front 1 The table to the right lists probabilities for the corresponding numbers of girls in three births. What is the random variable, what are its possible values, and are its values numerical? | back 1 The random variable is x, which is the number of girls in three births. The possible values of x are 0, 1, 2, and 3. The values of the random value x are numerical. The variable x is a random variable because its values depend on chance. The possible values of x are 0, 1, 2, and 3. Because these values are numbers, x is a numerical variable. |
front 2 Determine whether the following value is a continuous random variable, discrete random variable, or not a random variable.
| back 2 a. It is a continuous random variable. b. It is not a random variable. c. It is a continuous random variable. d. It is a continuous random variable. e. It is a continuous random variable. f. It is a discrete random variable. |
front 3 Determine whether the following value is a continuous random variable, discrete random variable, or not a random variable.
| back 3 a. It is a discrete random variable. b. It is not a random variable. c. It is a discrete random variable. d. It is a continuous random variable. e. It is a discrete random variable. f. It is a discrete random variable. |
front 4 Ted is not particularly creative. He uses the pickup line "If I could rearrange the alphabet, I'd put U and I together." The random variable x is the number of girls Ted approaches before encountering one who reacts positively. Determine whether the table describes a probability distribution. If it does, find its mean and standard deviation. | back 4 The table is not a probability distribution. Σ P(x) = 0.902 ≠ 1.00 |
front 5 Groups of people aged 15-65 are randomly selected and arranged in groups of six. In the accompanying table, the random variable x is the number in the group who say that their family and/or partner contribute most to their happiness (based on a survey). If the table is a probability distribution, what is its mean? | back 5 Its mean is 4.5. Its standard deviation is 1.1. |
front 6 The accompanying data table describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children. Use the range rule of thumb to identify a range of values containing the usual numbers of girls in 10 births. Based on the result, is 1 girl in 10 births an unusually low number of girls? Explain. | back 6 The maximum usual value is 8.4.
The range rule of thumb for the maximum usual value is given by
the formula below, where μ and σ are the mean and standard
deviation, respectively, for a probability distribution.
= 5.034 + 2(1.696126175) = 8.42625235 The minimum usual value is 1.6. minimum u sual value = μ – 2σ = 5.034 – 2(1.696126175) = 1.64174765 Yes, 1 girl is an unusually low number of girls, because 1 girl is outside of the range of usual values. |
front 7 The accompanying table describes results of roadworthiness tests of cars that are 3 years old. The random variable x represents the numbers of cars that failed among six that were tested for roadworthiness. Find the mean and standard deviation for the number of cars that failed among the six cars that are tested. | back 7 μ = 1.3 σ = 1.2 |
front 8 Assume that 12 jurors are selected from a population in which 80% of the people are Mexican-Americans. The random variable x is the number of Mexican-Americans on the jury.
| back 8 a. P(8) = 0.133 b. The probability of 8 or fewer Mexican-Americans among 12 jurors is 0.206. (0 + 0 + 0 + 0 + 0.001 + 0.003 + 0.016 + 0.053 + 0.133) = 0.206 c. The result from part (b), because it measures the probability of 8 or fewer successes. d. No, because the relevant probability is greater than 0.05. |
front 9 Assume that 12 jurors are selected from a population in which 80% of the people are Mexican-Americans. The random variable x is the number of Mexican-Americans on the jury.
| back 9 a. P(5) = 0.003 b. The probability of 8 or fewer Mexican-Americans among 12 jurors is 0.004. (0 + 0 + 0 + 0 + 0.001 + 0.003) = 0.004 c. The result from part (b), because it measures the probability of 8 or fewer successes. d. Yes, because the relevant probability is less than or equal to 0.05. |
front 10 A Gallup poll of 1236 adults showed that 12% of the respondents believe that it is bad luck to walk under a ladder. Consider the probability that among 30 randomly selected people from the 1236 who were polled, there are at least 2 who have that belief. Given that the subjects surveyed were selected without replacement, the events are not independent. Can the probability be found by using the binomial probability formula? Why or why not? | back 10 Yes. Although the selections are not independent, they can be treated as being independent by applying the 5% guideline. Only 30 people are randomly selected from the group of 1236, which is less than 5%. |
front 11 Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An experimental drug is administered to 120 randomly selected individuals, with the number of individuals responding favorably recorded. | back 11 Yes, because the experiment satisfies all the criteria for a binomial experiment. |
front 12 Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. Four cards are selected from a standard 52-card deck without replacement. The number of clubs selected is recorded. | back 12 No, because the trials of the experiment are not independent and the probability of success differs from trial to trial. |
front 13 Multiple-choice questions each have five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to three such questions.
| back 13 a. P(CWW) = 0.128 P(CWW) = P(C) x P(W) x P(W) = (0.20) x (0.80) x (0.80) = 0.128 b. P(WWC) = 0.128 P(WCW) = 0.128 P(WWC) = P(W) x P(W) x P(C) = 0.80 x 0.80 x 0.20 = 0.128 P(WCW) = P(W) x P(C) x P(W) = 0.80 x 0.80 x 0.20 = 0.128 c. 0.384 P(CWW) + P(WWC) + P(WCW) = 0.128 + 0.128 + 0.128 = 0.384 |
front 14 Assume that a procedure yields a binomial distribution with 6 trials and a probability of success of 0.70. Use a binomial probability table to find the probability that the number of successes is exactly 6. | back 14 0.118 n = 6 p = 0.70 x = 6 |
front 15 Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n = 20, x = 3, p = 0.25 | back 15 P(3) = 0.134 P(x=3) = ( ^{n} _{x} ) x ( p )^{x} x ( q )^{(n - x)} = (^{ 20} _{3} ) x (0.25)^{3} x (0.75)^{17} = 0.1338956152 |
front 16 Refer to the accompanying technology display. The probabilities were obtained by entering the values of n equals 5 and p equals 0.716. In a clinical test of a drug, 71.6% of the subjects treated with 10 mg of the drug experienced headaches. In each case, assume that 5 subjects are randomly selected and treated with 10 mg of the drug. Find the probability that at least four of the subjects experience headaches. Is it unusual to have fewer than four subjects experience headaches? | back 16 The probability that at least four of the subjects experience headaches is 0.5614. P(x ≥ 4) = P(x=4 or x=5) = 0.3732 + 0.1882 = 0.5614 No, because the probability that fewer than four subjects will experience headaches is not unlikely. |
front 17 In a region, there is a 0.9 probability chance that a randomly selected person of the population has brown eyes. Assume 10 people are randomly selected.
| back 17 a. The probability that all of the 10 selected people have brown eyes is 0.349. P(x = 10) = ( 0.9^{10 }) = 0.3486784401 b. The probability that exactly 9 of the selected people have brown eyes is 0.387. P(x = 9) = ( ^{n} _{x} ) x ( p )^{x} x ( q )^{(n - x)} = ( ^{10} _{9}) x (0.9^{9}) x (0.1^{1}) = 0.387420489 c. The probability that the number of selected people that have brown eyes is 8 or more is 0.930. P(x = 8) = ( ^{n} _{x} ) x ( p )^{x} x ( q )^{(n - x)} = ( ^{10} _{8}) x (0.9)^{8} x (0.1)^{2} = 0.1937102445 P(x ≥ 8) = P(x=8 or x=9 or x=10) = 0.194 + 0.387 + 0.349 = 0.930 d. No, because the probability that 8 or more of the selected people have brown eyes is greater than 0.05. |
front 18 Researchers conducted a study to determine whether there were significant differences in graduation rates between medical students admitted through special programs and medical students admitted through the regular admissions criteria. It was found that the graduation rate was 91% for the medical students admitted through special programs in all medical schools.
| back 18 a. The probability that at least 12 of the 13 students graduated is 0.670. P(x ≥ 12) = P(x=12 or x=13) = [ ( ^{13} _{12} ) x (0.91)^{12} x (0.09)^{1} ] + [ (0.91^{13}) ] = 0.377 + 0.293 = 0.6704 b. No, because the students admitted through a single special program at a specific medical school are not a random sample. |
front 19 When someone buys a ticket for an airline flight, there is a 0.0973 probability that the person will not show up for the flight. A certain jet can seat 17 passengers. Is it wise to book 19 passengers for a flight on the jet? Explain. | back 19 It is not a wise decision because the probability that there are not enough seats on the jet is 0.436. So, overbooking is not an unlikely event. * n=19, q=0.0973, p=0.9027 P(x > 17) = P(x=18 or x=19) = [ ( ^{19} _{18} ) x (0.9027)^{18} x (0.0973)^{1} ] + [ (0.9027)^{19} ] = 0.293 + 0.143 = 0.4358 |
front 20 The probability of a randomly selected adult in one country being infected with a certain virus is 0.003. In tests for the virus, blood samples from 26 people are combined. What is the probability that the combined sample tests positive for the virus? Is it unlikely for such a combined sample to test positive? Note that the combined sample tests positive if at least one person has the virus. | back 20 The probability that the combined sample will test positive is 0.075. P(1 or 2 or ... or 26) = 1 – P(0) = 1 – [ ( ^{26} _{} _{0} ) x (0^{}.003^{0}) x (0.997^{26}) ] = 1 – 0.925 = 0.0751440049 It is not unlikely for such a combined sample to test positive, because the probability that the combined sample will test positive is greater than 0.05. |
front 21 A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 23 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 3% rate of defects, what is the probability that this whole shipment will be accepted? | back 21 The probability that this whole shipment will be accepted is 0.849. P(x=0 or x=1) = [ ( ^{23} _{0} ) x (0.03)^{0} x (0.97)^{23} ] + [ ( ^{23} _{1} ) x (0.03)^{1} x (0.97)^{22} ] = 0.496 + 0.353 = 0.849 |
front 22 A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 19 tablets. The entire shipment is accepted if at most 2 tablets do not meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 3.0% rate of defects, what is the probability that this whole shipment will be accepted? | back 22 The probability that this whole shipment will be accepted is 0.982. P(x ≤ 2) = P(x=0 or x=1 or x=2) = [ (0.97)^{19} ] + [ ( ^{19} _{1} ) x (0.03)^{1} x (0.97)^{18} ] + [ (^{} ^{19} _{2} ) x (0.03)^{2} x (0.97)^{17} ] = 0.5606 + 0.3294 + 0.0917 = 0.9817 |
front 23 Which of the following is not a requirement of the binomial probability distribution?
| back 23 The trials must be dependent. "The trials must be dependent" is not a requirement of a binomial probability distribution. For a binomial distribution, the trials must be independent. |
front 24 Which of the following is NOT one of the three methods for finding binomial probabilities that is found in the chapter on discrete probability distributions?
| back 24 Use a simulation Using a simulation is not one of the suggested methods for finding binomial probabilities. |
front 25 In a clinical trial of a cholesterol drug, 427 subjects were given a placebo, and 21% of them developed headaches. For such randomly selected groups of 427 subjects given a placebo, identify the values of n, p, and q that would be used for finding the mean and standard deviation for the number of subjects who develop headaches. | back 25 The value of n is 427. The value of p is 0.21. The value of q is 0.79. |
front 26 Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ – 2σ and the maximum usual value μ + 2σ. n = 1588, p = 3 / 4 | back 26 μ = 1191 μ = np = (1588)(0.75) = 1191 σ = 17.3 σ = √npq = √(1588 x 0.75 x 0.25) = 17.25543393 μ – 2σ = 1156.5 μ + 2 σ = 1225.5 |
front 27 Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ – 2σ and the maximum usual value μ + 2σ. In an analysis of preliminary test results from the a gender-selection method, 19 babies are born and it is assumed that 50% of babies are girls, so n = 19 and p = 0.5. | back 27 The value of the mean is μ = 9.5. μ = np = (19)(0.5) = 9.5 The value of the standard deviation σ = 2.2 . σ = √npq = √(19 x 0.5 x 0.5) = 2.179449472 The minimum usual value is μ – 2σ = 5.1 The maximum usual value is μ + 2σ = 13.9 |
front 28 Several psychology students are unprepared for a surprise true/false test with 12 questions, and all of their answers are guesses.
| back 28 a. μ = 6 = (12)(0.5) = 6 σ = 1.7 σ = √npq = √(12 x 0.5 x 0.5) = 1.732050808 b. Yes, because 10 is greater than the maximum usual value. μ – 2σ = 2.6 μ + 2σ = 9.4 |
front 29 A headline in a certain newspaper states that "most stay at first job less than 2 years." That headline is based on an online poll of 390 college graduates. Among those polled, 76 % stayed at their first full-time job less than 2 years.
| back 29 a. The mean is 195 graduates. μ = np = (390)(0.50) = 195 The standard deviation is 9.9 graduates. σ = √npq = √(390 x 0.50 x 0.50) = 9.874208829 b. The numbers of graduates among 390 who stay at their first job less than two years should usually fall between 175.2 and 214.8. μ – 2σ = 175.2 μ + 2σ = 214.8 c. The actual number of surveyed graduates who stayed at their first job less than two years is 296. 390 x 76% = 296.4 (Round to the nearest whole number) Since the actual number is outside of the range of usual values found in part (b), it is unusual. The result suggests that the headline is justified. Because the actual number of surveyed graduates who stayed at their first job less than two years is larger than the maximum usual value under the assumption that 50% is the true percentage. Therefore, it is unlikely that the true percentage is 50% or less. d. Since the sample of 390 respondents is a voluntary response sample, the results are very questionable. |
front 30 A government agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. In standard English text, a particular letter is used at a rate of 6.8%.
| back 30 a. μ = 108.8 μ = np = (1600)(0.068) = 108.8 σ = 10.1 σ = √npq = √(1600 x 0.068 x 0.932) = 10.06983615 b. No, because 116 is within the range of usual values. μ – 2σ = 88.6 μ + 2σ = 129.0 |
front 31 If a gambler places a bet on the number 7 in roulette, he or she has a 1/38 probability of winning.
| back 31 a. The value of the mean is μ = 4.5 μ = np = (170)(1/38) = 4.473684211 The value of the standard deviation is σ = 2.1 σ = √npq = √(170 x 1/38 x 37/38) = 2.087092638 b. Yes, because 0 is below the minimum usual value. μ – 2σ = 0.3 μ + 2σ = 8.7 |
front 32
| back 32 a. The value of the mean is μ = 0.219178 μ = np = (80)(1/365) = 0.2191780822 The value of the standard deviation is σ = 0.467523 σ = √npq = √(80 x 1/365 x 364/365) = 0.4675228276 b. Yes, because 2 is greater than the maximum usual value. μ – 2σ = -0.715868 = 0 μ + 2σ = 1.154224 |
front 33 For the binomial distribution, which formula finds the standard deviation? | back 33 √npq The formula √npq is the formula that finds the standard deviation for the binomial distribution. |
front 34 Identify the expression for calculating the mean of a binomial distribution. | back 34 np The formula μ = np is the expression that is equal to the mean of a binomial probability distribution. |