CHAPTER 8 REVIEW



CHAPTER 8 REVIEW

46) Bob did not study for his multiple choice statistics test. Every problem has 5 options and he is guessing at random. There are 23 problems on the test. Assume he does this for every test. Define “success” to be “he gets it right” and “failure” to be “he gets it wrong”.

a) What is the probability of success in this experiment?

b) What is the mean number of problems he will get right?

c) What is the standard deviation of the number of problems he will get right?

d) What is the probability that he gets exactly 6 questions right? (Do this calculation using the formula.)

e) What is the probability that he gets fewer than 7 right?

BONUS: f) What is the probability that he gets from 7 to 12 questions right?

ANS:

a) P (he gets it right) = 1/5 = 0.2

b) mean = np = 0.2 (23) = 4.6

c) standard deviation = [pic]

d) P (X = 6) = [pic]

e) P (X < 7) = binomcdf (23, 0.2, 6) = 0.8402

f) P (7 ≤ X ≤ 12) = binomcdf (23, 0.2, 12) – binomcdf (23, 0.2, 6) = 0.9999 – 0.8402 = 0.1597

47) A wildlife biologist examines frogs for a genetic trait he suspects may be linked to sensitivity to industry toxins in the environment. Previous research had established that this trait is usually found in 1 of every 8 frogs. He collects and examines a dozen frogs. If the frequency of the trait has not changed, what’s the probability he finds the trait in

a) none of the 12 frogs?

b) at least 2 frogs?

c) 3 or 4 frogs?

d) no more than 4 frogs?

The biologist decides to collect and examine 150 frogs.

e) Determine the mean and standard deviation of the number of frogs with the trait he should expect to find in his sample.

f) Verify that he can use a Normal model to approximate the distribution of the number of frogs with the trait.

ANS:

a) P (x = 0) = bpdf (12, 0.125, 0) = 0.201

b) P ( x ≤ 2) = 1 – bcdf (12, 0.125, 1) = 0.453

c) P (x = 3) + P (x = 4) = bpdf (12, 0.125, 3) + bdf (12, 0.125, 4) = 0.171

d) P (x ≤ 4) = bcdf (12, 0.125, 4) = 0.989

e) µ = 150 (0.125) = 18.75, σ = 4.05

f) np = 18.75 ≥ 10 and n (1-p) = 150 (0.875) = 131.25 ≥ 10

48) An orchard owner knows that he’ll have to use about 6% of the apples he harvests for cider because they will have bruises or blemishes. He expects a tree to produce about 300 apples.

a) Describe an appropriate model for the number of cider apples that may come from that tree. Justify your mode.

b) Find the probability that there will be no more than a dozen cider apples.

ANS: a) Assuming apples fall and become blemished independently of each other, B (300, 0.6) is appropriate. Since np ≥ 10 and n(1-p) ≥ 10, N (18, 4.11) is also appropriate.

b) Normal approximate: 0.072 and Binomial approximate: 0.085

49) Police estimate that 80% of drivers now wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelts use.

a) How many cars do they expect to stop before finding a driver whose seatbelt is not buckled?

b) What’s the probability that the first unbelted driver is in the 6th car stopped?

c) What’s the probability that the first 10 drivers are all wearing their seatbelts?

d) If they stop 30 cars during the first hour, find the mean and standard deviation of the number of expected to be wearing seatbelts.

e) If they stop 120 cars during this safety check, what’s the probability they find at least 20 drivers not wearing their seatbelts?

ANS:

a) µ = 1/0.2 = 5

b) P (x = 6) = gpdf (0.2, 6) = 0.066

c) P (x = 1) ^ 10 = gpdf (0.8, 1) ^ 10 = 0.8 ^ 10 = 0.107

d) µ = 30 (0.8) = 24, σ = 2.19

e) P (x ≤ 20) = 1 – bcdf (120, 0.2, 19) = 0.848

50) Vitamin D is essential for strong, healthy bones. Our bodies produce vitamin D naturally when sunlight falls upon the skin, or it can be taken as a dietary supplement. Although the bone disease rickets was largely eliminated in England during the 1950s, some people there are concerned that this generation of children is at increased risk because they are more likely to watch TV or play computer games than spend time outdoors. Recent research indicated that about 20% of British children are deficient in vitamin D. Suppose doctors test a group of elementary school children.

a) What’s the probability that the first vitamin D-deficient child is the 8th one tested?

b) What’s the probability that the first 10 children tested are all okay?

c) How many kids do they expect to test before finding one who has this vitamin deficiency?

d) They will test 50 students at the third grade level. Find the mean and standard deviation of the number who may be deficient in vitamin D.

e) If they test 320 children at this school, what’s the probability that no more than 50 of them have the vitamin deficiency?

ANS:

a) P (x = 8) = gpdf (0.2, 8) = 0.042

b) P (x = 1) ^ 10 = gpdf (0.8, 1) ^ 10 = 0.8^10 = 0.107

c) µ = 1/0.2 = 5

d) µ = 50 (0.2) = 10, σ = 2.83

e) P (x ≤ 50) = bcdf (320, 0.2, 50) = 0.027

51) The Centers for Disease Control say that about 30% of high-school students smoke tobacco (down from a high of 38% in 1997). Suppose you randomly select high-school students to survey them on their attitudes toward scenes of smoking in the movies. What’s the probability that

a) none of the first 4 students you interview is a smoker?

b) the first smoker is the sixth person you choose?

c) there are no more than 2 smokers among 10 people you choose?

You randomly select 120 high-school students to survey them on their attitudes toward scenes of smoking in the movies.

d) What’s the expected number of smokers?

e) What’s the standard deviation of the number of smokers?

f) The number of smokers among 120 randomly selected students will vary from group to group. Explain why that number can be described with a Normal model.

ANS:

a) P (x = 0) = bpdf (4, 0.3, 0) = 0.240

b) P (x = 6) = gpdf (0.3, 6) = 0.050

c) P (x ≤ 2) = bcdf (10, 0.3, 2) = 0.383

d) µ = 120 (0.3) = 36

e) σ = [pic]5.02

f) Because both np = 36 ≥ 10 and n(1-p) = 84 ≥ 10

52) Suppose that in a certain metropolitan area, 9 out of 10 households have a VCR. Let x denote the number among four randomly selected households that have a VCR. What is the probability that

a) exactly 2 households have a VCR?

b) all four selected households have a VCR?

c) no more than 3 selected households have a VCR?

ANS:

a) P (x = 2) = bpdf (4, 0.9, 2) = 0.0486

b) P (x = 4) = bpdf (4, 0.9, 4) = 0.6561

c) P (x ≤ 3) = bcdf (4, 0.9, 3) = 0.3439

53) The article on polygraph testing for FBI agents indicated that the probability of a false-positive (a trustworthy person who nonetheless fails the test) is 0.15.

a) What is the probability that the first false-positive will occur when the third person is tested?

b) What is the probability that fewer than four are tested before the first false-positive occurs?

c) What is the probability that more than three agents are tested before the first false-positive occurs?

ANS:

a) P (x = 3) = gpdf (0.15, 3) = 0.1084

b) P (x < 4) = gcdf (0.15, 3) = 0.3859

c) P (x > 3) = 1 – gcdf (0.15, 3) = 6141

54) Sophie is a dog that loves to play catch. Unfortunately, she isn’t very good, and the probability that she catches a ball is only 0.1. Let x be the number of tosses required until Sophie catches a ball.

a) Does x have a binomial or geometric distribution?

b) What is the probability that it will take exactly two tosses for Sophie to catch a ball?

c) What is the probability that more than three tosses will be required?

ANS:

a) Geometric

b) P (x = 2) = gpdf (0.1, 2) = 0.09

c) P (x > 3) = 1 – gcdf (0.1, 3) = 0.729

55) Thirty percent of all automobiles undergoing an emission inspection at a certain inspection station fail the inspection.

a) Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection?

b) Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass the inspection?

c) Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection?

ANS:

a) P (x ≤5) = bcdf (15, 0.3, 5) = 0.722

b) P (5 ≤ x ≤ 10) = bcdf (15, 0.3, 10) – bcdf (15, 0.3, 4) = 0.484

c) µ = 25 (0.3) = 7.5, σ = [pic]2.29

56) You are to take a multiple-choice exam consisting of 100 questions with 5 possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let x represent the number of correct responses on the test.

a) What kind of probability distribution does x have?

b) What is your expected score on the exam?

c) Compute and variance and standard deviation of x.

d) Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

ANS:

a) Binomial distribution

b) µ = 100 (0.2) = 20

c) [pic]

σ = 4

d) normalcdf (50, 1E99, 20, 4) = 0%

57) Twenty percent of American households own three or more motor vehicles. You choose 12 households at random.

a) What is the probability that none of the chosen households owns three or more vehicles? What is the probability that at least one household owns three or more vehicles?

b) What are the mean and standard deviation of the number of households in your sample that own three or more vehicles?

c) What is the probability that your sample count is greater than the mean?

ANS:

a) P (x = 0) = bpdf (12, 0.2, 0) = 0.0687

P (x ≥ 1) = 1 – bpdf (12, 0.2, 0) = 0.9313

b) µ = 12 (0.2) = 2.4

σ = 1.39

c) P (x > 2.4) = 1 – bcdf(12, 0.2, 2) = 0.4417

58) A rapid test for the presence in the blood of antibodies to HIV, the virus that causes AIDS< gives a positive result with probability about 0.004 when a person who is free of HIV antibodies is tested. A clinic tests 1000 people who are all free of HIV antibodies.

a) What is the distribution of the number of positive tests?

b) What is the mean number of positive tests?

c) You cannot safely use the Normal approximation for this distribution. Explain why.

ANS:

a) Let x = number of positive tests; x is B (1000, 0.004)

b) µ = 1000 (0.004) = 4

c) We need np and n (1 – p) to be at least 10, and as we saw in (b), np = 4.

59) Dentists are increasingly concerned about the growing trend of local school districts to grant soft drink companies exclusive rights to install soda pop machines in schools in return for money—usually millions—that goes directly into school coffers. According to a recent study by the National Soft Drink Association, 62% of schools nationally already have such contracts. This comes at a time when dentists are seeing and alarming increase in horribly decayed teeth and eroded enamel in the mouth of teenagers and young adults. With ready access to soft drinks, children tend to drink them all day. That, combined with no opportunity to brush, leads to disaster, dentists say. Suppose that 20 schools around the country are randomly selected and asked if they have a soft drink contract. Find the probability that the number of “Yes” answers is

a) exactly 8.

b) at most 8.

c) at least 4.

d) between 4 and 12, inclusive.

e) Identify the random variable of interest, X. Then write the probability distribution table for X.

ANS:

a) P (x = 8) = bpdf (20, 0.62, 8) = 0.0249

b) P (x ≤ 8) = bcdf (20, 0.62, 8) = 0.0381

c) P (x ≥ 4) = 1 – bcdf (20, 0.62, 3) = 0.99998

d) P ( 4 ≤ x ≤ 12) = bcdf (20, 0.62, 12) – bcdf (20, 0.62, 3) = 0.51076

60) There are 20 red marbles, 10 blue marbles, and 5 white marbles in a jar. Select a marble without looking, note the color, and then replace the marble in the jar. We’re interested in the number of marbles you would have to draw in order to be sure you have a red marble.

a) Is this a binomial or a geometric setting?

b) Calculate the probability of drawing a red marble on the second draw. Calculate the probability of drawing a red marble by the second draw. Calculate the probability that it would take more than 2 draws to get a red marble.

ANS:

a) Geometric

b) P (x = 2) = gpdf (4/7, 2) = 0.2449

P (x ≤ 2) = gcdf (4/7, 2) = 0.8163

P (x ≥ 2) = 1 – gcdf (4/7, 1) = 0.1837

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