AP Statistics- Unit 4 Exam Review Part 1 – FRQ (Ch. 14 – 17) Do the work on a separate sheet of paper. 1. A new clothing store advertises that during its Grand Opening every customer that enters the store can throw a bouncy rubber cube onto a table that has squares labeled with discount amounts. The table is divided into ten regions. Five regions award a 10% discount, two regions award a 20% discount, two regions award a 30% discount, and the remaining region awards a 50% discount. a. What is the probability that a customer gets more than a 20% discount? b. What is the probability that a customer gets less than a 20% discount? c. What is the probability that the first two customers both get a 50% discount? d. What is the probability that none of the first three customers gets more than a 30% discount? e. What is the probability that the first customer to win a 30% discount is the sixth customer that enters the store? f. What is the probability that there is at least one customer to win a 50% discount among the first five customers that enter the store? g. As you enter the store you watch four people in front of you all win 50% discounts. The store manager tells you how lucky you are to be throwing the cube while it is on a hot streak, but a friend with you says you’re unlucky because the streak can’t continue. Comment on their statements. 2. According to the American Pet Products Manufacturers Association (APPMA) 2003-2004 National Pet Owners Survey, 39% of U.S. households own at least one dog and 34% of U.S. households own at least one cat. Assume 60% of U.S. households own a cat or a dog. a. Draw the Venn Diagram b. What is the probability that a randomly selected U.S. household owns neither a cat nor a dog? c. What is the probability that a randomly selected U.S. household owns both a cat and a dog? d. What is the probability that a randomly selected U.S. household owns a cat if the household has a dog? e. Is having a dog or a cat mutually exclusive? Explain. f. Is having a dog or a cat independent? 3. An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. X is the score. a. Find the missing probability from the following probability distribution X = score 1 2 3 4 5 P(X = x) 0.16 0.22 0.20 0.14 b. P(X ≤ 3) = c. P(1< X ≤4) = d. Find the mean and standard deviation. 4. If P(A) = 0.65 and P(B) = 0.23 and P(A∩B) = 0.15, find the following: a. P(A U B) = b. P(B|A) = c. Are A and B disjoint events? Why or why not? d. Are A and B independent? Why or why not? 5. If P(D) = 0.37, P(C) = 0.41 and D and C are disjoint, what is the probability of D or C? 6. If P(K)= 0.71, P(R)= 0.23 and K and R are independent, what is the probability of K and R? 7. If P(F) = 0.33 and P(H) = 0.28 and P(H|F) = 0.13, find the following: a. P(F and H) = b. P(F or H) = 8. If P(A) = 0.25, P(AB)=0.78, and P(AB) = 0.12, find P(B). 9. On a certain day, there is a 37% chance for rain. The probability of John remembering to bring an umbrella to work and it raining is 15%. What is the probability that John remembers to bring an umbrella given that it rains?
10. In a History class, 50% of the students are males. The probability of a student being a male and a junior is 20%. What is the probability that a student selected at random is a junior given that the student is a male?
Sex
11. A survey of an introductory statistics class in Autumn 2003 asked students whether or not they ate breakfast the morning of the survey. Results are as follows: Breakfast Yes No 66 66 Male 125 74 Female a. What is the probability that a randomly selected student is female? b. What is the probability that a randomly selected student ate breakfast? c. What is the probability that a randomly selected student is a female that ate breakfast? d. What is the probability that a randomly selected female ate breakfast? e. What is the probability that a student who ate breakfast is female? f. Does it appear that whether or not a student ate breakfast is independent of the student’s sex? Explain. 12. A manufacturing firm orders computer chips from three different companies: 10% from Company A; 20% from Company B; and 70% from Company C. Some of the computer chips that are ordered are defective: 4% of chips from Company A are defective; 2% of chips from Company B are defective; and 0.5% of chips from Company C are defective. A worker at the manufacturing firm discovers that a randomly selected computer chip is defective. What is the probability that the computer chip came from Company B? 13. At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. a. Create the probability model. b. What is the expected value for one play? c. Find the standard deviation for one play. d. Find the expected value and standard deviation for 3 plays. 14. The distribution of SAT scores for college-bound male seniors has mean of 1532 and a standard deviation of 312. The distribution of SAT scores for college-bound female seniors has a mean of 1506 and a standard deviation of 304. One male and one female are randomly selected. Assume their scores are independent. a. If their scores are added together what is the new mean and standard deviation? b. What is the mean and standard deviation of the differences of their scores? 15. Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of each of these variables: a. -2X b. 4Y – 7 c. X + Y
d. X – Y
e. X1 + X2
f. 2X – 4Y
X Y
Mean 12 18
SD 5 8
16. Suppose 40% of the drivers have jumper cables. You have driven to a large university campus but left your lights on. Now your car has a dead battery and you don’t have jumper cables. a. What is the probability that you would need to ask 7 people if they could jump your car? b. What is the probability that you would need to ask fewer than 7 if they could jump your car? c. How many people would you expect to have to ask before you found someone to jump your car? d. What is the probability that among 8 drivers 3 would have jumper cables? e. What is the probability that among 6 drivers at least 4 would have jumper cables? f. What is the probability that among 10 drivers more than half would have jumper cables? g. How many drivers would you expect to have jumper cables if there were 12 in the parking lot? h. After never having found someone with jumper cables, and calling you dad for help, you wonder if it isn’t true that 40% of the students of the university have jumper cables. You decide to randomly sample 80 drivers. What is the probability that only 30 have jumper cables?
Unit 4 Exam Review Answers 1. 5 – 10%; 2 – 20%; 2 – 30%; 1 – 50% X 10% 20% P(X = x) 5/10 2/10
30% 2/10
50% 1/10
a. P(>20%) = P(30%) + P(50%) = 2/10 + 1/10 = 0.30 b. P(<20%) = P(10%) = 5/10 = 0.50 2
1 c. P(50% 50%) 0.01 10 3
9 d. P(50% 50% 50% ) 0.729 10 OR: binomial B(3, 0.1) P(X = 0) = 0.729 C
C
C
C
C
C
5
e. P(30%
30%
C
30%
8 2 30%) 0.0655 10 10
C
30%
30% 5
OR: geometric
8 2 P(X = 6) = 0.0655 10 10 5
9 50% 50% 50% 50% ) 1 0.4095 f. 1- P(50% 10 OR: binomial B(5, 0.1) P(X>1) = 1- P(X<0) = 0.4095 C
C
C
C
C
g. Both are incorrect. Each roll of the rubber cube is independent of any other roll. So the probability of getting a 50% discount is the same no matter what the previous values were. 2. Dogs and cats a. D = Family owns at least one dog C = Family owns at least one cat D .26
C .13
(.39 + .34) - .60 = .13
.21 .40
b. P(CC DC ) 0.40 c. P(C D) 0.13 P(C D) 0.13 0.33 d. P(C|D) P(D) 0.39 e. No. P(C D) 0.13 0 . A household can own a cat and a dog at the same time. f. Yes. Knowing that a family has a dog doesn’t change the probability that they own a cat. P(C) = 0.34; P(C|D) = 0.33
3. a. P(X = 3) = 0.28 b. P(X ≤ 3) = 0.16 + 0.22 + 0.28 = 0.66 c. P(1
5. P(D U C) = P(D) + P(C) = 0.78
6. P(K n R) = P(K)*P(R) = 0.1633
7. (a) P(F n H) = P(H|F) * P(F) = 0.0429 (b) P(F U H) = P(F) + P(H) – P(F n H) = 0.5671 8. P(A U B) = P(A) + P(B) – P(A n B) 0.78 = 0.25 + P(B) – 0.12 P(B) = 0.65
9. P(R) = 0.37 P(R n U) = 0.15
P(U|R) = 0.15 = 0.405 0.37
10. P(M) = 0.5 P(M n J) = 0.20
P(J|M) = 0.20 = 0.40 0.50
Sex
Male Female
Breakfast Yes No 66 66 125 74 191 140
132 199 331
11. a. P(F) = 199/331 = 0.6012 b. P(B) = 191/331 = 0.5770 c. P(F ∩ B) = 125/331 = 0.3776 d. P(B|F) = 125/199 = 0.6281 e. P(F|B) = 125/191 = 0.6545 f. No it doesn’t appear that they are independent. Knowing that a student is female changes the probability that they ate breakfast. P(B|F) = 0.6281 ≠ P(B) = 0.5770
12. Tree Diagram
P(B | D)
P(B D) 0.004 0.348 P(A D) P(B D) P(C D) 0.004 0.004 0.0035
13. Probability model for the payout of a prize X = payout P(X=x)
-$2 1496/1500
$73 1/1500
$148 1/1500
$248 1/1500
$498 1/1500
For (b) and (c), just put X-values in L1, probabilities in L2 and then do 1-var stats L1,L2: b. E(X) = -$1.35 c. SD(X) = $15.06 d. E(X + X + X) = -1.35 + -1.35 + -1.35 = -$4.05 SD(X + X + X) =
15.062 15.062 15.062 $26.08
14. SAT Scores M – score for a randomly selected male F – score for a randomly selected female
E(M) = 1532 SD(M) = 312 E(F) = 1506 SD(F) = 304
a. M + F; E(M + F) = 1532 + 1506 = 3038 SD(M + F) =
3122 3042 435.61
b. M – F E(M – F) = 1532 – 1506 = 26 SD(M – F) =
3122 3042 435.61
DON’T WORRY ABOUT (C) for the test! c. Using the work from part b. We assume a normal model with N(26, 435.61). If a female scored higher than a male the value (M – F) would have to be less than zero. So we check the probability that the differences in the scores would be less than zero. P(M-F < 0) 0 26 First find the z-score: z 0.0597 435.61 P(z 0.0597) 0.4762 There is a 47.62% probability that a randomly selected female scored higher than a randomly selected male on the SAT.
15. Random Variables a. -2X E(-2X) = -2(12) = -24 SD(-2X) = 2(5) = 10
b. 4Y – 7 E(4Y – 7) = 4(18) – 7 = 65 SD(4Y – 7) = 4(8) = 32
c. X + Y E(X + Y) = 12 + 18 = 30
d. X – Y E(X – Y) = 12 – 18 = -6
SD(X + Y) =
52 82 9.43
e. X1 + X2 E(X1 + X2) = 12 + 12 = 24 SD(X1 + X2) =
52 52 7.07
SD(X – Y) =
X Y
Mean 12 18
52 82 9.43
f. 2X – 4Y E(2X – 4Y) = 2(12) – 4(18) = -48 E(2X – 4Y) =
22 (52 ) 42 (82 ) 33.53
16. Check: Bernoulli? 1. 2 Outcomes – has jumper cables or doesn’t have jumper cables 2. p = 0.40 3. p is constant 3. 10% Condition – we can assume independence as long as the number of drivers asked is less than 10% of all drivers. a. Geom(0.40) P(X = 7) = (0.60)6(0.40) = 0.0187 b. Geom(0.40) P(X < 7) = P(X ≤ 6) = P(X = 1) + … + P(X = 6) = 0.40 + (0.60)1(0.40) + … + (0.60)5(0.40) = 0.9533 c. Geom(0.40) 1 1 E(X) = 2.5 drivers p 0.40 d. Binom(8,0.40) P(X = 3) = binompdf(8, 0.4, 3) = 0.2787 e. Binom(6,0.40) P X 4 1 P( X 3 ) 1 binomcdf ( 6, 0.4, 3 ) 0.1792 f. Binom(10,0.40) P X 6 1 P( X 5 ) 1 binomcdf ( 10,0.4,5 ) 0.1662 g. Binom(12,0.40) E(X) = np = 12(0.40) = 4.8 drivers
SD 5 8
h. Binom(80,0.40) 1) CHECK: Success/Failure Condition np = 80(0.40) = 32 ≥ 10 nq = 80(0.60) = 48 ≥ 10 Since there were at least 10 expected successes and 10 failures we can use the normal model to approximate the binomial probabilities. N(32,4.382) P(X ≤ 30) = normalcdf(-E99, 30, (80*0.40), √(80 ∗ 0.40 ∗ 0.60) ) = 0.324 2) z = -2 x 32 2 x 23.236 4.382 If there was 23 or less drivers with jumper cables I would question whether the 40% estimate is valid.
