Test 2 SOLUTIONS (Chapters 5 7)

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1 Test 2 SOLUTIONS (Chapters 5 7) I have been sitting at my desk rolling a six-sided die (singular of dice), and counting how many times I rolled a 6. For example, after my first roll, I had rolled 0 6 s. After 10 rolls, I had rolled 1 6. A summary of the results of my first 100 rolls is shown in Attachment 1. Assume that I continued to roll the die for a total of 1000 trials. In the Attachment 1 table, complete the shaded boxes with the experimental probabilities for trial numbers 1, 10, 25, 50 and 100. Estimate the experimental probability after 1000 trials. So, fill in ALL of the shaded boxes in the table. In addition, complete the graph. Assume that you have the experimental data for all 1000 trials. What will the graph look like? Draw it in. Finally, give a brief discussion of how you estimated the experimental probability after 1000 trials, and how you completed the graph. Explain how you knew to do this, and include terms that you learned (i.e. specific types of probabilities, laws) in your explanation for full credit. You know that the theoretical probability of rolling a 6 on a single roll of a die is 1/6. Therefore, after many trials (say, 1000), the experimental probability should approach the theoretical probability, which is exactly what the Law of Large Numbers says. Therefore, to complete the rest of the graph, the probabilities will probably continue to bounce around a little more until you reach maybe 500 trials or so. Then they will probably settle down and really start targeting the theoretical probability of 1/6, or Note: this is exactly the same behavior as the graph that you made on Assignment A recent poll was conducted for CNN by SSRS, an independent research company. Interviews were conducted from November 2 5, 2017 among a sample of 1,021 American adults. One question was Do you approve or disapprove of the way Donald Trump is handling his job as president? Results of the question are shown in the following table, according to whether the respondent was white or non-white. Note: calculate ALL probabilities accurate to three significant figures. 9 Approve Disapprove Don t Know or Undecided White 45% 48% 7% Non-White 18% 77% 5% (a) If one White adult is randomly selected, what is the probability that they approve of the job that Trump is doing? (b) If two Non-White adults are randomly selected, what is the probability that they both disapprove of the job that Trump is doing? P(disapprove & disapprove) = (0.77)(0.77) = or (a) 45%, or 0.45 (b) or Page 1 of 10

2 (c) If four Non-White adults are randomly selected, what is the probability that at least one of them approves of the job that Trump is doing? (c) P(at least one) = 1 P(none) = 1 P(not & not & not & not) For the Non-Whites, the not approvers include the Disapprove and Don t Know or Undecided, or a total of = 82%. P(at least one) = 1 (0.82)(0.82)(0.82)(0.82) = *Source: Refer to Attachment 2, which contains the images of a standard deck of playing cards. Remember that hearts and diamonds are RED, and clubs and spades are BLACK. Also remember that the Jacks, Queens and Kings are called face cards. Note: calculate ALL probabilities accurate to three significant figures. (a) If one card is randomly selected, what is the probability that it is a heart? = 13/52 = ¼ = 0.25 (b) If one card is randomly selected, what is the probability that it is a diamond or a Jack? There are 13 diamonds and 4 Jacks, but one of the Jacks is the overlap (the Jack of diamonds), so a total of 16 ways. P = 16/52 = (c) If one card is randomly selected, what is the probability that it is a 10, given that it is NOT a face card. If you take out the 12 face cards, there are 40 cards remaining, and 4 10s. Therefore, P(10) = 4/40 = 1/10 = 0.1. (d) If two cards are selected without replacement, what is the probability that they are both face cards? There are 12 face cards to start with. P(face & face) = (12/52)(11/51) = (e) If three cards are selected without replacement, what is the probability that at least one of them is a club? P(at least one club) = 1 P(NO clubs) = 1 P(not club & not club & not club). There are 39 cards to begin with that are NOT clubs. 1 P(NO clubs) = 1 (39/52)(38/51)(37/50) = (a) 0.25 (b) (c) (d) (e) Find the number of ways each of the following can occur. (a) Using the digits 1, 2, 3 and 5, how many four digit numbers can be formed if the number must be divisible by 2, and repetition of digits is allowed? Use the Multiplication Rule, or the fill-in-the-blank method, with four blanks. There is only ONE way to choose the last digit, it has to be a 2. The other three digits can be any of the four given = 64 ways. Page 2 of 10 (a) 64

3 (b) Suppose an organization elects its officers from a board of trustees. If there are 30 trustees, how many possible ways could the board elect a president, vice-president, secretary, and treasurer? Choosing without replacement, and order DOES matter, because of the different positions to be filled. Use Permutations with n = 30 and x = 4. 30P 4 = 657,720. (b) 657,720 (c) 126 (c) Maria has five tickets for a concert. She s going to use one of the tickets for herself. She will offer the remaining four tickets to any of nine friends of hers. How many ways can she select four friends to go to the concert with her? Choosing without replacement, and order does NOT matter. Use Combinations with n = 9 and x = 4. 9C 4 = I took all of the final GPAs (grade point averages) for my Stats students in the previous calendar year, and rounded them off to the nearest integer. The random variable x represents the GPA. P(x) gives the associated probability for each value of x, from my student data. x (GPA) P(x) x P(x) Mean = (a) Use the probability distribution to calculate the mean of the random variable, or the mean GPA for my Statistics students in the last calendar year. Calculate accurate to one decimal place. Calculate xp(x), and sum the values, which equals the mean of the random variable. (b) Based on the probability distribution, what was the probability of a student getting a GPA of 3 or more? P(3 or more) = P(3) + P(4) = (c) Based on the probability distribution, would it be unusual for a student to get a GPA of 0? Explain why or why not. P(0) = , therefore it would NOT be unusual, because the probability is greater than (d) What is the shape of the probability distribution? Note: you do NOT need to calculate anything here, just look at the probabilities. If needed, you can make a rough sketch. Just by looking at the probabilities, you can see that it is clearly going to be very left-skewed. (a) 2.8 (b) (c) No, not unusual, P > (d) left-skewed (e) Answer to left. Page 3 of 10

For one thing, people are often trying to get into the nursing program, or other highly competitive programs. They work very hard, and more often than not, end up with a good grade. 16 6.

4 (e) Why do you think the distribution is shaped like that? Not the mechanics of the numbers, but what is the underlying reason? Well, it s because MORE of my students do well in the class, as opposed to those who do not do well. In general, people who take this class are very motivated to get a good grade. For one thing, people are often trying to get into the nursing program, or other highly competitive programs. They work very hard, and more often than not, end up with a good grade According to Pew Research, about 5% of the U.S. civilian workforce consists of unauthorized immigrants (just FYI, this is about 8 million people). This includes people who are working or are unemployed and looking for work. Assume that this probability remains constant. A random sample of 15 American civilian workers is obtained, and the number of people who are unauthorized immigrants is recorded. Notice that this represents a binomial distribution. *Source: Note: report probabilities to 4 decimal places. n = 15, and p = Therefore, just look up the binomial probabilities in Table III. (a) Find the probability that exactly 1 of the 15 American workers is an unauthorized immigrant. P(1) = (b) Find the probability that at least 1 of the 15 American workers is an unauthorized immigrant. At least 1 means 1 or more, so add up: P(1) + P(2) + P(3) + P(4) + P(5), the remaining values are 0. P = (a) (b) Page 4 of 10

(c) For such groups of 15 American workers, find the mean and standard deviation for the number of them who are unauthorized immigrants. (to two significant figures) (c) = 0.75 = np = (15)(0.05) = 0.

Explain your answer, and include the relevant probability. P(3 or more) = 0.0362, therefore it WOULD be unusual, because this probability is < 0.05.

5 (c) For such groups of 15 American workers, find the mean and standard deviation for the number of them who are unauthorized immigrants. (to two significant figures) (c) = 0.75 = np = (15)(0.05) = 0.75 = np(1 p) = 15(0. 05)(0. 95) = (d) Would it be unusual to find that 3 or more of the 15 American workers are unauthorized immigrants? Explain your answer, and include the relevant probability. P(3 or more) = , therefore it WOULD be unusual, because this probability is < Note: it is not appropriate to use the Empirical Rule here, because the distribution is very skewed. = (d) Yes, see explanation to left. 7. Determine the area under the standard normal curve that lies: (a) To the left of 1.36 From the table, area = (a) (b) To the right of 1.59 From the table, area to the left = Therefore, area to the right = = (b) (c) Between 0.15 and 2.89 Area to the left of 2.89 = Area to the left of = Area between = = (c) Page 5 of 10

6 (d) Obtain the z-score for which the area under the standard normal curve to its right is Therefore, the area to its left is = The closest area in the table is , which corresponds to a z-score of (d) 0.74 (e) Determine the two z-scores that divide the area under the standard normal curve symmetrically into a middle 0.85 area and two outside areas (the tails). The closest area in the table to is , which corresponds to a z-score of By symmetry, the z-score on the positive side is (e) -1.44, The website MLB.com lists a variety of statistics for the 2017 baseball season. The hitting statistics include the batting averages for the top 144 league batting leaders. The distribution of the batting averages is normal, with a mean of and a standard deviation of Note: do not round off any of the probabilities. (a) The highest ranked Seattle Mariner on the list is Jean Segura, with a batting average of What proportion of the players have a batting average greater than this? Calculate the z-score associated with a data value x = 0.300: z = x μ σ = = From Table V, the area to the left of z = 1.03 is , therefore the area to the right is = (a) Page 6 of 10

7 (b) The next highest ranked Seattle Mariner is Nelson Cruz, with a batting average of What proportion of the players have a batting average less than this? Calculate the z-score associated with a data value x = 0.288: z = x μ σ = = From Table V, the area to the left of z = 0.60 is (b) (c) Find the 90 th percentile of the batting averages, or the batting average that separates the bottom 90% of the data from the upper 10%. The closest area in the table to 0.90 is , which corresponds to a z-score of Use that z-score to calculate the data value, x: x = μ + zσ = (1. 28)( ) = (c) Page 7 of 10

8 (d) Jose Altuve, from the World Champion Houston Astros, has the highest batting average of Is this value unusual? Explain why or why not, and include a relevant calculation or probability. Because the distribution is normal, you can use the Empirical Rule. Use the mean and standard deviation to calculate the maximum usual: Max usual = µ + 2 = ( ) = Altuve s batting average is above that, so therefore YES, it is unusual. You could also use probabilities. Calculate the z-score associated with a data value x = 0.346: z = x μ σ = = From Table V, the area to the left of z = 2.67 is , therefore the area to the right is = Because this probability is < 0.05, the data value is unusual. (d) Yes, unusual, see explanation to the left. Source: hild&click_text=sortable+player+hitting&game_type='r'&season=2017&season _type=any&league_code='mlb'&sectiontype=sp&stattype=hitting&page=1&ts = &timeframe=&playerType=QUALIFIER Extra Credit #1 (4 points): Suppose that I flip a coin 100 times. What is the theoretical probability of getting exactly 50 Heads out of the 100 flips? This is binomial, with n = 100, and p = 0.5. Find P(50): P(50) = 100C 50(0.5) 50 (0.5) 50 = Page 8 of 10

9 Attachment Rolls of a Single Die INSTRUCTIONS: Fill in all of the shaded boxes below. Complete the graph through trial Trial Number, n Number of times I rolled a 6 P(x) = Experimental Probability of Rolling a = 0/1 = = 1/10 = = 3/25 = = 9/50 = = 16/100 = ? Estimated P(x) = 1/6 = Page 9 of 10

10 Attachment 2 Standard deck of playing cards Page 10 of 10

The point value of each problem is in the left-hand margin. You must show your work to receive any credit, except on problems 1 & 2. Work neatly.

The point value of each problem is in the left-hand margin. You must show your work to receive any credit, except on problems 1 & 2. Work neatly. Introduction to Statistics Math 1040 Sample Exam II Chapters 5-7 4 Problem Pages 4 Formula/Table Pages Time Limit: 90 Minutes 1 No Scratch Paper Calculator Allowed: Scientific Name: The point value of