quiz4_key - mean Y = 176 mg/dLi and standard deviation Y =...

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NAME: Stat 205 Quiz 4 1. In the United States, 42% of the population has type A blood. Consider taking a sample of size n = 4 . Let Y denote the number of persons in the sample with type A blood. (a) (3 points) What is P (0 Y 2) = P ( Y 2) , the probability that two or fewer have type A blood? Answer: You can compute this directly with your TI-84 using binomcdf(4,0.42,2) obtaining 0.800. Otherwise you need to compute P ( Y 2) = P ( Y = 0) + P ( Y = 1) + P ( Y = 2) = 4 C 0 (0 . 42) 0 (0 . 58) 4 + 4 C 1 (0 . 42) 1 (0 . 58) 3 + 4 C 2 (0 . 42) 2 (0 . 58) 2 = 1(0 . 42) 0 (0 . 58) 4 + 4(0 . 42) 1 (0 . 58) 3 + 6(0 . 42) 2 (0 . 58) 2 = 1(0 . 113) + 4(0 . 082) + 6(0 . 059) = 0 . 795 (b) (2 points) What is P ( Y = 0) , the probability that none of the four sampled have type A blood? Answer: binompdf(4,0.42,0) gives 0.113. Otherwise P ( Y = 0) = 4 C 0 (0 . 42) 0 (0 . 58) 4 = (0 . 58) 4 = 0 . 113 . (c) (2 points Extra Credit ): What is the mean and standard deviation of Y ? Answer: μ Y = E ( Y ) = np = 4(0 . 42) = 1 . 68 , σ Y = p np (1 - p ) = p 4(0 . 42)(0 . 58) = 0 . 987 . 2. The serum cholesterol level Y of a random 17-year old follows a normal distribution with
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Unformatted text preview: mean Y = 176 mg/dLi and standard deviation Y = 30 mg/dLi. High cholesterol is any level greater than 200 mg/dLi. (a) (3 points) What is the probability that a randomly selected 17-year old will have high cholesterol, P ( Y > 200) ? Answer: normalcdf(200,10 99,176,30) gives 0.212. Otherwise you can use a table and compute P ( Y > 200) = 1-P ( Y < 200) = 1-P Y-176 30 < 200-176 30 = 1-P ( Z < . 8) = 1-. 7881 = 0 . 212 (b) (2 points) What is the 78.8th percentile of cholesterol levels, i.e. the number k such that P ( Y < k ) = 0 . 788 ? (Hint: what is P ( Y < 200) from part (a)?) Answer: You could use invNorm(0.788,176,30) to get 200 or else use part (a) directly noting that P ( Y < 200) = 1-. 212 = 0 . 788 ....
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