MAT 122 Statistics

The data in the following table summarize results from 122 pedestrian deaths that were caused by accidents. If three different deaths are randomly selected without​ replacement, find the probability that they all involved intoxicated drivers. Driver intoxicated, pedestrian intoxicated = 28 Driver intoxicated, pedestrian sober = 18 Driver sober, pedestrian intoxicated = 89 Driver sober, pedestrian sober = 25

(sample size)/(population size) x 100 28 + 18 + 89 + 25 = 160 (3/122) x 100 = 1.9% P(A and B) = P(A) x P(B|A) P(A) = (number of times A occurred)/(number of times the procedure was performed) 39 + 27 = 46 66/122 = 46/160 46/160 x 46/160 x 46/160 = 0.023764 Is such an event unlikely? Yes, because its probability is less than 0.05.

The table below summarizes results from a study of people who refused to answer survey questions. A pharmaceutical company is interested in opinions of the elderly. What is the probability that the selected subject is someone 60 and over who​ responded? 18-21: 72 responded, 13 refused 22-29: 254 responded, 22 refused 30-39: 244 responded, 35 refused 40-49: 135 responded, 28 refused 50-59: 137 responded, 37 refused 60 and over: 201 responded, 59 refused

72+3+254+22+244+22+244+35+135+28+137+37+201+59= 1493 201/1493= 0.135

The table below summarizes results from a study of people who refused to answer survey questions. A market researcher is not interested in refusals or subjects below 22 years of age or over 59. Find the probability that the selected person refused to answer or is below 22 or is older than 59. 18-21: 79 responded, 13 refused 22-29: 261 responded, 22 refused 30-39: 251 responded, 35 refused 40-49: 142 responded, 28 refused 50-59: 144 responded, 37 refused 60 and over: 208 responded, 59 refused

All of those who responded = 1085 All who refused = 194 (1085) + (194) = (1279) Total under 22 who responded = 79 Total over 59 who have responded = 208 (194) + (79) + (208) = 481 (481) / (1279) = 0.376 The probability that the selected subject is someone 60 and over who responded is 0.363

Which word is associated with multiplication when computing probabilities?

And

Which of the following is NOT a principle of​ probability? Choose the correct answer below. A. All events are equally likely in any probability procedure. B. The probability of any event is between 0 and 1 inclusive. C. The probability of an event that is certain to occur is 1. D. The probability of an impossible event is 0.

B.

You are certain to get a number or a face card when selecting cards from a shuffled deck. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

If an event cannot occur then it has a probability of​ 0, and if it is certain to occur then it has a probability of 1. The probability is: 1

In a certain​ country, the true probability of a baby being a boy is 0.515. Among the next six randomly selected births in the​ country, what is the probability that at least one of them is a girl​?

P(A) = 1 - P (overbarA​) What is the complement of​ A, Upper overbarA​? Answer: The next seven births are all boys. P(all 6 are boys) = P(1st is a boy and 2nd is a boy...and last is a boy). Answer: P(1st is a boy and 2nd is a boy...and last is a boy) = 0.515 x 0.515 x 0.515 x 0.515 x 0.515 x 0.515 =0.019 P(A) = 1 - P(overbarA) = 1 - 0.019 =0.981

Testing for a disease can be made more efficient by combining samples. If the samples from five people are combined and the mixture tests​ negative, then all fiveall five samples are negative. On the other​ hand, one positive sample will always test​ positive, no matter how many negative samples it is mixed with. Assuming the probability of a single sample testing positive is 0.2​, find the probability of a positive result for fivefive samples combined into one mixture. Is the probability low enough so that further testing of the individual samples is rarely​ necessary?

P(A) = 1 - P(*A) ​P(all three are negative) = P(1st is negative and 2nd is negative and 3rd is negative) P(negative) = 1 - P (positive) = 1 - 0.05 = 0.95 P(*A) = P(1st is negative and 2nd is negative) = 0.95 x 0.95 =0.9025 P(A) = 1 - P(*a) = 1 - 0.9025 = 0.0975 Is the probability low enough so that further testing of the individual samples is rarely​ necessary? Answer: . The probability is not​ low, so further testing of the individual samples will not be a rarely necessary event.

The table below summarizes results from a study of people who refused to answer survey questions. A market researcher is interested in​ responses, especially from those between the ages of 22 and 39. Find the probability that a selected subject responds or is between the ages of 22 and 39. 18-21: 71 Responded, 12 refused 22-29: 253 responded, 22 refused 30-39: 243 responded, 35 refused 40-49: 134 responded, 28 refused 50-59: 136 responded, 37 refused 60 and over: 200 responded, 59 refused

P(A) = P(responded to survey) P(B) = (respondents that are between the ages of 22 and 39) P(A) = (total number of people that responded) / (total number of people) Total Number of People = 1231 P(A) = (71 + 253 + 243 + 134 + 136 + 200) = 1037 P(A) = (1037) / (1231) P(B) = (total number of people between the ages of 22 and 39) / (total number of people) P(B) = (253 + 22 + 243 + 35) / 1231 P(B) = 553/1231 P(A and B) = total number of people who responded and is between 22 and 39) / (total number of people) 496 / 1231 P(A or B)=P(A) + P(B) - P(A and B) = (1037/1231) + (553/1231) - (469/1231) =0.889

Among 500 randomly selected drivers in the 20−24 age​ bracket, 14 were in a car crash in the last year. If a driver in that age bracket is randomly​ selected, what is the approximate probability that he or she will be in a car crash during the next​ year? Is it unlikely for a driver in that age bracket to be involved in a car crash during a​ year? Is the resulting value high enough to be of concern to those in the 20−24 age​ bracket? Consider an event to be​ "unlikely" if its probability is less than or equal to 0.05.

P(A)=(number of times A occurred)/(number of times the trial was repeated) 14/500=0.028 Would it be unlikely for a driver in that age bracket to be involved in a car crash this​ year? Yes Is the probability high enough to be of concern to those in the 20−24 age​ bracket? No

Twelve of the 50 digital video recorders​ (DVRs) in an inventory are known to be defective. What is the probability you randomly select an item that is not​ defective?

P(A)=(number of ways A can occur)/(number of different simple events)=s/n 50-12=38 (38)/(50)=0.76 The probability is: 0.76

For the given pair of events A and​ B, complete parts​ (a) and​ (b) below. ​A: When a baby is​ born, it is a girl. ​B: When a 99​-sided die is​ rolled, the outcome is 88. a. Determine whether events A and B are independent or dependent.​ (If two events are technically dependent but can be treated as if they are independent according to the​ 5% guideline, consider them to be​ independent.) b. Find​ P(A and​ B), the probability that events A and B both occur.

The 5% guideline is not applicable. P(B|A) = (number of ways B|A can occur) / (number of different simple events) = s/n (number of ways B|A can occur) = The number of ways 1 can be rolled after a girl has already been born = 1 (number of different simple events) = 9 P(B|A) = 1/9 B = number of ways 1 can be rolled = 9 P(B) = 1/9 a.) The two events are independent because the occurrence of one does not affect the probability of the occurrence of the other. P(A) = (number of ways A can occur) / (number of different simple events) = 1/2 P(A and B) = P(A)xP(B) (1/2)x(1/9) = 0.0556

Determine whether the two events are disjoint for a single trial.​ (Hint: Consider​ "disjoint" to be equivalent to​ "separate" or​ "not overlapping".) Randomly selecting a piano from the instrument assembly line and getting one that is free of defects. Randomly selecting a piano from the instrument assembly line and getting one with a missing key.

The events are disjoint. They cannot occur at the same time.

Among respondents asked which is their favorite seat on a​ plane, 485 chose the window​ seat, 77 chose the middle​ seat, and 310 chose the aisle seat. What is the probability that a passenger prefers the middle​ seat? Is it unlikely for a passenger to prefer the middle​ seat? If​ so, why is the middle seat so​ unpopular?

The probability that a passenger prefers the middle seat is: 0.009. Is it unlikely for a passenger to prefer the middle​ seat? ​Yes, because the probability that a passenger prefers the middle seat is less than 0.05. If​ so, why is the middle seat so​ unpopular? The middle seat lacks an outside​ view, easy access to the​ aisle, and a passenger in the middle seat has passengers on both sides instead of on one side only.

A research center poll showed that 82​% of people believe that it is morally wrong to not report all income on tax returns. What is the probability that someone does not have this​ belief?

The probability that someone does not believe that it is morally wrong to not report all income on tax returns is 0.18. 100=82=18 18/100=0.18

For the given pair of events A and​ B, complete parts​ (a) and​ (b) below. ​A: When a page is randomly selected and ripped from a 5​-page document and​ destroyed, it is page 2. ​B: When a different page is randomly selected and ripped from the​ document, it is page 4. a. Determine whether events A and B are independent or dependent.​ (If two events are technically dependent but can be treated as if they are independent according to the​ 5% guideline, consider them to be​ independent.) b. Find​ P(A and​ B), the probability that events A and B both occur.

The sample size of pages ripped from the document is 2. The population size of pages in the document is 5. sample percentage of the population = ((2/5)x100)% =40% Page 2 and 4 can only be ripped once and ripping one increases the chances of ripping the other. The two events are dependent because the occurrence of one affects the probability of the occurrence of the other. P(A and B) = probability that both events A and B occur P(A) = probability that event A occurs) =1/5 P(B) = probability of event B occurring = 1/5 P(B|A) = probability of event B occurring after A occurred 1/4 P(A and B) = P(A) x P(B|A) (1/5)x(1/4) = 0.05

Which of the following values cannot be​ probabilities? 0.03, -0.44, 5/3, 0, sqrt2, 1.54, 3/5, 1

Values that cannot be probabilities: -0.44, 1.45, sqrt2, 5/3

Decide whether the following two events are disjoint. 1. Randomly selecting someone who smokes cigarsRandomly selecting someone who smokes cigars 2. Randomly selecting a maleRandomly selecting a male Are the two events​ disjoint?

Yes​, because the events cannot occur at the same time.

Refer to the table below. Given that 2 of the 136 subjects are randomly​ selected, complete parts​ (a) and​ (b) Type Rh+ = (0 = 53), (A = 38), (B = 16), (AB = 7) Type Rh- = (0 = 11), (A = 8), (B = 2), (AB = 1)

a. Assume that the selections are made with replacement. What is the probability that the 2 selected subjects are both group Upper B and type Rh+​? Because the selections are made with​ replacement, the probability of event B is the same as the probability of event A. How many subjects are both group Upper B and type Rh+​? = 16 How many total subjects are there? = 136 P(A) = (number of ways A can occur) / (number of simple events) = s/n = P(1 subject who is B and Rh+) = 16/136 P(1 subject who is B and Rh+ | the 1st subject was B and Rh+) = 16/116 P(2 subjects who are B and Rh+) = (16/136) x (16/136) = 0.0138 b. Now assume the selections are made without replacement. This means the probability of selecting the second subject who is group Upper B and Rh+ is not the same as the probability of selecting the first subject. P(1 subject who is B and Rh+) =16/136 Reduce by 1: subjects who are group Upper BB and type Rh = 16-1 = 15, total number of subjects = 136 - 1 = 135 P(1 subject who is B and Rh+ | the 1st subject was B and Rh+) = 15/135 P(2 subjects who are B and Rh+) = (16/136) x (15/135) = 0.0131

A modified roulette wheel has 40 slots. One slot is​ 0, another is​ 00, and the others are numbered 1 through 38​, respectively. You are placing a bet that the outcome is an odd number.​ (In roulette, 0 and 00 are neither odd nor​ even.)

a. P(odd)=40-2=38/2=19 =19/40 b.P(not odd) = 40-19=21 =21/40 21:19 c. 1:1 = $14 d. 21:19 (a*c)/b (21*14)/19 =15.47

To the right are the outcomes that are possible when a couple has three children. Refer to that​ list, and find the probability of each event. a. Among three​ children, there are exactly 22 girls. b. Among three​ children, there are exactly 33 girls. c. Among three​ children, there is exactly 11 boy. Groups: (boy-boy-boy),(boy,boy,birl),(boy-girl-boy),(boy-girl-girl),(girl-boy-boy),(girl-boy-girl),(girl-boy-girl),(girl-girl-girl)

a. What is the probability of exactly 2 girls out of three​ children? 3/8 b. What is the probability of exactly 3 girls out of three​ children? 1/8 c. What is the probability of exactly 1 boy out of three​ children? 3/8

A tire company produced a batch of 6,500 7,000 tires that includes exactly 210 220 that are defective. a. If 4 tires are randomly selected for installation on a​ car, what is the probability that they are all​ good? b. If 100 tires are randomly selected for shipment to an​ outlet, what is the probability that they are all​ good? Should this outlet plan to deal with defective tires returned by​ consumers?

a.) 7,000 - 220 = 6,780 good tires P(1 good tire) = 6780/7000 P(1 good tire | first tire was good) = 6779/6999 P( 1 good tire | first two tires were good) = 6778/6998 P(1 good tire | first three were good) 6777/6997 P(4 good tires) = (6780/7000) x (6779/6999) x (6778/6998) x (6777/6997) = =0.880 b.) If 100 tires are randomly selected for shipment to an​ outlet, what is the probability that they are all​ good? (6780/7000)^100 = = 0.041 Should this outlet plan to deal with defective ties returned by consumers? Yes, because there is a very small chance that all 100 tires are good.

The accompanying table contains the results from experiments with a polygraph instrument. Find the probabilities of the events in parts​ (a) and​ (b) below. Are these events​ unlikely? a. Four of the test subjects are randomly selected with​ replacement, and they all had true negative test results. b. Four of the test subjects are randomly selected without​ replacement, and they all had true negative test results. Positive test result and did not lie: 12 Positive test result and did lie: 30 Negative test result and did not lie: 70 Negative test result and did lie: 19

a.) P(A) = probability event A occurs P(B) = probability event B occurs P(A and B) = P(A) x P(B) P(A) = (number of times A occurred) / (number of times the procedure was performed) 12 + 30 + 70 + 19 = 131 P(negative test result and did not lie) = 70/131 (70/131) x (70/131) x (70/131) x (70/131) = 0.082 Is such an even unlikely? No, because the probability of the even is greater that 0.05. b.) P(A and B) = P(A) x P(B|A) 5757​, 109109​, 5656​, 108108​, 5555​, 107107 P(true negative for 1st selection) = 70/131 P(true negative for 2nd selection) = 69/130 P(true negative for 3rd selection) = 68/129 P(true negative for 4th selection) = 67/128 P(true negative for all 4) = P(1st) x P(2nd) x P(3rd) x P(4th) =(70/131) x (69/130) x (68/129) x (67/1128) = 0.077 Is such an event unlikely? No because the probability of the event is greater that 0.05.

The following data summarizes results from 10091009 pedestrian deaths that were caused by accidents. If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was intoxicated or the driver was intoxicated. ((Pedestrian intoxicated)+(driver intoxicated))=99 ((Pedestrian not intoxicated)+(driver intoxicated))=57 ((Pedestrian intoxicated )+(driver not intoxicated))=294 ((Pedestrian not intoxicated)+(driver not intoxicated))=559

​P(pedestrian was intoxicated or driver was ​intoxicated)equals= (99+57+294+559)=1004 (99+57+294) = 450 450/1004= =0.446

In a certain​ country, the true probability of a baby being a girl is 0.476. Among the next eight randomly selected births in the​ country, what is the probability that at least one of them is a boy​?

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