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homework_4_sol

# homework_4_sol - Course MTH 1210 Homework#4(solutions and...

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Course MTH 1210 Homework #4 ( solutions and answers ) ( Binomial and Poisson Distributions ) Mar 16’10 Use the MS Excel program Three_distributions.xls Binomial Probability Experiment is an experiment that is made up of repeated trials of the same basic experimental event. Such an experiment must possess the following properties: 1. Each trial has only two possible outcomes (usually labeled as a success or a failure ). 2. There are n repeated independent trials, so that the probability of success remains the same for each trial. 3. The number x of successful trials may take on any integer value from 0 to n , and may serve as the binomial discrete random variable . 1 . Drug Effectiveness . A new drug curies 70% of the people taking it. Suppose 20 people take the drug: find the probabilities of the following. This is a binomial probability problem: after taking a drug a person may be either curried or not (two possible outcomes); the probability of success is: p = 0.7 ; thus, q = 1 – 0.7 = 0.3 . a) Exactly 18 people are cured. P(x=18) = n C x p x q n–x = 20 C 18 0.7 18 0.3 20–18 = 0.028 b) Exactly 17 people are cured. P(x=17) = n C x p x q n–x = 20 C 17 0.7 17 0.3 20–17 = 0.0716 c) At least 17 people are cured. P(X 17) = P(X=17) + P(X=18) + P(X=19) + P(X=20) = = 1 – P(X ≤ 16) = 1 – 0.8929 = 0.107 16 The cumulative probability P(X ≤ 16) = Σ P(i) i=0 d) At least 18 people are cured. P(X 18) = P(X=18) + P(X=19) + P(X=20) = = 1 – P(X ≤ 17) = 1 – 0.9645 = 0.0354 2 . Vitamin A Deficiency . Six mice from the same litter, all suffering from a vitamin A deficiency, are fed a certain dose of carrots. If the probability of recovery under such a treatment is 0.70 , find the probabilities of the following results. Again, this is a binomial probability problem: after eating a carrot containing the vitamin A a mouse may be either curried or not (two possible outcomes); the probability of success is: p = 0.7 ; thus, q = 1 – 0.7 =

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