# QUIZ 4-1 - P X ≤ 8 = P X< 8.5 T P X< 8.5 =< 19 2 6...

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STAT 400 Spring 2011 Version A Name ANSWERS . Quiz 4 (10 points) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. (5) A recruiter is interviewing job candidates. From past data, the recruiter thinks that about 20% of the potential job candidates have the qualifications necessary to be hired for a certain middle management position. Assume independence. If the recruiter interviews 30 candidates, what is the probability that there will be at most 8 suitable job prospects? (Use Normal approximation.) Let X = the number of suitable job prospects out of 30 candidates. Then X has Binomial distribution, n = 30, p = 0.20. Need P ( X 8 ) = ? o μ = n × p = 30 × 0.20 = 6. σ 2 = n × p × ( 1 – p ) = 30 × 0.20 × 0.80 = 4.8. σ = 8 . 4 2.19. t 0.5 correction: want 8, do not want 9
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Unformatted text preview: P ( X ≤ 8 ) = P( X < 8.5 ) T P ( X < 8.5 ) = -< 19 . 2 6 5 . 8 Z P = P ( Z < 1.14 ) = 0.8729 . ( Binomial distribution, n = 30, p = 0.20, P ( X ≤ 8 ) = 0.87135. ) 2. (5) A communication system for a company has 40 outside lines. If the number X of line requests at a given time follows a Poisson distribution with mean 36, compute the probability that an incoming call cannot find an open line. That is, find the probability that the number of lines needed, X, exceeds 40. (Use Normal approximation.) Need P(X > 40) = ? o μ = λ = 36. σ = λ = 6. t 0.5 correction: do not want 40, want 41. P(X > 40.5) = ? T P(X > 40.5) = P -< 6 36 5 . 40 Z = P(Z > 0.75) = 1 – Φ (0.75) = 0.2266 . ( Poisson distribution, λ = 36, P(X > 40) = 0.2229. )...
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## This note was uploaded on 11/28/2011 for the course STAT 400 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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