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Practice Quiz 5 1) For a certain type of battery, the lifetimes of batteries under continuous use follow a Normal distribution with mean 17 hours and standard deviation 0.8 hours. a) What is the probability that a randomly chosen battery has a lifetime less than 16.7 hours? (Hint: this is not a chapter 7 problem (yet) because you donβt have a random sample of batteries, just one randomly chosen battery) b) What are the mean and standard deviation of the sampling distribution of π₯Μ , the average battery lifetime of a random sample of 60 batteries? Please use correct notation. c) What is the probability that the average battery lifetime of a random sample of 60 batteries is less than 16.7 hours? (Explain why you can use the Normal calculation) 2) Suppose that 47% of all adult women think they do not get enough time for themselves. An opinion poll interviews 1025 randomly chosen women and records the sample proportion, πΜ ,who feel they donβt get enough time for themselves. a) What is the mean of the sampling distribution of πΜ ? b) What is the standard deviation of the sampling distribution of πΜ ? c) Check that the distribution of πΜ is approximately Normal. d) Find the probability that πΜ takes on a value between 0.45 and 0.50. Solutions 1) a) P(X < 16.7) = P(z < -0.38) = 0.3520 b) ο x = 17, ο³ x = 0.10328 c) P( x < 16.7) = P(z < -2.90) = 0.0019, we can use the Normal calculation for TWO reasons. First, because we were told that the population distribution is Normal. Second, because the sample size is at least 30, and according to the Central Limit Theorem the distribution of π₯Μ follows approximately a Normal distribution if the sample is large enough (at least 30). We only need ONE of these conditions to be met, but in this case both of them were met. 2) a) ππΜ =0.47, b) ππΜ =0.01559 c) the distribution of πΜ follows a Normal distribution because both np and nq are at least 5. 1025(.47) = 482 >= 5, 1025(.53) = 543 >= 5 d) P(0.45 < πΜ < 0.50) = P(-1.28 < z < 1.92) = 0.9726 β 0.1003 = 0.8723