it will turn out on any single trial. Note that this is a somewhat different meaning of the word “experiment” than we developed in the last chapter.

example: If we roll a six-sided die, we know that we will get a 1, 2, 3, 4, 5, or 6, but we don”t know which one of these we will get on the next trial. Assuming a fair die, however, we do have a good idea of approximately what proportion of each possible outcome we will get over a large number of trials.

Outcome: One of the possible results of an experiment (random phenomenon).

example: the possible outcomes for the roll of a single die are 1, 2, 3, 4, 5, 6.

Sample Spaces and Events

Sample space: The set of all possible outcomes of an experiment.

example: For the roll of a single die, S = {1, 2, 3, 4, 5, 6}.

Event: A collection of outcomes or simple events. That is, an event is a subset of the sample space.

example : For the roll of a single die, the sample space (all outcomes) is S = {1, 2, 3, 4, 5, 6}. Let event A = “the value of the die is 6.” Then A = {6}. Let B = “the face value is less than 4.” Then B = {1, 2, 3}. Events A and B are subsets of the sample space.

example: Consider the experiment of flipping two coins and noting whether each coin lands heads or tails. The sample space is S = {HH, HT, TH, TT}. Let event B = “at least one coin shows a head.” Then B = {HH, HT, TH}. Event B is a subset of the sample space S.

Probability of an event: the relative frequency of the outcome. That is, it is the fraction of time that the outcome would occur if the experiment were repeated indefinitely. If we let E = the event in question, s = the number of ways an outcome can succeed, and f = the number of ways an outcome can fail, then

Note that s + f equals the number of outcomes in the sample space. Another way to think of this is that the probability of an event is the sum of the probabilities of all outcomes that make up the event.

For any event A, P (A) ranges from 0 to 1, inclusive. That is, 0 ≤ P (A) ≤ 1. This is an algebraic result from the definition of probability when success is guaranteed (f = 0, s = 1) or failure is guaranteed (f = 1, s = 0).

The sum of the probabilities of all possible outcomes in a sample space is one. That is, if the sample space is composed of n possible outcomes,

example: In the experiment of flipping two coins, let the event A = obtain at least one head. The sample space contains four elements ({HH, HT, TH, TT}). s = 3 because there are three ways for our outcome to be considered a success ({HH, HT, TH}) and f = 1.

Thus

example: Consider rolling two fair dice and noting their sum. A sample space for this event can be given in table form as follows:

Let B = “the sum of the two dice is greater than 4.” There are 36 outcomes in the samples space, 30 of which are greater than 4. Thus,

Furthermore,

Probabilities of Combined Events

P (A or B): The probability that either event A or event B occurs. (They can both occur, but only one needs to occur.) Using set notation, P (A or B) can be written P (A ∪ B). A ∪ B is spoken as, “A union B.”

P (A and B): The probability that both event A and event B occur. Using set notation, P (A and B) can be written P (A ∩ B). A ∩ B is spoken as, “A intersection B.”

example: Roll two dice and consider the sum (see table). Let A = “one die shows a 3,” B = “the sum is greater than 4.” Then P (A or B) is the probability that either one die shows a 3 or the sum is greater than 4. Of the 36 possible outcomes in the sample space, there are 32 possible outcomes that are successes [30 outcomes greater than 4 as well as (1, 3) and (3, 1)], so

There are nine ways in which a sum has one die showing a 3 and has a sum greater than 4: [(3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (2, 3), (4, 3), (5, 3), (6, 3)], so

Complement of an event A: events in the sample space that are not in event A. The complement of an event A is symbolized by , or Ac . Furthermore, P ( ) = 1 – P (A).

Mutually Exclusive Events

Mutually exclusive (disjoint) events : Two events are said to be mutually exclusive (some texts refer to mutually exclusive events as disjoint ) if and only if they have no outcomes in common. That is, (A ∩ B) = ∅. If A and B are mutually exclusive, then P (A and B) = P (A ∩ B) = 0.

example: In the two-dice rolling experiment, A = “face shows a 1” and B = “sum of the two dice is 8” are mutually exclusive because there is no way to get a sum of 8 if one die shows a 1. That is, events A and B cannot both occur.

Conditional Probability

Conditional probability: “The probability of A given B” assumes we have knowledge of an event B having occurred before we compute the probability of event A. This is symbolized by P (A|B). Also,

Although this formula will work, it”s often easier to think of a condition as reducing, in some fashion, the original sample space. The following example illustrates this “shrinking sample space.”

example: Once again consider the possible sums on the roll of two dice. Let A = “the sum is 7,” B = “one die shows a 5.” We note, by counting outcomes in the table, that P (A) = 6/36. Now, consider a slightly different question: what is P (A|B) (that is, what is the probability of the sum being 7 given that one die shows a 5)?

solution: Look again at the table:

The condition has effectively reduced the sample space from 36 outcomes to only 11 (you do not count the “10” twice). Of those, two are 7s. Thus, the P (the sum is 7|one die shows a 5) = 2/11.

alternate solution: If you insist on using the formula for conditional probability, we note that P (A and B) = P (the sum is 7 and one die shows a 5) = 2/36, and P (B) = P (one die shows a 5 ) = 11/36. By formula

Some conditional probability problems can be solved by using a tree diagram . A tree diagram is a schematic way of looking at all possible outcomes.

example: Suppose a computer company has manufacturing plants in three states. 50% of its computers are manufactured in California, and 85% of these are desktops; 30% of computers are manufactured in Washington, and 40% of these are laptops; and 20% of computers are manufactured in Oregon, and 40% of these are desktops. All computers are first shipped to a distribution site in Nebraska before being sent out to stores. If you picked a computer at random from the Nebraska distribution center, what is the probability that it is a laptop?

solution:

Note that the final probabilities add to 1 so we know we have considered all possible outcomes. Now, P (laptop ) = 0.075 + 0.12 + 0.12 = 0.315.

Independent Events

Independent events: Events A and B are said to be independent if and only if P (A) = P (A|B) or P (B) = P (B|A). That is, A and B are independent if the knowledge of one event having occurred does not change the probability that the other event occurs.

example: Consider drawing one card from a standard deck of 52 playing cards.

(i) Are A and B independent?

solution: P (A|B) = P (the card drawn is an ace | the card is a 10, J, Q, K, or A) = 4/20 = 1/5 (there are 20 cards to consider, 4 of which are aces). Since P (A) = 1/13, knowledge of B has changed what we know about A. That is, in this case, P (A) ≠ P (A|B), so events A and B are not independent.

(ii) Are A and C independent?

solution : P (A|C) = P (the card drawn is an ace | the card drawn is a diamond) = 1/13 (there are 13 diamonds, one of which is an ace). So, in this case, P (A) = P (A| C), so that the events “the card drawn is an ace” and “the card drawn is a diamond” are independent.

Probability of A and B or A or B

The Addition Rule: P (A or B) = P (A) + P(B) – P (A and B).

Special case of The Addition Rule : If A and B are mutually exclusive ,

P (A and B) = 0, so P (A or B) = P (A) + P (B).

The Multiplication Rule : P (A and B) = P (A) · P (B|A).

Special case of The Multiplication Rule: If A and B are independent ,

P (B|A) = P (B), so P (A and B) = P (A) · P (B).

example: If A and B are two mutually exclusive events for which P (A) = 0.3, P (B) = 0.25. Find P (A or B).

solution : P (A or B) = 0.3 + 0.25 = 0.55.

example: A basketball player has a 0.6 probability of making a free throw. What is his probability of making two consecutive free throws if

(a) he gets very nervous after making the first shot and his probability of making the second shot drops to 0.4.

solution : P (making the first shot) = 0.6, P (making the second shot | he made the first) = 0.4. So, P (making both shots) = (0.6)(0.4) = 0.24.

(b) the events “he makes his first shot” and “he makes the succeeding shot” are independent.

solution: Since the events are independent, his probability of making each shot is the same. Thus, P (he makes both shots) = (0.6)(0.6) = 0.36.

Random Variables

Recall our earlier definition of a probability experiment (random phenomenon) : An activity whose outcome we can observe and measure, but for which we can”t predict the result of any single trial. A random variable , X , is a numerical value assigned to an outcome of a random phenomenon. Particular values of the random variable X are often given lowercase names, such as x to represent a general value, or k to represent a specific value. It is common to see expressions of the form P (X = x ) or P (X = k ), which refers to the probability that the random variable X takes on the particular value x .

example: If we roll a fair die, the random variable X could be the face-up value of the die. The possible values of X are {1, 2, 3, 4, 5, 6}. P (X = 2) = 1/6.

example: The score a college-hopeful student gets on her SAT test can take on values from 200 to 800. These are the possible values of the random variable X , the score a randomly selected student gets on his/her test.

There are two types of random variables: discrete random variables and continuous random variables .

Discrete Random Variables

A discrete random variable (DRV) is a random variable with a countable number of outcomes. Although most discrete random variables have a finite number of outcomes, note that “countable” is not the same as “finite.” A discrete random variable can have an infinite number of outcomes. For example, consider f (n ) = (0.5) n . Then f (1) = 0.5, f (2) = (0.5)2 = 0.25, f (0.5)3 = 0.125, … There are an infinite number of outcomes, but they are countable in that you can identify f (n ) for any non-negative integer n .

example: the number of votes earned by different candidates in an election.

example: the number of successes in 25 trials of an event whose probability of success on any one trial is known to be 0.3.

Continuous Random Variables

A continuous random variable (CRV) is a random variable that assumes values associated with one or more intervals on the number line. The continuous random variable X has an infinite number of outcomes.

example: Consider the uniform distribution y = 3 defined on the interval 1 ≤ x ≤ 5. The area under y = 3 and above the x axis for any interval corresponds to a continuous random variable. For example, if 2 ≤ x ≤ 3, then X = 3. If 2 ≤ x ≤ 4.5, then X = (4.5 – 2)(3) = 7.5. Note that there are an infinite number of possible outcomes for X .

Probability Distribution of a Random Variable

A probability distribution for a random variable is the possible values of the random variable X together with the probabilities corresponding to those values.

A probability distribution for a discrete random variable is a list of the possible values of the DRV together with their respective probabilities.

example: Let X be the number of boys in a three-child family. Assuming that the probability of a boy on any one birth is 0.5, the probability distribution for X is

The probabilities P i of a DRV satisfy two conditions:

(1) 0 ≤ P i ≤ 1 (that is, every probability is between 0 and 1).

(2) ΣP i = 1 (that is, the sum of all probabilities is 1).

(Are these conditions satisfied in the above example?)

The mean of a discrete random variable, also called the expected value , is given by

The variance of a discrete random variable is given by

The standard deviation of a discrete random variable is given by

example: Given that the following is the probability distribution for a DRV,

find P (X = 3).

solution: Since Σ Pi = 1, P (3) = 1 – (0.15 + 0.2 + 0.2 + 0.35) = 0.1.

example: For the probability distribution given above, find μ x and s x .

solution:

Calculator Tip: While it”s important to know the formulas given above, in practice it”s easier to use your calculator to do the computations. The TI-83/84 can do this easily by putting the x -values in, say, L1 , and the values of P(X ) in, say, L2 . Then, entering 1-Var Stats L1,L2 and pressing ENTER will return the desired mean and standard deviation. Note that the only standard deviation given is σx —the Sx is blank. Your calculator, in its infinite wisdom, recognizes that the entries in L2 are relative frequencies and assumes you are dealing with a probability distribution (if you are taking measurements on a distribution , there is no such thing as a sample standard deviation).

example: Redo the previous example using the TI-83/84, or equivalent, calculator.

solution: Enter the x values in a list (say, L1 ) and the probabilities in another list (say, L2 ). Then enter “1-Var Stats L1, L2 ” and press ENTER. The calculator will read the probabilities in L2 as relative frequencies and return 4.5 for the mean and 1.432 for the standard deviation.

Probability Histogram

A probability histogram of a DRV is a way to picture the probability distribution. The following is a TI-83/84 histogram of the probability distribution we used in a couple of the examples above.

Probability Distribution for a Continuous Random Variable (CRV) . The probability distribution of a continuous random variable has several properties.

  • There is a smooth curve, called adensity curve (defined by a density function ), that describes the probability distribution of a CRV (sometimes called a probability distribution function). A density curve is always on or above the horizontal axis (that is, it is always nonnegative) and has a total area of 1 underneath the curve and above the axis.
  • The probability of any individual value is 0. That is, ifa is a point on the horizontal axis, P (X = a ) = 0.
  • The probability of a given event is the probability thatx will fall in some given interval on the horizontal axis and equals the area under the curve and above the interval. That is, P (a < X < b ) equals the area under the graph of the curve and above the horizontal axis between X = a and X = b .
  • The previous two bulleted items imply thatP (a < X < b ) = P (a ≤ X ≤ b ).

In this course, there are several CRVs for which we know the probability density functions (a probability distribution defined in terms of some density curve). The normal distribution (introduced in Chapter 4 ) is one whose probability density function is the normal probability distribution . Remember that the normal curve is “bell-shaped” and is symmetric about the mean, μ , of the population. The tails of the curve extend to infinity, although there is very little area under the curve when we get more than, say, three standard deviations away from the mean. (The 68-95-99.7 rule stated that about 99.7% of the terms in a normal distribution are within three standard deviations of the mean. Thus, only about 0.3% lie beyond three standard deviations of the mean.)

Areas between two values on the number line and under the normal probability distribution correspond to probabilities. In Chapter 4 , we found the proportion of terms falling within certain intervals. Because the total area under the curve is 1, in this chapter we will consider those proportions to be probabilities.

Remember that we standardized the normal distribution by converting the data to z -scores

We learned in Chapter 4 that a standardized normal distribution has a mean of 0 and a standard deviation of 1.

A table of Standard Normal Probabilities for this distribution is included in this book and in any basic statistics text. We used these tables when doing some normal curve problems in Chapter 6 . Standard normal probabilities, and other normal probabilities, are also accessible on many calculators. We will use a table of standard normal probabilities as well as technology to solve probability problems, which are very similar to the problems we did in Chapter 6 involving the normal distribution.

Using the tables, we can determine that the percentages in the 68-95-99.7 rule are, more precisely, 68.27%, 95.45%, 99.73%. The TI-83/84 syntax for the standard normal is normalcdf(lower bound, upper bound) . Thus, the area between z = –1 and z = 1 in a standard normal distribution is normalcdf(-1,1)= 0.6826894809 .

Normal Probabilities

When we know a distribution is approximately normal, we can solve many types of problems.

example: In a standard normal distribution, what is the probability that z < 1.5? (Note that because z is a CRV, P (X = a ) = 0, so this problem could have been equivalently stated “what is the probability that z ≤ 1.5?”)

solution: The standard normal table gives areas to the left of a specified z -score. From the table, we determine that the area to the left of z = 1.5 is 0.9332. That is, P (z < 1.5) = 0.9332. This can be visualized as follows:

Calculator Tip: The above image was constructed on a TI-83/84 graphing calculator using the ShadeNorm function in the DISTR DRAW menu. The syntax is ShadeNorm (lower bound, upper bound,[mean, standard deviation])—only the first two parameters need be included if we want standard normal probabilities. In this case we enter ShadeNorm(–100,1.5) and press ENTER (not GRAPH ). The lower bound is given as – 100 (any large negative number will do—there are very few values more than three or four standard deviations from the mean). You will need to set the WINDOW to match the mean and standard deviation of the normal curve being drawn. The WINDOW for the previous graph is [–3.5, 3.5, 1, –0.15, 0.5, 0.1, 1].

example: It is known that the heights (X ) of students at Downtown College are approximately normally distributed with a mean of 68 inches and a standard deviation of 3 inches. That is, X has N (68,3). Determine

(a) P (X < 65).

solution: (the area to the left of z = – 1 from Table A). On the TI-83/84, the corresponding calculation is normalcdf (-100,-1) = normalcdf(-1000,65,68,30) = 0.1586552596.

(b) P (X > 65).

solution : From part (a) of the example, we have P (X < 65) = 0.1587. Hence, P (X > 65) = 1 – P (X < 65) = 1 – 0.1587 = 0.8413. On the TI-83/84, the corresponding calculation is normalcdf(-1,100)= normalcdf(65,1000,68,3) =0.8413447404.

(c) P (65 < X < 70).

solution: P (65 < X < 70) = = P (–1 > z >0.667) = 0.7486 – 0.1587 = 0.5899 (from Table A, the geometry of the situation dictates that the area to the left of z = – 1 must be subtracted from the area to the left of z = 0.667). Using the TI-83/84, the calculation is normalcdf( -1,0.67) = normalcdf (65,70,68,3) =0.5889. This situation is pictured below.

Note that there is some rounding error when using Table A (see Appendix). In part (c), z = 0.66667, but we must use 0.67 to use the table.

(d) P (70 < X < 75)

solution: Now we need the area between 70 and 75. The geometry of the situation dictates that we subtract the area to the left of 70 from the area to the left of 75. This is pictured below.

We saw from part (c) that the area to the left of 70 is 0.7486. In a similar fashion, we find that the area to the left of 75 is 0.9901 (based on z = 2.33). Thus P (70 < X < 75) = 0.9901 – 0.7486 = 0.2415. The calculation on the TI-83/84 is: normalcdf (70,75,68,3) = 0.2427 . The difference in the answers is due to rounding.

example: SAT scores are approximately normally distributed with a mean of about 500 and a standard deviation of 100. Laurie needs to be in the top 15% on the SAT in order to ensure her acceptance by Giant U. What is the minimum score she must earn to be able to start packing her bags for college?

solution: This is somewhat different from the previous examples. Up until now, we have been given, or have figured out, a z -score, and have needed to determine an area. Now, we are given an area and are asked to determine a particular score. If we are using the table of normal probabilities, it is a situation in which we must read from inside the table out to the z -scores rather than from the outside in. If the particular value of X we are looking for is the lower bound for the top 15% of scores, then there are 85% of the scores to the left of x . We look through the table and find the closest entry to 0.8500 and determine it to be 0.8508. This corresponds to a z -score of 1.04. Another way to write the z -score of the desired value of X is

Thus,

Solving for x , we get x = 500 + 1.04(100) = 604. So, Laurie must achieve an SAT score of at least 604. This problem can be done on the calculator as follows: invNorm(0.85,500,100).

Most problems of this type can be solved in the same way: express the z -score of the desired value in two different ways (from the definition, finding the actual value from Table A, or by using the invNorm function on the calculator), then equate the expressions and solve for x .

Simulation and Random Number Generation

Sometimes probability situations do not lend themselves easily to analytical solutions. In some situations, an acceptable approach might be to run a simulation . A simulation utilizes some random process to conduct numerous trials of the situation and then counts the number of successful outcomes to arrive at an estimated probability. In general, the more trials, the more confidence we can have that the relative frequency of successes accurately approximates the desired probability. The law of large numbers states that the proportion of successes in the simulation should become, over time, close to the true proportion in the population.

One interesting example of the use of simulation has been in the development of certain “systems” for playing Blackjack. The number of possible situations in Blackjack is large but finite. A computer was used to conduct thousands of simulations of each possible playing decision for each of the possible hands. In this way, certain situations favorable to the player were identified and formed the basis for the published systems.

example: Suppose there is a small Pacific Island society that places a high value on families having a baby girl. Suppose further that every family in the society decides to keep having children until they have a girl and then they stop. If the first child is a girl, they are a one-child family, but it may take several tries before they succeed. Assume that when this policy was decided on that the proportion of girls in the population was 0.5 and the probability of having a girl is 0.5 for each birth. Would this behavior change the proportion of girls in the population? Design a simulation to answer this question.

solution: Use a random number generator, say a fair coin, to simulate a birth. Let heads = “have a girl” and tails = “have a boy.” Flip the coin and note whether it falls heads or tails. If it falls heads, the trial ends. If it falls tails, flip again because this represents having a boy. The outcome of interest is the number of trials (births) necessary until a girl is born (if the third flip gives the first head, then x = 3). Repeat this many times and determine how many girls and how many boys have been born.

If flipping a coin many times seems a bit tedious, you can also use your calculator to simulate flipping a coin. Let 1 be a head and let 2 be a tail. Then enter MATH PRB randInt(1,2) and press ENTER to generate a random 1 or 2. Continue to press ENTER to generate additional random integers 1 or 2. Enter randInt(1,2,n) to generate n random integers, each of which is a 1 or a 2. Enter randInt(a,b,n) to generate n random integers X such that a ≤ X ≤ b .

The following represents a few trials of this simulation (actually done using the random number generator on the TI-83/84 calculator):

In this limited simulation the number of boys and girls in the population of 15 families are equal. Based on our simulation, if every familiy kept having children until they have a girl, it would not change the proportion of girls in the population. If the simulation were to be run again, it is unlikely that it would turn out exactly the same. For more reliable estimates, do more trials.

Exam Tip: If you are asked to do a simulation on the AP Statistics exam (and there have been such questions), you will probably be asked to use a table of random numbers rather than the random number generator on your calculator. This is to make your solution understandable to the person reading your solution. A table of random numbers is simply a list of the whole numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 appearing in a random order. This means that each digit should appear approximately an equal number of times in a large list and the next digit should appear with probability 1/10 no matter what sequence of digits has preceded it.

The following gives 200 outcomes of a typical random number generator separated into groups of 5 digits:

example: A coin is known to be biased in such a way that the probability of getting a head is 0.4. If the coin is flipped 50 times, how many heads would you expect to get?

solution: Let 0, 1, 2, 3 be a head and 4, 5, 6, 7, 8, 9 be a tail. If we look at 50 digits beginning with the first row, we see that there are 18 heads (bold-faced below), so the proportion of heads is 18/50 = 0.36. This is close to the expected value of 0.4.

Sometimes the simulation will be a wait-time simulation . In the example above, we could have asked how long it would take, on average, until we get five heads. In this case, using the same definitions for the various digits, we would proceed through the table until we noted five numbers with digits 0–3. We would then write down how many digits we had to look at. Three trials of that simulation might look like this (individual trials are separated by \\):

So, it took 13, 14, and 7 trials to get our five heads, or an average of 11.3 trials (the theoretical expected number of trials is 12.5).

Calculator Tip: There are several random generating functions built into your calculator, all in the MATH PRB menu: rand, randInt, randNorm , and randBin. rand(k) will return k random numbers between 0 and 1; randInt(lower bound, upper bound, k) will return k random integers between lower bound and upper bound inclusive; randNorm(mean, standard deviation, k) will return k values from a normal distribution with mean meanand standard deviation standard deviation ; randBin(n,p,k) returns k values from a binomial random variable having n trials each with probability of success p .

Remember that you will not be able to use these functions to do a required simulation on the AP exam, although you can use them to do a simulation of your own design.

Exam Tip: You may see probability questions on the AP exam that you choose to do by a simulation rather than by traditional probability methods. As long as you explain your simulation carefully and provide the results for a few trials, this approach is usually acceptable. If you do design a simulation for a problem where a simulation is not required , you can use the random number generating functions on your calculator. Just explain clearly what you have done—clearly enough that the reader could replicate your simulation if needed.

Transforming and Combining Random Variables

If X is a random variable, we can transform the data by adding a constant to each value of X , multiplying each value by a constant, or some linear combination of the two. We may do this to make numbers more manageable. For example, if values in our dataset ranged from 8500 to 9000, we could subtract, say, 8500 from each value to get a dataset that ranged from 0 to 500. We would then be interested in the mean and standard deviation of the new dataset as compared to the old dataset.

Some facts from algebra can help us out here. Let μ x and σ x be the mean and standard deviation of the random variable X . Each of the following statements can be algebraically verified if we add or subtract the same constant, a , to or from each term in a dataset (X ± a ), or multiply each term by the same constant b (bX ), or some combination of these (a ± bX ):

  • μa ± bX = a ± bμ x .

example: Consider a distribution with μ X = 14, σ X = 2. Multiply each value of X by 4 and then add 3 to each. Then μ 3+4X = 3 + 4(14) = 59, σ 3+4X = 4(2) = 8.

Rules for the Mean and Standard Deviation of Combined Random Variables

Sometimes we need to combine two random variables. For example, suppose one contractor can finish a particular job, on average, in 40 hours (μ x = 40). Another contractor can finish a similar job in 35 hours (μ y = 35). If they work on two separate jobs, how many hours, on average, will they bill for completing both jobs? It should be clear that the average of X + Y is just the average for X plus the average for Y . That is,

  • μX ±Y = μ X ± μ Y .

The situation is somewhat less clear when we combine variances. In the contractor example above, suppose that

Does the variance of the sum equal the sum of the variances? Well, yes and no. Yes, if the random variables X and Y are independent (that is, one of them has no influence on the other, i.e., the correlation between X and Y is zero). No, if the random variables are not independent, but are dependent in some way. Furthermore, it doesn”t matter if the random variables are added or subtracted, we are still combining the variances. That is,

  • , if and only if X and Y are independent.
  • if and only if X and Y are independent.

Digression: If X and Y are not independent, then , where Σ is the population correlation between X and Y. Σ = 0 if X and Y are independent. You do not need to know this for the AP exam.

Exam Tip: The rules for means and variances when you combine random variables may seem a bit obscure, but there have been questions on more than one occasion that depend on your knowledge of how this is done.

The rules for means and variances generalize. That is, no matter how many random variables you have: μ X1 ±X2± … ±Xn = μ X 1 ± μ X 2±…+ μXn and, if X 1 , X 2 , ..., X n are all independent, .

example: A prestigious private school offers an admission test on the first Saturday of November and the first Saturday of December each year. In 2008, the mean score for hopeful students taking the test in November (X ) was 156 with a standard deviation of 12. For those taking the test in December (Y ), the mean score was 165 with a standard deviation of 11. What are the mean and standard deviation of the total score X + Y of all students who took the test in 2008?

solution: We have no reason to think that scores of students who take the test in December are influenced by the scores of those students who took the test in November. Hence, it is reasonable to assume that X and Y are independent. Accordingly,

Rapid Review

  1. A bag has 8 green marbles and 12 red marbles. If you draw one marble from the bag, what isP (draw a green marble)?

Answer: Let s = number of ways to draw a green marble.

Let f = number of ways to draw a red marble.

  1. A married couple has three children. At least one of their children is a boy. What is the probability that the couple has exactly two boys?

Answer: The sample space for having three children is {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}. Of these, there are seven outcomes that have at least one boy. Of these, three have two boys and one girl. Thus, P (the couple has exactly two boys | they have at least one boy) = 3/7”.

  1. Does the following table represent the probability distribution for a discrete random variable?

Answer: No, because

  1. In a standard normal distribution, what isP (z > 0.5)?

Answer: From the table, we see that P (z < 0.5) = 0.6915. Hence, P (z > 0.5) = 1 – 0.6915 = 0.3085. By calculator, normalcdf (0.5,100) = 0.3085375322.

  1. A random variableX has N (13, 0.45). Describe the distribution of 2 – 4X (that is, each datapoint in the distribution is multiplied by 4, and that value is subtracted from 2).

Answer: We are given that the distribution of X is normal with μ X = 13 and σ X = 0.45. Because μ a ±bX = a ± bμ X , μ 2 –4 X = 2 – 4μ X = 2 – 4(13) = –50. Also, because σ a ±bX = bσ X , σ 2 –4 X = 4σ X = 4(0.45) = 1.8.

Practice Problems

Multiple-Choice

In the table above what are P (A and E) and P (C| E)?

(a) 12/125, 28/125

(b) 12/63, 28/60

(c) 12/125, 28/63

(d) 12/125, 28/60

(e) 12/63, 28/63

For the tree diagram pictured above, what is P (B | X)?

(a) 1/4

(b) 5/17

(c) 2/5

(d) 1/3

(e) 4/5

  1. It turns out that 25 seniors at Fashionable High School took both the AP Statistics exam and the AP Spanish Language exam. The mean score on the Statistics exam for the 25 seniors was 2.4 with a standard deviation of 0.6, and the mean score on the Spanish Language exam was 2.65 with a standard deviation of 0.55. We want to combine the scores into a single score. What are the correct mean and standard deviation of the combined scores?

(a) 5.05; 1.15

(b) 5.05; 1.07

(c) 5.05; 0.66

(d) 5.05; 0.81

(e) 5.05; you cannot determine the standard deviation from this information.

  1. The GPAs (grade point averages) of students who take the AP Statistics exam are approximately normally distributed with a mean of 3.4 and a standard deviation of 0.3. Using Table A, what is the probability that a student selected at random from this group has a GPA lower than 3.0?

(a) 0.0918

(b) 0.4082

(c) 0.9082

(d) -0.0918

(e) 0

  1. The 2000 Census identified the ethnic breakdown of the state of California to be approximately as follows: White: 46%, Latino: 32%, Asian: 11%, Black: 7%, and Other: 4%. Assuming that these are mutually exclusive categories (this is not a realistic assumption), what is the probability that a randomly selected person from the state of California is of Asian or Latino descent?

(a) 46%

(b) 32%

(c) 11%

(d) 43%

(e) 3.5%

  1. The students in problem #4 above were normally distributed with a mean GPA of 3.4 and a standard deviation of 0.3. In order to qualify for the school honor society, a student must have a GPA in the top 5% of all GPAs. Accurate to two decimal places, what is the minimum GPA Norma must have in order to qualify for the honor society?

(a) 3.95

(b) 3.92

(c) 3.75

(d) 3.85

(e) 3.89

  1. The following are the probability distributions for two random variables, X and Y :

If X and Y are independent, what is P (X = 5 and Y = 4)?

(a)

(b)

(c)

(d)

(e)

  1. The following table gives the probabilities of various outcomes for a gambling game.

What is the player”s expected return on a bet of $1?

(a) $0.05

(b) –$0.60

(c) –$0.05

(d) –$0.10

(e) You can”t answer this question since this is not a complete probability distribution.

  1. You own an unusual die. Three faces are marked with the letter “X,” two faces with the letter “Y,” and one face with the letter “Z.” What is the probability that at least one of the first two rolls is a “Y”?

(a)

(b)

(c)

(d)

(e)

  1. You roll two six-sided dice. What is the probability that the sum is 6 given that one die shows a 4?

(a)

(b)

(c)

(d)

(e)

Free-Response

  1. Find μ X and σ X for the following discrete probability distribution:
  2. Given that P (A) = 0.6, P (B) = 0.3, and P (B|A) = 0.5.

(a) P (A and B) = ?

(b) P (A or B) = ?

(c) Are events A and B independent?

  1. Consider a set of 9000 scores on a national test that is known to be approximately normally distributed with a mean of 500 and a standard deviation of 90.

(a) What is the probability that a randomly selected student has a score greater than 600?

(b) How many scores are there between 450 and 600?

(c) Rachel needs to be in the top 1% of the scores on this test to qualify for a scholarship. What is the minimum score Rachel needs?

  1. Consider a random variable μX = 3, . Find

(a) μ 3+6 X

(b) σ 3+6 X

  1. Harvey, Laura, and Gina take turns throwing spitwads at a target. Harvey hits the target 1/2 the time, Laura hits it 1/3 of the time, and Gina hits the target 1/4 of the time. Given that somebody hit the target, what is the probability that it was Laura?
  2. Consider two discrete, independent, random variables X and Y with μX = 3, , μY = 5, and . Find μX–Y and σX–Y .
  3. Which of the following statements is (are) true of a normal distribution?
  4. Exactly 95% of the data are within two standard deviations of the mean.
  5. The mean = the median = the mode.

III. The area under the normal curve between z = 1 and z = 2 is greater than the area between z = 2 and z = 3.

  1. Consider the experiment of drawing two cards from a standard deck of 52 cards. Let event A = “draw a face card on the first draw,” B = “draw a face card on the second draw,” and C = “the first card drawn is a diamond.”

(a) Are the events A and B independent?

(b) Are the events A and C independent?

  1. A normal distribution has mean 700 and standard deviation 50. The probability is 0.6 that a randomly selected term from this distribution is above x . What is x ?
  2. Suppose 80% of the homes in Lakeville have a desktop computer and 30% have both a desktop computer and a laptop computer. What is the probability that a randomly selected home will have a laptop computer given that it has a desktop computer?
  3. Consider a probability density curve defined by the line y = 2x on the interval [0,1] (the area under y = 2x on [0,1] is 1). Find P (0.2 ≤ X ≤ 0.7).
  4. Half Moon Bay, California, has an annual pumpkin festival at Halloween. A prime attraction to this festival is a “largest pumpkin” contest. Suppose that the weights of these giant pumpkins are approximately normally distributed with a mean of 125 pounds and a standard deviation of 18 pounds. Farmer Harv brings a pumpkin that is at the 90th percentile of all the pumpkins in the contest. What is the approximate weight of Harv”s pumpkin?
  5. Consider the following two probability distributions for independent discrete random variable X and Y :

If P (X = 4 and Y = 3) = 0.03, what is P (Y = 5)?

  1. A contest is held to give away a free pizza. Contestants pick an integer at random from the integers 1 through 100. If the number chosen is divisible by 24 or by 36, the contestant wins the pizza. What is the probability that a contestant wins a pizza?

Use the following excerpt from a random number table for questions 15 and 16:

  1. Men and women are about equally likely to earn degrees at City U. However, there is some question whether or not women have equal access to the prestigious School of Law. This year, only 4 of the 12 new students are female. Describe and conduct five trials of a simulation to help determine if this is evidence that women are underrepresented in the School of Law.
  2. Suppose that, on a planet far away, the probability of a girl being born is 0.6, and it is socially advantageous to have three girls. How many children would a couple have to have, on average, until they had three girls? Describe and conduct five trials of a simulation to help answer this question.
  3. Consider a random variable X with the following probability distribution:

(a) Find P (X ≤ 22).

(b) Find P (X > 21).

(c) Find P (21 ≤ X < 24).

(d) Find P (X ≤ 21 or X > 23).

  1. In the casino game of roulette, a ball is rolled around the rim of a circular bowl while a wheel containing 38 slots into which the ball can drop is spun in the opposite direction from the rolling ball; 18 of the slots are red, 18 are black, and 2 are green. A player bets a set amount, say $1, and wins $1 (and keeps her $1 bet) if the ball falls into the color slot the player has wagered on. Assume a player decides to bet that the ball will fall into one of the red slots.

(a) What is the probability that the player will win?

(b) What is the expected return on a single bet of $1 on red?

  1. A random variable X is normally distributed with mean μ , and standard deviation s (that is, X has N (μ ,σ)). What is the probability that a term selected at random from this population will be more than 2.5 standard deviations from the mean?
  2. The normal random variable X has a standard deviation of 12. We also know that P (x > 50) = 0.90. Find the mean μ of the distribution.

Cumulative Review Problems

  1. Consider the following histogram:

Which of the following statements is true and why?

  1. The mean and median are approximately the same value.
  2. The mean is probably greater than the median.

III. The median is probably greater than the mean.

  1. You are going to do an opinion survey in your school. You can sample 100 students and desire that the sample accurately reflects the ethnic composition of your school. The school data clerk tells you that the student body is 25% Asian, 8% African American, 12% Latino, and 55% Caucasian. How could you sample the student body so that your sample of 100 would reflect this composition, and what is such a sample called?
  2. The following data represent the scores on a 50-point AP Statistics quiz: 46, 36, 50, 42, 46, 30, 46, 32, 50, 32, 40, 42, 20, 47, 39, 32, 22, 43, 42, 46, 48, 34, 47, 46, 27, 50, 46, 42, 20, 23, 42

Determine the five-number summary for the quiz and draw a boxplot of the data.

  1. The following represents some computer output that can be used to predict the number of manatee deaths from the number of powerboats registered in Florida.

(a) Write the least-square regression line for predicting the number of manatee deaths from the number of powerboat registrations.

(b) Interpret the slope of the line in the context of the problem.

  1. Use the 68-95-99.7 rule to state whether it seems reasonable that the following sample data could have been drawn from a normal distribution: 12.3, 6.6, 10.6, 9.4, 9.1, 13.7, 12.2, 9, 9.4, 9.2, 8.8, 10.1, 7.0, 10.9, 7.8, 6.5, 10.3, 8.6, 10.6, 13, 11.5, 8.1, 13.0, 10.7, 8.8.

Solutions to Practice Problems

Multiple-Choice

  1. The correct answer is (c). There are 12 values in the A and E cell out of the total of 125. When we are given column E, the total is 63. Of those, 28 are C.
  2. The correct answer is (b).

(This problem is an example of what is known as Bayes”s rule. It”s still conditional probability, but sort of backwards. That is, rather than being given a path and finding the probability of going along that path—P (X | B ) refers to the probability of first traveling along B and then along X—we are given the outcome and asked for the probability of having gone along a certain path to get there—P (B | X) refers to the probability of having gotten to X by first having traveled along B. You don”t need to know Bayes”s rule by name for the AP exam, but you may have to solve a problem like this one.)

  1. The correct answer is (e). If you knew that the variables “Score on Statistics Exam” and “Score on Spanish Language Exam” were independent, then the standard deviation would be given by

However, you cannot assume that they are independent in this situation. In fact, they aren”t because we have two scores on the same people. Hence, there is not enough information.

  1. The correct answer is (a).

The calculator answer is normalcdf(–100,3,3.4,0.3) = 0.0912. Note that answer (d) makes no sense since probability values must be nonnegative (and, of course, less than or equal to 1).

  1. The correct answer is (d). Because ethnic group categories are assumed to be mutually exclusive, P (Asian or Latino) = P (Asian) + P (Latino) = 32% + 11% = 43%.
  2. The correct answer is (e). The situation is as pictured below:

From Table A, z x = 1.645 (also, invNorm(0.95) = 1.645) .

Hence, x = 3.4 + (0.3)(1.645) = 3.89. Norma would need a

minimum GPA of 3.89 in order to qualify for the honor society.

  1. The correct answer is (c). . Since they are independent, P (X = 5 and Y = 4) = P(X = 5) · .
  2. The correct answer is (c). The expected value is (–1)(0.6) + (1)(0.25) + (2)(0.15) = –0.05.
  3. The correct answer is (d). P (at least one of the first two rolls is “Y”) = P (the first roll is “Y”) + P (the second roll is “Y”) – P (both rolls are “Y”)

Alternatively, P (at least one of the first two rolls is “Y”) = 1 – P (neither roll is “Y”) =

  1. The correct answer is (b). The possible outcomes where one die shows a 4 are highlighted in the table of all possible sums:

There are 11 cells for which one die is a 4 (be careful not to count the 8 twice), 2 of which are 6s.

Free-Response

1.

This can also be done on the TI-83/84 by putting the X values in L1 and the probabilities in L2 . Then 1-Var Stats L1,L2 will give the above values for the mean and standard deviation.

  1. (a) P (A and B) = P (A) · P (B | A) = (0.6)(0.5) = 0.30.

(b) P (A or B) = P (A) + P (B) – P (A and B) = 0.6 + 0.3 – 0.3 = 0.6
(Note that the 0.3 that is subtracted came from part (a).)

(c) P (B) = 0.3, P (B|A) = 0.5. Since P (B) ≠ P (B|A), events A and B are not independent.

  1. (a) Let X represent the score a student earns. We know that X has approximately N (500, 90). What we are looking for is shown in the following graph.

P (X > 600) = 1 – 0.8667 = 0.1333.

The calculator answer is normalcdf(600,10,000,500.900) =0.1333 (remember that the upper bound must be “big”—in this exercise, 10,000 was used to get a sufficient number of standard deviations above 600).

(b) We already know, from part (a), that the area to the left of 600 is 0.8667. Similarly we determine the area to the left of 450 as follows:

Then

P (450 < X < 600) = 0.8667 – 0.2877 = 0.5790.

There are 0.5790 (9000) ≈ 5211 scores.

[This could be done on the calculator as follows: normalcdf (450,600, 500,90) = 0.5775.]

(c) This situation could be pictured as follows.

The z -score corresponding to an area of 0.99 is 2.33 (invNorm (0.99) on the calculator). So, z x = 2.33. But, also,

Thus,

.

Solving algebraically for x , we get x = 709.7. Rachel needs a score of 710 or higher.

Remember that this type of problem is usually solved by expressing z in two ways (using the definition and finding the area) and solving the equation formed by equating them.

On the TI-83/84, the answer could be found as follows: invNorm(0.99,500, 90) = 709.37.

  1. (a) μ 3 + 6 X = 3 + 6μ X = 3 + 6(3) = 21.

(b) Because .

Thus,

Because we are given that the target was hit, we only need to look at those outcomes. P (the person who hit the target was Laura|the target was hit)

  1. μX–Y = μX – μY = 3 –5 = –2.

Since X and Y are independent, we have . Note that the variances add even though we are subtracting one random variable from another.

  1. I is not true. This is an approximation based on the 68-95-99.7 rule . The actual value proportion within two standard deviations of the mean, to 4 decimal places, is 0.9545.

II is true. This is a property of the normal curve.

III is true. This is because the bell shape of the normal curve means that there is more area under the curve for a given interval length for intervals closer to the center.

  1. (a) No, the events are not independent. The probability of B changes depending on what happens with A · Because there are 12 face cards, if the first card drawn is a face card, then P (B) = 11/51. If the first card is not a face card, then P (B) = 12/51. Because the probability of B is affected by the outcome of A, A and B are not independent.

(b) P (A) = 12/52 = 3/13. P (A| C) = 3/13 (3 of the 13 diamonds are face cards). Because these are the same, the events “draw a face card on the first draw” and “the first card drawn is a diamond” are independent.

  1. The area to the right of x is 0.6, so the area to the left is 0.4. From the table of Standard Normal Probabilities, A = 0.4 ⇒ z x = – 0.25. Also

So,

60% of the area is to the right of 687.5. The calculator answer is given by invNorm(0.4,700,50) =687.33.

  1. Let D = “a home has a desktop computer”; L = “a home has a laptop computer.” We are given that P (D) = 0.8 and P (D and L) = 0.3. Thus,
  2. The situation can be pictured as shown below. The shaded area is a trapezoid whose area is
  3. The fact that Harv”s pumpkin is at the 90th percentile means that it is larger than 90% of the pumpkins in the contest. From the table of Standard Normal Probabilities, the area to the left of a term with a z -score of 1.28 is about 0.90. Thus,

So, Harv”s pumpkin weighed about 148 pounds (for your information, the winning pumpkin at the Half Moon Bay Pumpkin Festival in 2008 weighed over 1528 pounds!). Seven of the 62 pumpkins weighed more than 1000 pounds!

  1. Since , we have P (X = 4) = 1 – P (X = 2) – P (X = 3) = 1 – 0.3 – 0.5 = 0.2. Thus, filling in the table for X , we have

Since X and Y are independent, P (X = 4 and Y = 3) = P (X = 4) · P (X = 3). We are given that P (X = 4 and Y = 3) = 0.03. Thus, P (X = 4) · P (Y = 3) = 0.03. Since we now know that P (X = 4) = 0.2, we have (0.2). P (Y = 3) = 0.03, which gives us .

Now, since ΣP (Y ) = 1, we have P (Y = 5) = 1 – P (Y = 3) – P (Y = 4) – P (Y = 6) = 1 – 0.15 – 0.1 – 0.4 = 0.35.

  1. Let A = “the number is divisible by 24” = {24, 48, 72, 96}. Let B = “the number is divisible by 36” = {36, 72}.

Note that P (A and B) = (72 is the only number divisible by both 24 and 36 ).

P (win a pizza) = P (A or B) = P (A) + P (B) – P (A and B) = .

  1. Because the numbers of men and women in the school are about equal (that is, P (women) = 0.5), let an even number represent a female and an odd number represent a male. Begin on the first line of the table and consider groups of 12 digits. Count the even numbers among the 12. This will be the number of females among the group. Repeat five times. The relevant part of the table is shown below, with even numbers underlined and groups of 12 separated by two slanted bars (\\):

In the five groups of 12 people, there were 3, 6, 3, 8, and 6 women. (Note: The result will, of course, vary if a different assignment of digits is made. For example, if you let the digits 0 – 4 represent a female and 5 – 9 represent a male, there would be 4, 7, 7, 5, and 7 women in the five groups.) So, in 40% of the trials there were 4 or fewer women in the class even though we would expect the average to be 6 (the average of these 5 trials is 5.2). Hence, it seems that getting only 4 women in a class when we expect 6 really isn”t too unusual because it occurs 40% of the time. (It is shown in the next chapter that the theoretical probability of getting 4 or fewer women in a group of 12 people, assuming that men and women are equally likely, is about 0.19.)

  1. Because P (girl) = 0.6, let the random digits 1, 2, 3, 4, 5, 6 represent the birth of a girl and 0, 7, 8, 9 represent the birth of a boy. Start on the second row of the random digit table and move across the line until you find the third digit that represents a girl. Note the number of digits needed to get three successes. Repeat 5 times and compute the average. The simulation is shown below (each success, i.e., girl, is underlined and separate trials are delineated by \\).

It took 4, 7, 3, 6, and 5 children before they got their three girls. The average wait was 5. (The theoretical average is exactly 5—we got lucky this time!) As with Exercise 15, the result will vary with different assignment of random digits.

  1. (a) P (x ≤ 22) = P (x = 20) + P (x = 21) + P (x = 22) = 0.2 + 0.3 + 0.2 = 0.7.

(b) P (x > 21) = P (x = 22) + P (x = 23) + P (x = 24) = 0.2 + 0.1 + 0.2 = 0.5.

(c) P (21 ≤ x < 24) = P (x = 21) + P (x = 22) + P (x = 23) = 0.3 + 0.2 + 0.1 = 0.6.

(d) P (x ≤ 21 or x > 23) = P (x = 20) + P (x = 21) + P (x = 24) = 0.2 + 0.3 + 0.2 = 0.7.

  1. (a) 18 of the 38 slots are winners, so P (win if bet on red)

(b) The probability distribution for this game is

The player will lose 5.2¢, on average, for each dollar bet.

  1. From the tables, we see that P (z < – 2.5) = P (z > 2.5) = 0.0062. So the probability that we are more than 2.5 standard deviations from the mean is 2(0.0062) = 0.0124. (This can be found on the calculator as follows: 2 normalcdf (2.5,1000).)
  2. The situation can be pictured as follows:

If 90% of the area is to the right of 50, then 10% of the area is to the left. So,

Solutions to Cumulative Review Problems

  1. II is true: the mean is most likely greater than the median. This is because the mean, being nonresistant, is pulled in the direction of outliers or skewness. Because the given histogram is clearly skewed to the right, the mean is likely to be to the right of (that is, greater than) the median.
  2. The kind of sample you want is a stratified random sample . The sample should have 25 Asian students, 8 African American students, 12 Latino students, and 55 Caucasian students. You could get a list of all Asian students from the data clerk and randomly select 25 students from the list. Repeat this process for percentages of African American, Latino, and Caucasian students. Now the proportion of each ethnic group in your sample is the same as its proportion in the population.
  3. The five-number summary is [20, 32, 42, 46, 50]. The box plot looks like this:
  4. (a) The LSRL line is: = – 41.430 + 0.12486(#boats).

(b) For each additional registered powerboat, the number of manatee deaths is predicted to increase by 0.12. You could also say that the number increases, on average, by 0.12.

  1. For this set of data, and s = 2.0. Examination of the 25 points in the dataset yields the following.

The actual values in the dataset (16, 25, 25) are quite close to the expected values (17, 23.8, 24.9) if this truly were data from a normal population. Hence, it seems reasonable that the data could have been a random sample drawn from a population that is approximately normally distributed.

CHAPTER 9

Probability and Random Variables

  1. Organizers of a high school class reunion conducted a survey of the 385 graduates. Among the questions asked were “How many children do you have?” and “Do you have any pets?” The data are summarized in the table.

What is the probability that a randomly selected graduate owns a pet, given that the person has 3 or 4 children?

(A) 0.070

(B) 0.175

(C) 0.216

(D) 0.424

(E) 0.784

  1. A particular game awards points with the following probabilities:

E (x ) = 10 and s (x ) = 5. What is the standard deviation of the total number of points after 3 independent turns at this game?

(A) 3.87

(B) 5.00

(C) 8.66

(D) 15.00

(E) 75.00

  1. A tire manufacturer is testing the stopping distance of a new model of tire. The company uses the same vehicle and driver and runs the test on dry pavement. The car travels at 40 miles per hour, and the driver is instructed to stop as quickly as possible when given a signal. The stopping distances were approximately normally distributed, with a mean of 190 feet and a standard deviation of 7 feet. What is the approximate probability that a particular test results in a stopping distance of more than 200 feet?

(A) 0.05

(B) 0.08

(C) 0.15

(D) 0.50

(E) 0.92

  1. A restaurant is starting a promotion in which each customer gets to throw a dart at a specially designed dartboard. The prizes are gift certificates, and the value is determined by the region of the dartboard hit by the dart. (There is a substantial chance of not winning any prize.) The restaurant owner has estimated, based on past performance, that the average of the amounts awarded per customer is $0.75 and the standard deviation of the amounts is $3.15. If the owner decides to double all prize values and add $1.00 to the amount won (even if the customer hits $0), what would be the expected value and standard deviation of the amounts awarded, X , assuming customer skills stay the same?

(A) E (X ) = $1.50, s (X ) = $3.15

(B) E (X ) = $1.50, s (X ) = $6.30

(C) E (X ) = $1.50, s (X ) = $7.30

(D) E (X ) = $2.50, s (X ) = $6.30

(E) E (X ) = $2.50, s (X ) = $7.30

  1. In a particular game, a player has a chance of getting 0, 1, or 2 points on a turn, according to the following probability distribution:

What is the probability that, on two independent turns of the game, a player scores exactly 3 points?

(A) 0.03

(B) 0.06

(C) 0.07

(D) 0.12

(E) 0.40

Answers

  1. C
  2. C
  3. B
  4. D
  5. B