Instructions for Side by Side Printing
- Print the notecards
- Fold each page in half along the solid vertical line
- Cut out the notecards by cutting along each horizontal dotted line
- Optional: Glue, tape or staple the ends of each notecard together
General Statistics: Ch 5 Quiz
front 1  The accompanying table describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods.
| back 1 a. 0.268 b. 0.364 (0.268 + 0.096) c. The result from part (b) d. No, since the appropriate probability is greater than 0.05, it is not an unusually high number. |
front 2  Based on data from a car bumper sticker study, when a car is randomly selected, the number of bumper stickers and the corresponding probabilities are as shown below.
| back 2 a. Yes b. The mean is 0.6. The standard deviation is 1.5. c. The maximum usual value is 3.5. max: μ + 2σ = 0.578 + 2(1.462161414) = 3.502322828 The minimum usual value is 0. min: μ – 2σ = 0.578 – 2(1.462161414) = -2.346322828 d. No, because the probability of more than 1 bumper sticker is 0.120, which is greater than 0.05. 1 – (0.772 + 0.108) = 0.12 |
front 3 Determine whether the value is a discrete random variable, continuous random variable, or not a random variable.
| back 3 a. It is a discrete random variable. b. It is a discrete random variable. c. It is not a random variable. d. It is a discrete random variable. e. It is a continuous random variable. f. It is a discrete random variable. |
front 4  In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation. | back 4 Its mean is 2.9. Its standard deviation is 1.3. |
front 5  Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.45. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? | back 5 μ = 1.35 σ = 0.86 No, because the probability of having 3 girls is greater than 0.05. |
front 6  Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.31. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? | back 6 μ = 0.93 σ = 0.80 Yes, because the probability of having 3 girls is less than or equal to 0.05. |
front 7 In a state's Pick 3 lottery game, you pay $1.45 to select a sequence of three digits (from 0 to 9), such as 422. If you select the same sequence of three digits that are drawn, you win and collect $439.13.
| back 7 a. 1,000 b. 0.001 c. $437.68 ($439.13 – $1.45) d. $(1.01) (–$1.45 + $439.13/1000) = -1.45 + 0.44 = -1.01 e. Neither bet is better because both games have the same expected value. |
front 8  Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. | back 8 μ = 2.9 σ = 1.1 |
front 9 Determine whether or not the procedure described below results in a binomial distribution. If it is not binomial, identify at least one requirement that is not satisfied. Six hundred different voters in a region with two major political parties, A and B, are randomly selected from the population of 3.7 million registered voters. Each is asked if he or she is a member of political party A. | back 9 Yes, the result is a binomial probability distribution. |
front 10 A certain TV show recently had a share of 85, meaning that among the TV sets in use, 85% were tuned to that show. Assume that an advertiser wants to verify that 85% share value by conducting its own survey, and a pilot survey begins with 8 households having TV sets in use at the time of the TV show broadcast.
| back 10 a. The probability that all of the households are tuned to the TV show is 0.272. P(x=8) = (0.85)8 = 0.272490525 b. The probability that exactly 7 households are tuned to the TV show is 0.385. P(x=7) = ( 8 7 ) x (0.85)7 x (0.15)1 = 0.3846925059 c. The probability that at least 7 households are tuned to the TV show is 0.657. P(x ≥ 7) = P(x=7 or x=8) = 0.27249 + 0.38469 = 0.65718 d. No, because 7 households tuned to the TV show is not unusually high if the share is 85%. |
front 11  Refer to the accompanying technology display. The probabilities in the display were obtained using the values of n = 5 and p = 0.738. In a clinical test of a drug, 73.8% 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 more than one subject experiences headaches. Is it reasonable to expect that more than one subject will experience headaches? | back 11 The probability that more than one subject experiences headaches is 0.9814. P(x > 1) = P(x=2 or x=3 or ... or x=5) = 1 – P(x=0 or x=1) = 1 – (0.0012 + 0.0174) = 0.9814 Yes, because the event that the number of subjects that experience headaches is less than or equal to one is unlikely. |
front 12 Nine peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 9 offspring peas, at least 8 have green pods. Is it unusual to get at least 8 peas with green pods when 9 offspring peas are generated? Why or why not? | back 12 The probability that at least 8 of the 9 offspring peas have green pods is 0.300. P(x ≥ 8) = P(x=8 or x=9) = [ ( 9 8 ) x (0.75)8 x (0.25)1 ] + [ (0.75)9 ] = 0.225 + 0.075 = 0.3003387451 No, because the probability of this occurring is not small. |
front 13 A brand name has a 60% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 5 randomly selected consumers.
| back 13 a. The probability that exactly 4 of the 5 consumers recognize the brand name is 0.259. P(x=4) = ( 5 4 ) x (0.60)4 x (0.40)1 = 0.2592 b. The probability that all of the selected consumers recognize the brand name is 0.078. P(x=5) = (0.60)5 = 0.07776 c. The probability that at least 4 of the selected consumers recognize the brand name is 0.337. P(x ≥ 4) = P(x=4 or x=5) = 0.259 + 0.078 = 0.33696 d. No, because the probability that 4 or more of the selected consumers recognize the brand name is greater than 0.05. |
front 14  Assume that a procedure yields a binomial distribution with n = 2 trials and a probability of success of p = 0.50. Use a binomial probability table to find the probability that the number of successes x is exactly 1. | back 14 P(1) = 0.500 |
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=6, x=4, p=0.25 | back 15 P(4) = 0.033 P(x=4) = ( 6 4 ) x (0.25)4 x (0.75)2 = 0.0329589844 |
front 16 Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Determining whether each of 50 mp3 players is acceptable or defective | back 16 Yes, because all 4 requirements are satisfied. |
front 17 A TV show, Lindsay and Tobias, recently had a share of 15, meaning that among the TV sets in use, 15% were tuned to that show. Assume that an advertiser wants to verify that 15% share value by conducting its own survey, and a pilot survey begins with 18 households having TV sets in use at the time of a Lindsay and Tobias broadcast.
| back 17 a. 0.054 P(x=0) = (0.85)18 = 0.0536464098 b. 0.946 P(x ≥ 1) = P(x=1 or x=2 or ... or x=18) = 1 – P(x=0) = 1 – 0.054 = 0.9463535902 c. 0.224 P(x ≤ 1) = P(x=0 or x=1) = 0.054 + [ ( 18 1 ) x (0.15)1 x (0.75)17 ] = 0.054 + 0.170 = 0.2240526527 d. No, because with a 15% rate, the probability of at most one household is greater than 0.05. |
front 18 Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Treating 150 bald men with a special shampoo and asking them how their scalp feels | back 18 No, because there are more than two possible outcomes. |
front 19 Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied. Recording the genders of 150 people in a statistics class | back 19 Yes, because all 4 requirements are satisfied. |
front 20 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 = 1550, p = 3 / 5 | back 20 μ = 930 μ = np = (1550)(0.60)
= 930 σ = 19.3 σ = √npq = √(1550 x 0.60 x 0.40) = 19.28730152 μ – 2σ = 891.4 μ + 2 σ = 968.6 |