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QUIZ 6.3B 1. a) This is neither binomial nor geometric, because the probability of success (selecting a female student) is probably different in each trial, unless every classroom has exactly the same proportion of females, which is unlikely. b) This is a geometric setting: BINARY outcomes (more than two occupants or not), INDEPENDENT trials (the number of occupants of one car does not influence the number of occupants in the next randomly-selected car), we are counting the number of TRIALS to the first car with more than two occupants, and the probability of SUCCESS—finding a car with more than two occupants—is always the same. 20 (0.02)1 (0.98)19 = 0.2725 1 2. a) X is binomial with n = 20 and p = 0.02. P(X = 1) = b) P(X 2 ) c) µx = binomcdf (20, 0.02, 2) = 0.9929 = np = (20)(0.02) = 0.4 3. a) (0.8)2(0.2) = 0.128 σx = np(1 p) = (20)(0.02)(0.98) = 0.63 (b) P(X 4) = 1-P(X 3) = 1-geomcdf (0.2, 3) = 0.512 QUIZ 7.1C 1. a) The parameter is the proportion of people in the entire community who would answer “Yes” to the question. It’s equal to 0.40. The statistic is the proportion of people in the sample of 100 who would answer “Yes” to the question. b) The sampling distribution describes the distribution of the proportion of people who would answer “Yes” to this question in all possible samples of size 100 from this population. c) If a statistic is an unbiased estimator of the parameter, then the mean of the stastic’s sampling distribution is equal to the parameter. d) No. As long as the sample is less than 10% of the population, the size of the population from which the sample is taken does not influence the sampling distribution. e) Yes. The standard deviation of the sampling distribution would be larger if the sample size were smaller. 3. a) The population is all the guppies in the pet store. We’ve been given the population mean, 5 cm, and the population standard deviation, 0.5 cm. b) The sample mean is x = 4.8 cm and the sample size is n = 10. c) No, it’s merely an approximation of a sampling distribution generated by simulating 200 sample means. The actual sampling distribution includes the means from all possible samples of size 10 from the population—many more than the 200 values shown! d) 21 out of 200, or 10.5%, of the sample means in our simulation are as far or farther below 5.0 as our sample was. Our sample is not sufficiently unusual to arouse suspicions about the store’s claim. QUIZ 7.2B 1. a) p-hat = p = 0.22 ; σ p-hat = p(1 p) = n (.22)(.78) = 0.034 150 b) Since np = (150)(.22) = 33 10 AND n(1-p) = (150)(.78) = 117 10, the distribution is approximately normal. c) If the sample size were 36 instead of 150, the mean, µ p-hat would not change. The standard deviation, σ p-hat would be larger ( (.22)(.78) = 0.069), and the distribution 36 would be non-Normal, since np = (36)(0.22) = 7.92, which is less than 10. d) The largest sample we can take is 60 (n 1 N , where N = 600), otherwise the sample 10 would be more than 10% of the population, and sampling without replacement would require a finite population correction to calculate standard deviation. (That is not in the scope of this course!) 2.a) µ p-hat = p =0.22; σ p-hat = (.22)(.78) = 0.0589 50 b) State: What is the probability that more than 30% of the 50 randomly-selected songs on George’s mp3 player are Beatles songs: Plan: P (p-hat > 0.30) Check the Normal condition: np = (50)(.22) = 11, and n(1-p) = (50)(.78) = 39. Both are greater than 10, so the Normal approximation will be quite accurate. So. . .Do: Standardize! z 0.30 0.22 0.0589 = 1.36 . DRAW and LABEL the Normal curve! Shade to the right. Using Table A, P (z > 1.36) = 1- .9131 = .0869. Conclude: Therefore, there is approximately an 8.69% chance that more than 30% of George’s randomly selected songs are Beatles songs played on his mp3 player. QUIZ 7.3A 1. a) µ x = µ = 12 minutes b) Yes. It seems reasonable to assume that the sample of 10 is less than 10% of the entire population calls. σ x = 5 10 = 1.58 c) No. The population distribution is skewed right, and n = 10, which is not large enough for the CLT to apply. 2. a) P( x>400) = P z 400 380 = P( z > 0.71) = 1 - .7611 = 0.2389 28 There is a 23.89% chance that a single randomly selected apple will weigh more than 400 gm. 400 380 b) P( x > 400) = P z = P( z > 1.24) = 1 - .8925 = .1075 16.165 There is a 10.75% chance that the mean weight of the SRS of 3 apples will weigh more than 400 gm. c) The mean weight of a random sample of three apples is less variable than the weight of a single randomly-selected apple, so we are less likely to get a mean weight that is 20 gm above the mean when we take a sample of three apples.