Practice Exam1 Solutions

# Practice Exam1 Solutions - Practice Exam 1 Solutions 1...

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Practice Exam 1 Solutions 1. P(Bill applies) = 1 P(Tina applies) = .6 P(Bill awarded grant | Tina applies) = .3 P(Bill awarded grant | Tina NOT applies) = .65 Find P(Tina NOT applies | Bill awarded grant): P(Tina NOT applies | Bill awarded grant) = P(Tina NOT applies ∩ Bill awarded grant) / P(Bill awarded grant) = P(Bill awarded grant | Tina NOT applies) * P(Tina NOT applies) / P(Bill awarded grant) = .65*.4 / (P(Bill awarded grant | Tina not applies)*P(Tina NOT applies) + P(Bill grant|Tina applies)*P(Tina applies)) = .65*.4 / (.65*.4 + .3*.6) = .591 2. a. Use formulas: Mean=7.73, median=7.75, variance=1.83, sd=1.35 b. Easy. c. n = 23, p = .4 np = 9.2 10 th number is 40 th percentile = 7 (Wrong! n should be 22, so, np=8.8 -> 9 th number) d. Boxplot (all you need is max, min, Q1, Q2, Q3) e. 7 classes. Values 5 to 11. Distribution looks like this: Class Frequency [5, 6) 2 [6,7) 4 [7,8) 6 [8,9) 7 [9,10) 2 [10,11] 2 Now draw histogram… 3. sample of television sets from 10 with 3 defective. Let X = # of defectives in sample.

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• Spring '08
• Mendell
• Probability theory, Binomial distribution, Discrete probability distribution, Hypergeometric Distribution, grant | Tina

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Practice Exam1 Solutions - Practice Exam 1 Solutions 1...

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