QUIZ 4-3 - (0.45) = 0.6736 . ( Binomial distribution, n =...

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STAT 400 Spring 2011 Version D 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) Nielsen ratings show that 20% of all television viewers watch a particular program. A random sample of 625 TV viewers is taken. Use Normal approximation to find the probability that fewer than 130 viewers in the sample watch this program. Binomial, p = 0.20, n = 625. Need P(X < 130) = ? o μ = n p = 625 0.20 = 125. σ 2 = n p ( 1 – p ) = 625 0.20 0.80 = 100. σ = 10. t 0.5 correction: do not want 130, want 129. P(X < 129.5) = ? T P(X < 129.5) = P - < 10 125 5 . 129 Z = P(Z < 0.45) =
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Unformatted text preview: (0.45) = 0.6736 . ( Binomial distribution, n = 625, p = 0.20, P(X &lt; 130) = 0.6765. ) 2. (5) The number of customers who visit a small local book shop follows a Poisson distribution with the average rate of 2.7 customers per day. Find the probability that more than 85 customers would visit the shop in one month (30 days). (Use Normal approximation.) Need P(X &gt; 85) = ? o = = 30 2.7 = 81. = = 9. t 0.5 correction: do not want 85, want 86. P(X &gt; 85.5) = ? T P(X &gt; 85.5) = P -&lt; 9 81 5 . 85 Z = P(Z &gt; 0.50) = 1 (0.50) = 0.3085 . ( Poisson distribution, = 81, P(X &gt; 85) = 0.3037. )...
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