Stat 118 - Chapter 5 Homework, Quiz, & Excel Lab
Solve the following: a. 21!/16! b. 7P3 c. 10C9
a) 2,441,880 b) 210 c) 10
Thirty-six percent of all light emitting diode (LED) displays are manufactured by Samsung. What is the probability that in a collection of five independent LED HDTV purchases, at least one is a Samsung? (Round your answer to 3 decimal places.)
1 - (1 - 0. 36)^5 = 0.8926 0.893
A computer password consists of four characters. The characters can be one of the 26 letters of the alphabet. Each character may be used more than once. How many different passwords are possible?
26^4 = 456,976
Some people are in favor of reducing federal taxes to increase consumer spending and others are against it. Two persons are selected and their opinions are recorded. Assume no one is undecided.Find the number of possible outcomes.
4
An overnight express company must include seven cities on its route. How many different routes are possible, assuming that it does not matter in which order the cities are included in the routing?
7! = 5,040 MATH --> PRB then #4 !
Several years ago, Wendy's Hamburgers advertised that there are 256 different ways to order your hamburger. You may choose to have, or omit, any combination of the following on your hamburger: mustard, ketchup, onion, pickle, tomato, relish, mayonnaise, and lettuce. Is the advertisement correct?
8 categories = mustard, ketchup, onion, pickle, tomato, relish, mayonnaise and lettuce 2^8 = 256 Yes, Wendy's calculations were correct.
A quality control inspector selects a part to be tested. The part is then declared acceptable, repairable, or scrapped. Then another part is tested. Find the number of possible of outcomes of this experiment regarding two parts.
Acceptable - A Repairable - R Scrapped - S Outcomes 1. AA 2. AR 3. AS 4. RR 5. RA 6. RS 7. SS 8. SA 9. SR 9 possible outcomes
The marketing research department at PepsiCo plans a national survey of 2,500 teenagers regarding a newly developed soft drink. Each teenager will be asked to compare it with his or her favorite soft drink. What is the experiment?
Asking teenagers their reactions to the newly developed soft drink.
A firm will promote two employees out of a group of six men and three women. What probability concept would be used to assign probabilities to the outcomes?
Classical probability The firm is selecting the group of employees to be randomly selected. Each person is equally likely to be promoted.
An investor buys 100 shares of AT&T stock, and records its change in price daily. Which concept of probability would you most closely associate with recording and tracking the daily change in the price of the stock?
Empirical
ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these three groups in the proportions 30%, 50%, and 20%, respectively. The probability a "good" driver will have an accident is 0.01, the probability a "medium" risk driver will have an accident is 0.03, and the probability a "poor" driver will have an accident is 0.10. The company sells Mr. Brophy an insurance policy and he has an accident. What is the probability Mr. Brophy is: a. A "good" driver? (Round your answer to 3 decimal places.) b. A "medium" risk driver? (Round your answer to 3 decimal places.) c. A "poor" driver? (Round your answer to 3 decimal places.)
Given that P(good)=0.3 P(medium)=0.5 P(poor)=0.2 P(accident | good)=0.01 P(accident | medium)=0.03 P(accident | poor)=0.1 First find P(accident): P(accident)=P(accident | good) * P(good) + P(accident | medium) * P(medium) + P(accident | poor) * P(poor) 0.3*0.01 + 0.5*0.03 + 0.2*0.1 = 0.038 a)P(good/accident) = P(accident/good) * P(good) / P(accident) =(0.3*0.01)/0.038=0.003/0.038=0.079 b)P(medium/accident)=P(accident/medium)*P(medium)/P(accident) =(0.5*0.03)/0.038=0.015/0.038=0.395 c)P(poor/accident)=P(accident/poor)*P(poor)/P(accident) =(0.2*0.1)/0.038=0.02/0.038=0.526
The credit department of Lion's Department Store in Anaheim, California, reported that 22% of their sales are cash, 30% are paid with a credit card, and 48% with a debit card. Twenty percent of the cash purchases, 83% of the credit card purchases, and 58% of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the probability that she paid cash? (Round your answer to 3 decimal places.)
Given: P(cash) = 0.22 P(credit card) = 0.30 P(debit card) = 0.48 P(more than $50 | cash) = 0.20 P(more than $50 | credit card) = 0.83 P(more than $50 | debit card) = 0.58 P(more than $50) = P(more than $50 | cash)*P(cash) + P(more than $50 | credit card)*P(credit card) + P(more than $50 | debit card)*P(debit card) P(more than $50) = (0.20 * 0.22) + (0.83 * 0.30) + (0.58 * 0.30) = 0.5714 P(cash | more than $50) = [P(more than $50 | cash) * P(cash)] / P(more than $50) P(cash | more than $50) = 0.20 * 0.22 / 0.5714 = 0.077
The credit department of Lion's Department Store in Anaheim, California, reported that 27% of their sales are cash, 26% are paid with a credit card, and 47% with a debit card. Twenty percent of the cash purchases, 81% of the credit card purchases, and 63% of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the probability that she paid cash? (Round your answer to 3 decimal places.)
Given: P(cash) = 0.27 P(credit card) = 0.26 P(debit card) = 0.47 P(more than $50 | cash) = 0.20 P(more than $50 | credit card) = 0.81 P(more than $50 | debit card) = 0.63 P(more than $50) = P(more than $50 | cash)*P(cash) + P(more than $50 | credit card)*P(credit card) + P(more than $50 | debit card)*P(debit card) P(more than $50) = (0.20 * 0.27) + (0.81 * 0.26) + (0.47 * 0.63) = 0.5607 P(cash | more than $50) = [P(more than $50 | cash) * P(cash)] / P(more than $50) P(cash | more than $50) = 0.20 * 0.27 / 0.5607 = 0.096
All Seasons Plumbing has two service trucks that frequently need repair. If the probability the first truck is available is 0.80, the probability the second truck is available is 0.55, and the probability that both trucks are available is 0.44What is the probability neither truck is available? (Round your answer to 2 decimal places.)
P(First Truck) = 0.80 P(Second Truck) = 0.55 P(First AND Second) = term-90.44 Either Truck P(first or second) = P(first) + P(second) - P(first AND second) P(first or second) = 0.80 + 0.55 - 0.44 = 0.91 Neither Truck P(neither First nor Second) = 1 - P(First or Second) 1 - 0.91 = 0.09
Flashner Marketing Research Inc. specializes in providing assessments of the prospects for women's apparel shops in shopping malls. Al Flashner, president, reports that he assesses the prospects as good, fair, or poor. Records from previous assessments show that 60% of the time the prospects were rated as good, 30% of the time fair, and 10% of the time poor. Of those rated goods, 80% made a profit the first year; of those rated fair, 60% made a profit the first year; and of those rated poor, 20% made a profit the first year. Connie's Apparel was one of Flashner's clients. Connie's Apparel made a profit last year. What is the probability that it was given an original rating of poor? (Round your answer to 3 decimal places.)
P(good) = 0.60 P(fair) = 0.30 P(poor) = 0.10 P(first year profit | good) = 0.80 P(first year profit | fair) = 0.60 P(first year profit | poor) = 0.20 P(first year profit) = P(good)*P(first year profit | good) + P(fair)*P(first year profit | fair) + P(poor)*P(first year profit | poor) P(first year profit) = (0.60*0.80) + (0.30*0.60) + (0.10*0.20) = 0.68 P(poor | profit first year) = [P(poor)*P(first year profit | poor)] / P(first year profit) P(poor | profit first year) = [ 0.10 * 0.20 ] / 0.68 = 0.029
A student is taking two courses, history and math. The probability the student will pass the history course is 0.55, and the probability of passing the math course is 0.64. The probability of passing both is 0.43. What is the probability of passing at least one? (Round your answer to 2 decimal places.)
P(history) = 0.55 P(math) = 0.64 P(history AND math) = 0.43 At least one / either one --> Probability (history OR math) P(history or math) = P(history) + P(math) - P(history AND math) P(history or math) = 0.55 + 0.64 - 0.43 = 0.76
An Internet company located in Southern California has season tickets to the Los Angeles Lakers basketball games. The company president always invites one of the six vice presidents to attend games with him, and claims he selects the person to attend at random. One of the six vice presidents has not been invited to attend any of the last seven Lakers home games. What is the likelihood this could be due to chance? (Round your answer to 3 decimal places.)
P(invited) = 1/6 P(not invited) = 1 - (1/6) = 5/6 P(not invited 7 times in succession) = (5/6)^7 = 0.2790816472 0.279
A computer-supply retailer purchased a batch of 1,000 CD-R disks and attempted to format them for a particular application. There were 857 perfect CDs, 112 CDs were usable but had bad sectors, and the remainder could not be used at all. a. What is the probability a randomly chosen CD is not perfect? (Round your answer to 3 decimal places.) b. If the disk is not perfect, what is the probability it cannot be used at all? (Round your answer to 3 decimal places.)
P(perfect CD) = 857/1000 = 0.857 P(CDs were usable but had bad sectors) = 112/1000 or 0.112 P(could not be used at all) 1000 - (857 + 112) = 31/1000 or 0.031 a) 0.143 P(not perfect) = 1 - 0.857 = 0.143 b) 0.217 P(could not be used at all | not perfect) = P(could not be used at all) / P(not perfect) 0.031 / 0.143 = 0.217
Four women's college basketball teams are participating in a single-elimination holiday basketball tournament. If one team is favored in its semifinal match by odds of 1.55 to 1.45 and another squad is favored in its contest by odds of 2.35 to 1.65, what is the probability that: a. Both favored teams win their games? (Round your answer to 2 decimal places.) b. Neither favored team wins its game? (Round your answer to 4 decimal places.) c. At least one of the favored teams wins its game? (Round your answer to 4 decimal places.)
Probability of odds of 1.55 to 1.45 1.55 / (1.55 + 1.45) = 0.5166 or 31/60 P(win1) = 31/60 or 0.5166 P(lose1) = 29/60 or 0.4833 Probability of odds of 2.35 to 1.65 2.35 / (2.35 + 1.65) = 0.5875 or 47/80 P(win2) = 47/80 P(lose2) = 33/80 a) 0.3035 Probability both favored teams win their games : P(win1) * P(win2) (31/60) * (47/80) = 0.3035 b) 0.1994 P(both lose) = P(lose1) * P(lose2) P(both lose) = (29/60) * (33/80) = 0.199375 or 319/1600 in fraction c) 0.8006 P(at least one win) = 1 - P(both lose) 1 - 0.1994 = 0.8006
The events A and B are mutually exclusive. Suppose P(A)=0.29 and P(B)=0.32 a. What is the probability of either A or B occuring? (Round your answer to 2 decimal places.) b. What is the probability that neither A nor B will happen?
The events X and Y are mutually exclusive, therefore P(A and B) = 0 a) 0.61 P(A or B) = P(A) + P(B) - P(A and B) 0.29 + 0.32 - 0 = 0.47 b) 0.39 P(neither A nor B) = 1 - P(A or B) 1 - 0.61 = 0.39
The events X and Y are mutually exclusive. Suppose P(X) = 0.25 and P(Y) = 0.22. a. What is the probability of either X or Y occurring? b. What is the probability that neither X nor Y will happen?
The events X and Y are mutually exclusive, therefore P(X and Y) = 0 a) 0.47 P(X or Y) = P(X) + P(Y) - P(X and Y) 0.25 + 0.22 - 0 = 0.47 b) 0.53 P(neither X nor Y) = 1 - P(X or Y) 1 - 0.47 = 0.53
Berdine's Chicken Factory has several stores in the Hilton Head, South Carolina, area. When interviewing applicants for server positions, the owner would like to include information on the amount of tip a server can expect to earn per check (or bill). A study of 500 recent checks indicated the server earned the following amounts in tips per eight-hour shift. Amount of Tip Number $0 up to $20 200 20 up to 50 100 50 up to 100 75 100 up to 200 75 200 or more 50 Total 500 a. What is the probability of a tip of $200 or more? b. Are the categories "$0 up to $20," "$20 up to $50," and so on considered mutually exclusive? c. If the probabilities associated with each outcome were totaled, what would that total be? d. What is the probability of a tip of up to $50? e. What is the probability of a tip of less than $200?
a) 0.1 P(200 or more) = Number of times event occurs / Total number of observations P(200 or more) = 50 / 500 = 0.1 b) Yes, they are mutually exclusive. The occurrence of one event means that none of the other events can occur at the same time. c) By definition if the probabilities associated with each outcome were totaled, the total would be 1. d) 0.6 P(up to $50) = [ P($0 up to $20) + P($20 up to $50) ] / Total number of observations (200/500) + (100/500) = 0.6 e) 0.9 P(less than $200) = [ P(0 up to 20) + P(20 up to 50) + P(50 up to 100) + P(100 up to 200) ] / Total number ob observations (200 + 100 + 75 + 75) / 500 = 0.9
A recent survey reported in Bloomberg Businessweek dealt with the salaries of CEOs at large corporations and whether company shareholders made money or lost money. CEO Paid CEO Paid More Than Less Than $1 Million $1 Million TOTAL Shareholders made money 2 11 13 Shareholders lost money 4 3 7 TOTAL 6 14 20 If a company is randomly selected from the list of 20 studied, what is the probability: a. The CEO made more than $1 million? b. The CEO made more than $1 million or the shareholders lost money? c. The CEO made more than $1 million given the shareholders lost money? d. Of selecting two CEOs and finding they both made more than $1 million?
a) 0.3 By empirical probability, P(CEO made more than $1 million) = number of times the evens occurs / Total number of observations P(CEO made more than $1 million) = 6/20 = 0.3 b) 0.45 Using the special rule of addition, P(A or B) = P(A) + P(B) - P(A and B) P(CEO made more than $1 million OR the shareholders lost money) = P(CEO made more than $1 million) + P(the shareholders lost money) - P(CEO made more than $1 million AND the shareholders lost money (6/20) + (7/20) - (4/20) = 0.45 c) 0.5714 By empirical probability, P(CEO made more than $1 million) = P(CEO made more than $1 million AND shareholders lost their money) / Total Shareholders lost money 4 / 7 = 0.5714285714 d) 0.0789 First, use empirical probability. Determine the probability of one CEO making more than $1 million which is 11/32 = 0.34 from question a) Then, use the general rule of multiplication, P(A and B) = P(A) * P(B | A) P(A and B) = (6/20)(5/19) = 0.0789473684 **note** (6-1) /(20 - 1) = 5/19
Refer to the following table. First Event Second Event A1 A2 A3 Total B1 2 1 3 6 B2 1 2 1 4 Total 3 3 4 10 a. Determine P(A1). (Round your answer to 2 decimal places.) b. Determine P(B1 | A2). (Round your answer to 2 decimal places.) c. Determine P(B2 and A3). (Round your answer to 2 decimal places.)
a) 0.30 P(A1) = Total in A1 / Grand Total P(A1) = 3 / 10 = 0.30 b) 0.33 P(B1 | A2) = Number in both / Total in A2 or P(B1 | A2) = P(B1 and A2) / P(A2) 1 / 3 = 0.33 c) 0.10 P(B2 and A3) = Number in both / Grand Total 1/10 = 0.10
A survey of 545 college students asked: What is your favorite winter sport? And, what type of college do you attend? The results are summarized below: Favorite Winter Sport College Type Snowboarding Skiing Ice Skating TOTAL Junior College 68 41 46 155 Four-Year College 84 56 70 210 Graduate School 59 74 47 180 TOTAL 211 171 163 545 Using these 545 students as the sample, a student from this study is randomly selected. a.What is the probability of selecting a student whose favorite sport is skiing? (Round your answer to 4 decimal places.) b.What is the probability of selecting a junior-college student? (Round your answer to 4 decimal places.) c. If the student selected is a four-year-college student, what is the probability that the student prefers ice skating? (Round your answer to 4 decimal places.) d. If the student selected prefers snowboarding, what is the probability that the student is in junior college? (Round your answer to 4 decimal places.) e. If a graduate student is selected, what is the probability that the student prefers skiing or ice skating? (Round your answer to 4 decimal places.)
a) 0.3138 P(Skiing) = Total number of students whose favorite sport is Skiing / Total College students 171 / 545 = 0.3138 b) 0.2844 P(Junior College) = Total Junior College / Total College Students 155 / 545 = 0.2844 Of the 545 College students, 155 are Junior college students. c) 0.3333 P(Ice Skating | Four-Year College student) = P(Four-Year College AND Ice skating) / P(Total Four Year College Student) P(Ice Skating | Four-Year College student) = Number in both / Total Four Year college students college students 70 / 210 = 0.3333 Of the 210 Four-year college students, 70 prefer Ice Skating. d) 0.3223 P(Junior College | Snowboarding ) = P(Junior College AND Snowboarding) / P(Total of Snowboarding) Number in both / Total college students 68 / 211 = 0.3223 Of the 211 students who like snowboarding, 68 are Junior college students. e) 0.6722 Of the 180 Grad students, 74 likes Skiing and 47 likes Ice Skating Probability( Skiing or Ice Skating | Grad Students) = (74/180) + (47+180) = 0.6722
A recent survey reported in Bloomberg Businessweek dealt with the salaries of CEOs at large corporations and whether company shareholders made money or lost money. If a company is randomly selected from the list of 32 studied, what is the probability: a. The CEO made more than $1 million? (Round your answer to 2 decimal places.) b. The CEO made more than $1 million or the shareholders lost money? (Round your answer to 2 decimal places.) c. The CEO made more than $1 million given the shareholders lost money? (Round your answer to 4 decimal places.) d. Of selecting two CEOs and finding they both made more than $1 million? (Round your answer to 4 decimal places.)
a) 0.34 By empirical probability, P(CEO made more than $1 million) = number of times the evens occurs / Total number of observations P(CEO made more than $1 million) = 11/32 = 0.34375 b) 0.53 Using the special rule of addition, P(A or B) = P(A) + P(B) - P(A and B) P(CEO made more than $1 million OR the shareholders lost money) (11/32) + (14/32) - (8/32) = 0.53125 c) 0.5714 By empirical probability, P(CEO made more than $1 million) = 8/14 = 0.5714285714 d) 0.1109 First, use empirical probability. Determine the probability of one CEO making more than $1 million which is 11/32 = 0.34 from question a) Then, use the general rule of multiplication, P(A and B) = P(A)P(B|A) P(A and B) = (11/32)(10/31) = 0.1108870968
Refer to the following table. First Event Second Event A1 A2 A3 Total B1 4 10 12 26 B2 11 22 9 42 Total 15 32 21 68 a.DetermineP(B1).(Round your answer to 2 decimal places.) b.DetermineP(A2|B2).(Round your answer to 2 decimal places.) c.DetermineP(B2andA2).(Round your answer to 2 decimal places.)
a) 0.38 P(B1) = Total in B1 / Grand Total 26 / 68 = 0.38235 b) 0.52 P(A2 | B2) = Number in both / Total in B1 or P(A2 | B2) = P(A2 and B1) / P(B1) 22 / 42 = 0.523 c) 0.32 P(B2 and A2) = 22/68 = 0.32
Refer to the following table. First Event Second Event A1 A2 A3 Total B1 5 9 11 25 B2 8 18 9 35 Total 13 27 20 60 a. Determine P(B1). (Round your answer to 2 decimal places.) b. Determine P(A2 | B1). (Round your answer to 2 decimal places.) c. Determine P(B2 and A2). (Round your answer to 2 decimal places.)
a) 0.42 P(B1) = Total in B1 / Grand Total 25 / 60 = 0.4166 b) 0.36 P(A2 | B1) = Number in both / Total in B1 or P(A2 | B1) = P(A2 and B1) / P(B1) 9 / 25 = 0.36 c) 0.30 P(B2 and A2) = 18/60 = 0.30
A study of 200 advertising firms revealed their income after taxes: Income after Taxes Number of Firms Under $1 million 102 $1 million to $20 million 61 $20 million or more 37 a.What is the probability an advertising firm selected at random has under $1 million in income after taxes?(Round your answer to 2 decimal places.) b-1.What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more?(Round your answer to 2 decimal places.) b-2.What rule of probability was applied? Special rule of addition only Rule of complements only Both
a) 0.51 P(under $1 million) = number of firms / total P(under $1 million) = 102 / 200 = 0.51 b-1) 0.49 P(EITHER $1 million and $20 million OR $20 millin or more) (61/200) + (37/200) = 0.49 b-2) BOTH
A study of 268 advertising firms revealed their income after taxes: Income after Taxes ----- Number of Firms Under $1 million 152 $1 million to $20 million 68 $20 million or more 48 term-10 a.What is the probability an advertising firm selected at random has under $1 million in income after taxes? (Round your answer to 2 decimal places.) term-10 b-1. What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more?(Round your answer to 2 decimal places.) b-2. What rule of probability was applied?
a) 0.57 152 / 268 = 0.567 b-1) 0.43 P(either an income between $1 million and $20 million, or an income of $20 million or more) (68 / 268) + (48 / 268) = 0.4328 b-2) Either
The events A and B are mutually exclusive. Suppose P(A)=0.28 and P(B) = 0.31. a. What is the probability of either A or B occurring? (Round your answer to 2 decimal places.) b. What is the probability that neither A nor B will happen? (Round your answer to 2 decimal places.)
a) 0.59 The events A and B are mutually exclusive, therefore P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) 0.28 + 0.31 - 0 = 0.59 b) 0.41 P( neither A nor B) = 1 - P(A or B) 1 - 0.59 = 0.41
A survey of 27 students at the Wall College of Business showed the following majors: Accounting 9 Finance 3 Economics 5 Management 3 Marketing 7 From the 27 students, suppose you randomly select a student. a.What is the probability he or she is a management major?(Round your answer to 3 decimal places.) b.Which concept of probability did you use to make this estimate?
a) 3 / 27 = 0.111 b) Empirical
A survey of 30 students at the Wall College of Business showed the following majors: Accounting 10 Finance 3 Economics 4 Management 3 Marketing 10 From the 30 students, suppose you randomly select a student. a.What is the probability he or she is a management major? (Round your answer to 3 decimal places.) b.Which concept of probability did you use to make this estimate?
a) 3 / 30 = 0.100 b) Empirical
Solve the following: a. 20!/17! b. 6P4 c. 9C6
a) 6,840 b) 360 c) 84 on calculator, MATH down right to PRB to find !, nPr, nCr
Solve the following: a.40!/35! b.7P4 c.5C2
a) 78,960,960 b) 840 c) 10
Mookie Betts of the Boston Red Sox had the highest batting average for the 2018 Major League Baseball season. His average was 0.388. So, the likelihood of his getting a hit is 0.388 for each time he bats. Assume he has two times at bat tonight in the Red Sox-Yankee game. a. This is an example of what type of probability? b. What is the probability of getting two hits in tonight's game? (Round your answer to 3 decimal places.) c. Are you assuming his second at bat is independent or mutually exclusive of his first at bat? d. What is the probability of not getting any hits in the game? (Round your answer to 3 decimal places.) e. What is the probability of getting at least one hit? (Round your answer to 3 decimal places.)
a) Empirical P(hit) = 0.388 n = 2 b) 0.151 Probability of hit per at = 0.388 Probability of two hits in two bats = 0.388 x 0.388 = 0.150544 or P(x = 2) 0.388^2 c) Independent d) 0.375 Probability of not getting any hits P(x = 0) = (1 - 0.388)^2 = 0.374544 e) 0.625 P(1 <= x <= 2) = 1 - P(x = 0) 1 - 0.375 = 0.625
In each of the following cases, indicate whether classical, empirical, or subjective probability is used. a. A baseball player gets a hit in 36 out of 114 times at bat. The probability is 0.32 that he gets a hit in his next at bat. b. A six-member committee of students is formed to study environmental issues. What is the likelihood that any one of six are randomly chosen as the spokesperson. c. You purchase a ticket for the Lotto Canada lottery. Over twelve million tickets were sold. What is the likelihood you will win the $2 million jackpot? d. The probability of an earthquake in northern California in the next 11 years above 12.0 on the Richter Scale is 0.83.
a) Empirical b) Classical c) Classical d) Empirical
A sample of 45 oil industry executives was selected to test a questionnaire. One question about environmental issues required a yes or no answer. a. What is the experiment? b. Which of the following are possible events. (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.) check all that apply: 27 people respond "Yes." 36 people respond "Yes." 35 people respond "No." 49 people respond "No." The questionnaire fails to reach one executive. c. 30 of the 45 executives responded yes. Based on these sample responses, what is the probability that an oil industry executive will respond yes? d. What concept of probability does this illustrate? e. Are each of the possible outcomes equally likely and mutually exclusive?
a) The experiment is the survey of 45 oil industry employees. b) Checkmarks: 27 people respond "Yes." 36 people respond "Yes." 35 people respond "No." The questionnaire fails to reach one executive. no checkmarks: 49 people respond "No." - 49 is more than 45 c) 0.66 30 / 45 = 0.66 d) Empirical Empirical probability is the probability of an event happening is the fraction of the time similar events happened in the past. e) The outcomes are equally likely and are also mutually exclusive
PART 1 Each salesperson at Puchett, Sheets, and Hogan Insurance Agency is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated with respect to his or her potential for advancement—either fair, good, or excellent. These traits for the 500 salespeople were cross-classified into the following table. Potential for Advancement Sales Ability Fair Good Excellent TOTAL Below average 16 12 22 50 Average 45 60 45 150 Above average 93 72 135 300 TOTAL 154 144 202 500 a.What is this table called? Bayesian table Contingency table Probability table b.What is the probability a salesperson selected at random will have above average sales ability and excellent potential for advancement?(Round your answer to 2 decimal places.)
a) contingency table b) 0.27 P(Above average AND Excellent) = Number in both / TOTAL 135 / 500 = 0.27
A sample of 2,000 licensed drivers revealed the following number of speeding violations. Number of Violations Number of Drivers 0 1,910 1 46 2 18 3 12 4 9 5 or more 5 Total 2,000 c.What is the probability that a particular driver had exactly two speeding violations?(Round your answer to 3 decimal places.) d.What concept of probability does this illustrate?
c) 0.009 18 / 2000 = 0.009 d) Empirical
A representative of the Environmental Protection Agency (EPA) wants to select samples from 10 landfills. The director has 15 landfills from which she can collect samples. How many different samples are possible?
n = 15 landfills r = 10 landfills use nCr 3,003
A company uses four backup servers to secure its data. The probability that a server fails is 0.17. Assuming that the failure of a server is independent of the other servers, what is the probability that one or more of the servers is operational? (Round your answer to 6 decimal places.)
n = 4 P(server fails) = 0.17 P(all 4 server fails) = 0.17^4 P(one or more server is operational) = 1 - P(all server fails) = 1 - (0.17)^4 = 0.99916479 0.999165
PART 2 Each salesperson at Puchett, Sheets, and Hogan Insurance Agency is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated with respect to his or her potential for advancement—either fair, good, or excellent. These traits for the 500 salespeople were cross-classified into the following table. Potential for Advancement Sales Ability Fair Good Excellent TOTAL Below average 16 12 22 50 Average 45 60 45 150 Above average 93 72 135 300 TOTAL 154 144 202 500 c. Fill in the blanks to provide details for a tree diagram.
see attached photo for answer