Homework_4_Solutions

Homework_4_Solutions - Homework 4 Solutions 7.2 1/16 =...

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Homework 4 Solutions 7.2 1/16 = .0625. After four students have been selected, sixteen remain as candidates, each with an equal chance to be picked. 7.7 a. Relative frequency probability; the proportion of times the outcome occurs in the long run. b. Personal probability. c. Relative frequency probability; proportion of a “large” random sample that falls into the category of interest. 7.16 Of the children who slept in darkness, the number with myopia or high myopia is 15 + 2 = 17. So, the probability = 17/172 = .0988 that a randomly selected child who slept in darkness would develop some degree of myopia. 7.17 a. BY, BS, BA, YS, YA, SA. b. 1/6. 7.19 a. Yes. The outcome for one coin does not affect the probabilities for the other coin. b. No. The outcomes for the two coins apply to separate random circumstances (the outcome for each coin is one random circumstance) and complementary events are defined only for the same random circumstance. c.
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This note was uploaded on 12/06/2009 for the course STAT 200 at Penn State.

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Homework_4_Solutions - Homework 4 Solutions 7.2 1/16 =...

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