Ch. 4-5 HW & Test

Each of the digits 0,1,2,3,4,5,6,7,8,9 are written on a slip of paper, and the slips are placed in a hat. If 3 slips of paper are selected at random, find the probability that the 3 numbers are greater than 4.

1/12 5C3 / 10C3 = 10/120

Evaluate the given expression and express the result using the usual format for writing numbers (instead of scientific notation). 32C2

496 mCr = m! / r!(m-r)! = 32! / 2!(32-2)! = 32! / 2•(30!) (everything 30•29 and beyond gets cancelled out) = 32•31 / 2

Suppose 11 cars start at a car race. In how many ways can the top 3 cars finish the race?

990 Permutations rule; order matters! 11P3 = (11!) / (11-3)!

Evaluate the given expression and express the result using the usual format for writing numbers (instead of scientific notation). 5!

= 120

Evaluate the given expression and express the result using the usual format for writing numbers (instead of scientific notation). 45P2

= 1980 mPr = m! / (m-r)! = 45! / (45-2)! = (45•44•43•42...) / (43•42...) (everything gets cancelled out) = 45•44

Simplify: (12) (7) (giant set of parentheses around both numbers)

= 792 Binomial Coefficient: (n) = n! / x!(n-x)! (x) = 12! / 7!(12-7)!

Suppose you hold 23 out of a total of 200 tickets sold for a lottery. The grand-prize winner is determined by the random selection of 1 of the 200 tickets. Let G be the event that you win the grand prize. Find the probability that you will win the grand prize. P(G) = ______

P(G) = 23/200 = 0.115

Decide if the events are mutually exclusive. Event A: Randomly selecting an animal with green eyes. Event B: Randomly selecting an animal with brown eyes. a. Yes, b/c an animal w/ green eyes can't have brown eyes. b. No, b/c an animal w/ green eyes can't have brown eyes. c. No, b/c an animal w/ green eyes can have brown eyes. d. Yes, b/c an animal w/ green eyes can have brown eyes.

a. Yes, b/c an animal w/ green eyes can't have brown eyes.

What's the difference b/w selecting a member at random from a finite population and taking a simple random sample of size 1? a. A random member can't be selected from a finite population. b. Each member of the finite population doesn't have the same probability of being selected. c. The simple random sample is repeatable while the selection from the finite population isn't. d. There isn't a difference.

d. There isn't a difference.

Suppose 80% of kids who visit a doctor have a fever, and 40% of kids with a fever have sore throats. What's the probability that a kid who goes to the doctor has a fever and a sore throat?

0.32 General Multiplication Rule P(A&B) = P(A) • P(B|A) = 0.8 • 0.4

If P(E) = 0.60, P(E or F) = 0.80, and P(E and F) = 0.15, find P(F).

0.35 P(E) = 0.6 P(E & F) = 0.15 P(E or F) = 0.8 Gen. Add'n Rule: P(E or F) = P(E) + P(F) - P(E&F) 0.8 = 0.6 + P(F) - 0.15

For the month of February in a certain city, 91% of the days are cloudy. Also in the month of February in the same city, 65% of the days are cloudy and foggy. What is the probability that a randomly selected day in February will be foggy if it's cloudy?

0.714 P(C) = 0.91 P(C & F) = 0.65 Conditional P Rule: P(F|C) = P(C&F) / P(C) = 0.65/0.91

The owner of a stereo store wants to advertise that he has many different sound systems in stock. The store carries 8 different CD players, 7 different receivers, and 10 different speakers. Assuming a sound system consists of one of each, how many different sound systems can he advertise?

560 8•7•10

The table provides a frequency distribution for the number of rooms in this country's housing units (in thousands). Rooms // # of units 1 - 5825 2 - 1,446 3 - 10,975 4 - 23,378 5 - 27,931 6 - 24,660 7 - 14,682 8+ - 17,247 A = event that the unit has at most 4 rooms B = event that the unit has at least 2 rooms C = event that the unit has b/w 4-6 rooms, inclusive D = event that the unit has more than 5 rooms Among the events A, B, C, D, identify the collections of events that are mutually exclusive. Are each of the following pairings of events mutually exclusive? - A & B: - A & C: - A & D: - B & C: - B & D: - C & D: Are each of the following given events mutually exclusive? - A,B, & C: - A,B, & D: - A,C & D: - B,C, & D: A,B,C, & D:

A. - A & B: no - A & C: no - A & D: yes - B & C: no - B & D: no - C & D: no B. - A,B, & C: No - A,B, & D: No - A,C & D: No - B,C, & D: No A,B,C, & D: No

A recent census found that 52.2% of adults are female, 10.6% are divorced, and 6.2% are divorced females. For an adult selected at random, let F be the event that the person is female, and D be the event that the person is divorced. A. Obtain P(F), P(D), and P(F&D). B. Determine P(F or D). C. Find the probability that a randomly selected adult is male.

A. P(F) = 0.522 --- 52.2% are female, change to a decimal P(D) = 0.106 --- 10.6% are divorced, change to a decimal P(F&D) = 0.055332 --- Special Multiplication Rule; events are independent; P(A&B) = P(A)•P(B) RIGHT ANSWER: 0.062 B. P(F or D) = 0.572668 --- General Addition Rule: P(E or F) = P(E) + P(F) - P(E&F) RIGHT ANSWER: 0.566 C. P(male) = 0.478 --- Complementation Rule: P(not E) = 1 - P(E)

The probability is 0.45 that a traffic fatality involves an intoxicated or alcohol-impaired driver or non-occupant. In 10 traffic fatalities, find the probability that the number, Y, which involve an intoxicated or alcohol-impaired driver or non-occupant is... Aa. The probability that exactly 3 traffic fatalities involve an intoxicated or alcohol-impaired driver or non-occupant is _____. Ab. The probability that at least 3 traffic fatalities involve an intoxicated or alcohol-impaired driver or non-occupant is _____. Ac. The probability that at most 3 traffic fatalities involve an intoxicated or alcohol-impaired driver or non-occupant is _____. B. The probability that between 2-4 traffic fatalities, inclusive, involve an intoxicated or alcohol-impaired driver or nonoccupant is _____. Ca. The mean of Y is _____. Cb. Interpret the mean. a. In every 10 traffic fatalities, exactly this many will involve an intoxicated or alcohol-impaired driver or nonoccupant. b. On average, of 10 traffic fatalities, this many will involve an intoxicated or alcohol-impaired driver or nonoccupant. c. On average, of this many traffic fatalities, 1 will involve an intoxicated or alcohol-impaired driver or nonoccupant. D. The standard deviation of Y is _____.

A. a) Exactly 3: p(Y=3) = 0.1665 "Success" = fatality, p=0.45, n=10 Binomial Probability Formula: p(Y=y)= (n...y) • p^y • (1-p)^n-y = [10! / 3!(10-3)!] • (0.45)^3 • (1-0.45)^10-3 = 0.166478293 b) At least 3: p(Y=3+) = 0.9004 Find p(Y=0,1,2) then subtract from 1. p(Y=2) = 0.076302551 p(Y=1) = 0.02072415 p(Y=0) = 0.002532952 c) At most 3: p(Y=0-3) = 0.2660 Use the values calculated above, and add. ∑p(Y=3,2,1,0) B. Between 2-4: p(∑[Y=2,3,4]) = 0.4811 Find p(Y=4) = 0.238366647 Add p(Y=2,3,4) Ca. µ of Y = 4.5 µ of BRV = (n trials)(p of successes) = (10)(0.45) Cb. b. On average, of 10 traffic fatalities, this many will involve an intoxicated or alcohol-impaired driver or nonoccupant. D. σ = 1.57 Standard Deviation of Binomial Random Variable: σ = sq rt{n • p • (1-p)} σ = sq rt{10 • 0.45 • 0.55)}

According to an article, 47% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly selected adults, find the probability that the number, X, who have experienced a breakup at least once during the last 10 years is: A. exactly 5; at most 5; at least 5 B. at least 1; at most 1 C. Between 2-4, inclusive D. Determine the probability distribution of the random variable X. Complete the table: x | P(X=x) 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

A. a) Exactly 5: 0.0640 RIGHT ANSWER: 0.2280 --- P(X=x) = [n! / (x!(n-x)!)] • p^x • q^n-x Find P(X=5), p = 0.47, q = 0.53, n = 9 b) At most 5: 0.6375 RIGHT ANSWER: 0.8015 --- Find P(X=0,1,2,3,4,5) and ∑ --- P(X=0) = 0.003299764 --- P(X=1) = 0.026335849 --- P(X=2) = 0.093417729 --- P(X=3) = 0.193298319 --- P(X=4) = 0.257123236 c) At least 5: 0.2626 --- Find P(X=5,6,7,8,9) and ∑ --- P(X=6) = 0.134801288 --- P(X=7) = 0.051231756 --- P(X=8) = 0.011357984 --- P(X=9) = 0.00111913 RIGHT ANSWER: 0.4265 B. a) At least 1: 0.9967 --- Do 1 - P(X=0) = 0.996700236 b) At most 1: 0.0296 -- Do P(X=0) + P(X=1) = 0.029635613 C. 0.5438 ∑P(X=2,3,4) D. x | P(X=x) 0 | 0.0033 1 | 0.0263 2 | 0.0934 3 | 0.1933 4 | 0.2571 5 | 0.0640 RIGHT ANSWER: 0.2280 6 | 0.1348 7 | 0.0512 8 | 0.0114 9 | 0.0011

A construction company builds new-home communities in several areas. In one subdivision, it offers 3 models (S, P, V) each in 3 different elevations (A, B, C). A. Draw a tree diagram depicting the possible choices for the selection of a home, including both model and elevation. a. common point, branches off to bond to S, P, V; each is bonded to A, B, C Outcome: SA, SB, SC, PA, PB, PC, VA, VB, VC b. common point, branches off to bond to A, B, S; each is bonded to C, V, P Outcome: AC, AV, AP, BC, BV, BP, SC, SV, SP B. Use the tree diagram to determine the total number of choices for the selection of a home, including both model and elevation.

A. a. common point, branches off to bond to S, P, V; each is bonded to A, B, C Outcome: SA, SB, SC, PA, PB, PC, VA, VB, VC B. 9

Suppose A and B are 2 events. A. What does it mean for event B to be independent of event A? a. Event B is independent of A if P(A&B)=P(A or B) b. Event B is independent of A if P(A&B)=0 c. Event B is independent of A if P(A or B)=1 d. Event B is independent of A if P(B|A)=P(B) B. If event A and event B are independent, how can their joint probability be obtained from their marginal probabilities? a. The joint probability equals the product of the marginal properties; that is, P(A&B) = P(A)•P(B) b. The joint probability equals the sum of the marginal probabilities; that is, P(A&B)=P(A)+P(B) c. The joint probability equals the sum of the marginal probabilities minus the probability that either event will occur; that is, P(A&B)=P(A)+P(B)-P(A or B) d. The joint probability equals the product of the marginal probabilities minus the sum of the marginal probabilities; that is, P(A&B)=P(A)•P(B) - [P(A)+P(B)]

A. d. Event B is independent of A if P(B|A)=P(B) B. Special Multiplication Rule (for 2 independent events) a. The joint probability equals the product of the marginal properties; that is, P(A&B) = P(A)•P(B)

Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable. A. Is the # of people w/ blood type A in a random sample of 31 people discrete or continuous? a. The random variable is continuous. The possible values are 0<=x<=31 b. The random variable is continuous. The possible values are x=0,1,2,...,31 c. The random variable is discrete. The possible values are 0<=x<=31. d. The random variable is discrete. The possible values are x=0,1,2,...,31. B. Is the time it takes for a light bulb to burn out discrete or continuous? a. The random variable is continuous. The possible values are t>0 b. The random variable is continuous. The possible values are t=1,2,3,... c. The random variable is discrete. The possible values are t>0. d. The random variable is discrete. The possible values are t=1,2,3...

A. d. The random variable is discrete. The possible values are x=0,1,2,...,31. B. a. The random variable is continuous. The possible values are t>0

According to a health publication, 15.7% of persons under the age of 65 have no health insurance coverage. Suppose that 4 persons under the age of 65 are randomly selected. A. Complete the table to determine the probability distribution for the number, X, who have no health insurance coverage. x | P(X=x) 0 | 1 | 2 | 3 | 4 | B. The mean of X is ____. C. Choose the correct interpretation of the mean: a. On average, we'd expect this many people out of 4 under the age of 65 to have health insurance coverage. b. On average, we'd expect this many people out of 4 under the age of 65 to have no health insurance coverage. c. On average, we'd expect this many people out of 1 under the age of 65 to have no health insurance coverage.

A. p(X=0) = 0.5050 p(X=1) = 0.3762 p(X=2) = 0.1051 p(X=3) = 0.0130 p(X=4) = 0.0006 Binomial Probability Formula "success" = have no insurance, p = 0.157, q=0.843 B. µ = 0.628 C. b. On average, we'd expect this many people out of 4 under the age of 65 to have no health insurance coverage.

A. What is a permutation? a. A permutation of r objects from a collection of m objects is any ordered arrangement of r of the m objects. b. A permutation of r objects from a collection of m objects is any unordered arrangement of r of the m objects. c. A permutation, NPn, is the # of possible samples of size n from a population of size N. d. A permutation is the product of the first k positive integers (counting numbers) and is denoted k!. B. What is a combination? a. A combination of r objects from a collection of m objects is any unordered arrangement of r of the m objects. b. A combination of r objects from a collection of m objects is any ordered arrangement of r of the m objects. c. A combination of a counting number is obtained by successively multiplying it by the next smaller counting number until reaching 1. d. A combination, NCn, is the number of possible samples of size n from a population of size N. C. What is the major distinction between the 2? a. Order matters in combinations but not in permutations. b. More combinations of r objects from a collection of m objects can be made than permutations. c. Order matters in permutations but not in combinations. d. Permutations can only be used when r is greater than 10.

A. a. A permutation of r objects from a collection of m objects is any ordered arrangement of r of the m objects. B. a. A combination of r objects from a collection of m objects is any unordered arrangement of r of the m objects. C. c. Order matters in permutations but not in combinations.

Rob, Dom, Clarice, Marco and John work for a publishing company. The company wants to send 2 employees to a stats conference. To be fair, the company decides that the 2 who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. A. Determine the sample space of the experiment. List all possible simple random samples of size n=2. Choose the correct answer. a. RD, RC, RM, RJ, DC, DM, DJ, CM, CJ, MJ b. RD, RC, RM, RJ, DC, DM, DJ, CM, CJ, MJ, RR, DD, CC, MM, JJ c. RD, RC, RM, RJ d. RD, RC, RM, RJ, DC, DM, DJ, CM, CJ, MJ, DR, CR, MR, JR, CD, MD, JD, MC, JC, JM B. What is the probability that Dom and Clarice attend the conference? C. What is the probability that Roberto attends the conference?

A. a. RD, RC, RM, RJ, DC, DM, DJ, CM, CJ, MJ B. 0.1 C. 0.4

2 cards are drawn at random from an ordinary deck of 52 cards. Determine the probability that both cards are spades if A. The 1st card is replaced before the 2nd card is drawn. B. the 1st card isn't replaced before the 2nd card is drawn.

A. 0.063 (13/52)•(13/52) B. 0.059 (13/52)•(12/51)

A bag of 33 tulip bulbs contains 12 red ones, 11 yellow ones and 10 purple ones. A. What is the probability that 2 randomly selected tulip bulbs are both red? B. What is the probability that the 1st bulb selected is red, and the 2nd yellow? C. What is the probability that the 1st bulb selected is yellow and the 2nd red? D. What's the probability that 1 bulb is red and the other yellow?

A. 0.125 This is a Conditional Probability Gen. Multiplication Rule [dependent events] = P(E&F)=P(E)•P(F|E) P(both red) = P(1st red) • P(2nd red|1st red) = (12/33) • (11/32) = 0.125 b/c there's 12 red roses so 12/33 for 1st draw. after that, there's only 11 red roses left and a total of 32 roses left, so 11/32 for 2nd draw B. 0.125 Also a Conditional Probability Gen. Multiplication Rule [dependent events] P(red then yellow) = P(1st red) • P(2nd yellow|1st red) = (12/33) • (11/32) b/c there's 12 red roses, so 12/33 for 1st draw. after that, there's still all 11 yellow roses left, but only 32 total roses left, so 11/32 for 2nd draw C. 0.125 Gen. Multiplication Rule P(yellow then red) = P(1st yellow) • P(2nd red|1st yellow) = (11/33) • (12/32) D. 0.25 There's 2 ways to pick 1 yellow and 1 red; yellow or red can be 1st. These events are MUTUALLY EXCLUSIVE. Addition Rule for ME events: P(1 red & 1 yellow): P(red then yellow) + P(yellow then red) = 0.125 + 0.125 = 0.25

Info on the weights and years of experience for a football team was obtained. The given contingency table provides a cross-classification of those data. Compute the following conditional probabilities directly; that is, don't use the conditional probability rule. A player on the football team is selected at random. Complete a-e. .... | Rookie (Y1) | 1-5 (Y2) | 6-10 (Y3) | 10+ (Y4) | Total <200 (W1) 3 | 5 | 1 | 0 | 9 2-300 (W2) 8 | 13 | 15 | 8 | 44 300+ (W3) 0 | 8 | 6 | 0 | 14 Total 11 | 26 | 22 | 8 | 67 A. Find the probability that the player selected is a rookie. P(rookie) = _____ B. Find the probability that the player selected weighs under 200 lbs. P(weighs under 200) = ____ C. Find the probability that the player selected is a rookie, given that he weighs under 200 lbs. P(rookie|weighs under 200) = _____ D. Find the probability that the player selected weighs under 200 lbs, given that he is a rookie. P(weighs under 200|rookie) = ____ E. Interpret your answers in a-d in terms of percentages. Part A indicates that ___% of players are rookies. Part B indicates that ___% of players weigh under 200 lbs. Part C indicates that ___% of the players who weigh under 200 lbs are rookies. Part D indicates that ___% of the players who are rookies weigh under 200 lbs.

A. 0.164 B. 0.134 C. 0.333 D. 0.273 E. 16.4%, 13.4%, 33.3%, 27.3%

The table provides a freq. dist. for the # of rooms (in thousands) in this country's housing units. Rms | # of units 1 | 559 2 | 1,489 3 | 10,915 4 | 23,347 5 | 27,949 6 | 24,612 7 | 14,672 8+ | 17,230 A housing unit is selected at random. Find: A. the probability that the housing unit obtained has 4 rooms. B. the probability that the housing unit obtained has more than 4 rooms. C. the probability that the housing unit obtained has 1 or 2 rooms. D. the probability that the housing unit obtained has fewer than 1 room. E. the probability that the housing unit obtained has 1 or more rooms.

A. 0.193 23347 / 120773 B. 0.699 (27949 + 24612 + 14672 + 17230)/120773 C. 0.017 (559 + 1489) /120773 D. 0 E. 1

The table provides a frequency distribution for the number of rooms in this country's housing units (in thousands). Rooms // # of units 1 - 526 2 - 1,453 3 - 10,964 4 - 23,325 5 - 27,905 6 - 24,630 7 - 14,699 8+ - 17,277 A housing unit is selected at random. Find the following probabilities. A. Find the probability that the housing unit obtained has 4 rooms. B. Find the probability that the housing unit obtained has more than 4 rooms. C. Find the probability that the housing unit obtained has 1-2 rooms. D. Find the probability that the housing unit obtained has fewer than 1 room. E. Find the probability that the housing unit obtained has 1+ rooms.

A. 0.193 B. 0.700 C. 0.016 D. 0 E. 1

The data represent the # of live multiple-delivery births (3+ babies) in a particular year for women 15-54 years old. Age // # Multiple Births 15-19: 94 20-24: 515 25-29: 1620 30-34: 2834 35-39: 1848 40-44: 371 45-54: 111 A. Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother 30-39 years old. P(30-39) = ______ B. Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who wasn't 30-39 years old. P(not 30-39) = ______ C. Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who was less than 45 years old. P(less than 45) ~= _____ D. Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother who was at least 20 years old. P(at least 20) =

A. 0.633 B. 0.367 C. 0.985 D. 0.987

A space shuttle is made up of 755 "critically 1" item must have 99.99% reliability, according to standards, meaning that the probability of failure for such an item is 0.0001. Determine the probability that A. None of the "critically 1" items would fail. B. At least 1 "critically 1" item would fail.

A. 0.927 99.99% reliability = P(reliable) = 0.9999 These are INDEPENDENT events, Multiplication Rule: P(A)•P(B)•P(C)... = .9999^755 B. 0.073 P(not reliable) = 1 - P(reliable) = 1-0.927

A student takes a T/F test consisting of 8 questions. Assume the student guesses at each question and find the probability that A. The student gets at least 1 question correct. B. The student gets a 50% or better on the exam.

A. 0.996094 Complementation Rule: P(at least 1 correct) = 1 - P(none correct) Find P(none correct); use Combo Rule P(none) = 0 objects from a collection of 8 objects, so 8C0 = 8! / 0!(8-0)! = 1 Now use BCR to find total # poss. results of questions. There's 2 poss. results (correct/incorrect) for 8 questions, so 2^8 = 256 P(none) = 8C0 / 256 = 1/256 = 0.003906 P(at least one) = 1 - 0.003906 = 0.996094 B. 0.6368 1. Find P(getting exactly 4 questions correct); use Combo Rule: 8C4 = 8! / 4!(8-4)! = 70 Then divide by total # poss. for results of 8 questions -- 70/256 = 0.2734 = P(getting 4 questions correct) 2. Find P(5 ?s) = 8C5 = 8! / 5!(8-5)! = 56 Then divide...56/256 = 0.2188 3. Repeat: P(6) = 0.1094 P(7) = 0.0313 P(8) = 0.0039 4. Find the sum of all of those probabilities, and that's the probability of getting a 50% or better. 0.2734 + 0.2188 + 0.1094 + 0.0313 + 0.0039 = 0.6368

These are the possible outcomes when a couple has 3 children. 1st | 2nd | 3rd B - B - B B - B - G B - G - B B - G - G G - B - B G - B - G G - G - B G - G - G Find the probability of: A. Among 3 children, there are exactly 3 boys. B. Among 3 children, there're exactly 0 boys. C. Among 3 children, there's exactly 1 boy.

A. 1/8 B. 1/8 C. 3/8

A hand of 5-card draw poker consists of an unordered arrangement of 5 cards from an ordinary deck of 52 playing cards. A. How many 5 card draw poker hands are possible? B. How many different hands consisting of three 9's and two 8's are possible? C. The hand in B is an example of a 4 of a kind. How many different 4 of a kinds are possible? D. Calculate the probability of being dealt a 4 of a kind.

A. 2,598,960 UNordered = combo 52C5 = 52! / 5!(52-5)! B. 24 First find # of different ways to get three 9's: 4C3 = 4! / 3!(4-3)! = 4 Next, find # of diff ways to get two 8's: 4C2 = 4! / 2!(4-2)! = 6 Then, use basic counting rule --> 4•6 = 24 C. 3744 There's 13 different ways to pick the 3 card denomination • 12 different ways to pick the 2 card denomination • 24 ways to pick the 3 cards of one denomination and 2 cards of another domination = 3744 D. 0.00144 Use f/N rule: 3744/2598960

The table displays frequency distribution for the number of crew members on each shuttle mission over a 10 year period. Let X denote the crew size on a randomly selected shuttle from this time period. Crew size: 4 | 5 | 6 | 7 | 8 Frequency: 3 | 4 | 34|18|19 A. What are the possible values of the random variable X? B. Use random-variable notation to represent the event that the shuttle mission obtained has a crew size of 7. {___} C. Find P(X=5) D. Obtain the probability distribution of X.

A. 4,5,6,7,8 B. {X = 7} C. 0.051 Suppose experiment has N poss. outcomes, all equally likely. N = ∑(all frequencies). An event that can occur in f ways has P f/N of occurring. N=78, f=4. P(X=5) = 4/78 = 0.051 D. P(X=4) = 0.038 P(X=5) = 0.051 P(X=6) = 0.436 P(X=7) = 0.231 P(X=8) = 0.244 A probability distribution is a listing of all the poss. values & corresponding P's of a DRV. To get the P distribution of X, determine the P of each value of the random variable X (f/N)

The table provides a freq. dist. for the # of rooms (in thousands) in this country's housing units. Rms | # of units 1 | 534 2 | 1,424 3 | 10,959 4 | 23,365 5 | 27,907 6 | 24,694 7 | 14,603 8+ | 17,221 A = event that the unit has at most 4 rooms B = event that the unit has at least 2 rooms C = event that the unit has 5-7 rooms, inclusive D = event that the unit has 7+ rooms A. Determine the # of outcomes (housing units) that constitute the event (not A). B. Determine the # of outcomes (housing units) that constitute the event (A&B). C. Determine the # of outcomes (housing units) that constitute the event (C or D).

A. 84,425 B. 35,748 C. 84,425

The frequency distribution provies a certain condition prevalence frequency distribution for 61 regions. Condition (%) // Frequency 4-under 5: 7 5-under 6: 14 6-under 7: 18 7-under 8: 12 8-under 9: 5 9-under 10: 4 10-under 11: 1 A = event w/ prevalence of at least 8% B = event w/ prevalence of less than 7% C = event w/ prevalence of at least 5%, but less than 10% D = event that has a prevalence of less than 9% A. Determine the # of outcomes (regions) that constitute the event (not C). B. Determine the # of outcomes (regions) that constitute the event (A&B). C. Determine the # of outcomes (regions) that constitute the event (C or D). D. Determine the # of outcomes (regions) that constitute the event (C&B).

A. Event (not C) has 10 regions. B. Event (A&B) has 0 regions. C. Event (C or D) has 63 regions. D. Event (C&B) has 32 regions.

Shown are the # of trials and success probability for some Bernoulli trials. Let X denote the total number of successes. n=4, and p=0.7 A. Determine P(X=2) using the binomial probability formula. B. Determine P(X=2) using a table of binomial probabilities. C. Compare this answer to part A. a. The 2 probabilities are ~ equal at 3 decimal places. b. The probability from part A is much larger than the probability from part B. c. The 2 probabilities are exactly equal at 3 decimal places. d. The probability from part B is much larger than the probability from part A.

A. P(X=2) = 0.265 Binomial Probability Formula: P(X=x) = (n)p^x•(1-o)^[n-x] (x) B. 0.265 C. c. The 2 probabilities are exactly equal at 3 decimal places.

For each of the probability histograms of binomial distributions, specify whether the success probability is less than, equal to, or greater than 0.5 a) [right skewed] points at: (0,0.125) (1,0.3) (2,0.33) (3,0.19) (4,0.05) (5,0.01) b) [symmetrical] points at: (0,0.25) (1,0.15) (2,0.32) (3,0.32) (4,0.15) (5,0.025)

A. The success probabiity, p, is less than 0.5 B. The success probability, p, is equal to 0.5

A roulette wheel contains 38 numbers (18 red, 18 black, 2 green). When the roulette wheel is spun, the ball is equally likely to land on any of the 38 numbers. Suppose that you bet $7 on green. If the ball lands on a green number, you win $7; otherwise you lose your $7. Let X be the amount you win on your $7 bet. Then X is the random variable whose probability distribution is as follows. x | 7 | -7 P(X=x)|0.053|0.947 A. Find the expected value of the random variable X. b. Approximately how much would you expect to lose if you bet $7 on green 100 times? 1000 times?

A. µ = -6.258 µ = ∑x•P(X=x) µ = (7•0.053) + (-7•0.947) = -6.258 Thus, on avg., you'll lose $6.258 per play, assuming you bet $7 each time. B. Multiply expected loss per game • # of times you play Bet $7 on green 100 times: expect to lose $625.80 Bet $7 on green 1000 times: expect to lose $6258.00

The random variable X is the crew size of a randomly selected shuttle mission. Its probability distribution is shown. x | 2 | 3 | 4 | 5 | 6 | 7 | 8 P(X=x) |0.063|0.036|0.039|0.343|0.184|0.279|0.056 A. Find and interpret the mean of the random variable. µ = ______ B. Interpret the mean. a. As the # of obs., n, DEC, the mean of the obs. will approach the mean of the random variable. b. As the # of obs., n, INC, the mean of the obs. will approach the mean of the random variable. c. The observed value of the random variable will be less than the mean of the random variable in most obs. d. The observed value of the random variable will be equal to the mean of the random variable in most obs. C. Obtain the standard deviation of the random variable. σ = ______

A. µ = 5.61 µ = ∑x•P(X=x) (2•0.063) + (3•0.036) + (4•0.039) + (5•0.343) + (6•0.184) + (7•0.279) + (8•0.056) B. b. As the # of obs., n, INC, the mean of the obs. will approach the mean of the random variable. C. σ = 1.477 sq rt{∑[x^2•P(x)] - µx^2} = sq rt{ [(2^2•0.063) + (3^2•0.036) + (4^2•0.039) + (5^2•0.343) + (6^2•0.184) + (7^2•0.279) + (8^2•0.056)] - (5.61^2)}

In a sporting event, the championship is won by the 1st team to win 4 games. The lengths of the championship games are given: # of games // Freq // Rel Freq 4 - 19 - 0.198 5 - 22 - 0.229 6 - 23 - 0.240 7 - 32 - 0.333 Let X denote the number of games that it takes to complete the championship games, and Y denote the number of games that it took to complete a randomly selected championship from among those considered in the table. A. Determine the mean and standard deviation of the random variable Y.

A. µY = 5.708 µ = ∑x•P(X=x) First, multiply each possible value by its probability (relative frequency)(P(Y=y)) 4•0.198 = 0.792 5•0.229 = 1.145 6•0.240 = 1.44 7•0.333 = 2.331 Sum of them = 5.708 B. σY = 1.126 σ = sq rt{∑[y^2 • P(Y)] - µY^2} = sq rt{[(4^2•0.198) + (5^2•0.229) + (6^2•0.240) + (7^2•0.333)] - 5.708^2} Thus, the mean # of games played to win the championship is 5.708, with a standard deviation of 1.126.

A factory manager collected data on the number of equipment breakdowns per day. From those data, she derived the probability distribution shown, where W denotes the number of breakdowns on a given day. w | 0 | 1 | 2 P(W=w)|0.75|0.15|0.10 A. Determine µw and σW. B. On average, how many breakdowns occur per day? C. About how many breakdowns are expected during a 1 year period, assuming 250 work days per year?

A. µw = 0.35 σw = 0.654 B. 0.35 C. 87.5 (0.35•250)

The random variable X is the # of color TV sets owned by a randomly selected household with an annual income between $15k-$29,999k. Its probability distribution: x | P(X=x) 0 | 0.065 1 | 0.278 2 | 0.214 3 | 0.226 4 | 0.217 Aa. Find the mean. Ab. Interpret: a. The observed value of the random variable will be less than the mean of the random variable in most observations. b. The observed value of the random variable will be equal to the mean of the random variable in most observations. c. As the # of observations, n, DEC, the mean of the observations will approach the mean of the random variable. d. As the # of observations, n, INC, the mean of the observations will approach the mean of the random variable. B. Obtain the standard deviation of the random variable. C. Draw a probability histogram for the random variable. Choose the correct graph. a. (0,.05) (1,0.29) (2,.2) (3,.21) (4,.2) b. (0,.29) (1,.22) (2,.05) (3,.21) (4,.21) c. (0,.21) (1,.22) (2,.21) (3,.29) (4,.07) d. (0,.21) (1, .22) (2,.21) (3,.07) (4,.29)

Aa. µx = 2.252 Mean of Discrete Random Variable: µ = ∑(x•P(X=x)) µ = (0•0.065)+(1•0.278)+(2•0.214)+(3•0.226)+(4•0.217) Ab. d. As the # of observations, n, INC, the mean of the observations will approach the mean of the random variable. B. σ = 1.252 = √{∑[x²•P(x)] - µx²} = √{[(0²•0.065) + (1²•-.278)+(2²•0.214) + (3²•0.226) + (4²•0.217)] - (2.252²)} C. a. (0,.05) (1,0.29) (2,.2) (3,.21) (4,.2)

You draw a card at random from a standard deck of 52 cards. Find the following conditional probabilities. A. The card's a diamond, given that it's red. B. The card's red, given that it's a diamond. C. The card is a 10, given that it's red. D. The card is a king, given that it's a face card.

Conditional Probability Rule: P(B|A) = P(A&B) / P(A) a. 1/2 (13/52) / (26/52) b. 1 (13/52) / (13/52) c. 1/13 (2/52) / (26/52) d. 1/3 (4/52) / (12/52)