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# 400Hw09ans - STAT 400 Spring 2011 Homework#9(due Friday...

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STAT 400 Spring 2011 Homework #9 (due Friday, April 1, by 3:00 p.m.) 1. a) Suppose that the actual amount of instant coffee a filling machine puts into "6-ounce" cans varies from can to can and that the actual fill may be considered a random variable having a normal distribution, with a mean of 6.03 ounces and a standard deviation of 0.04 ounces. Suppose a case contains 12 cans and can be considered as a random sample of cans. What is the probability that the average fill in a case is below 6 ounces? μ = 6.03, σ = 0.04, n = 12. Want P( X < 6 ) = ? Since the distribution we sample from is normal, . n Z X = - σ μ P( X < 6 ) = - < 12 04 . 0 03 . 6 6 Z P = P( Z - 2.60 ) = 0.0047 . b) Suppose that the actual amount of instant coffee a filling machine puts into "6-ounce" cans varies from can to can and that the actual fill may be considered a random variable having a normal distribution, with a standard deviation of 0.04 ounces. If on average only 3 out of every 200 cans contain less than 6 ounces of coffee, what must be the mean fill of these cans? Normal distribution, μ = ? Know that P( X < 6 ) = 200 3 = 0.015. Find z such that P( Z < z ) = 0.015. The area to the left is 0.015 (less than 0.50). Using the standard normal table, z = 2.17 . x = μ + σ z . 6 = μ + 0.04 (– 2.17). μ = 6.0868 oz .

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2. Suppose that 60% of all voters in a certain county favor a tax increase to pay for a better school system. A random sample of 400 voters is selected. Use Normal approximation to find the following: Binomial( 400, 0.60 ). μ = 240, σ = 9.798. a) What is the probability that more than 250 of the sample members favor the tax increase? P( X > 250 ) = P( X > 250.5 ) = P( Z > 1.07 ) = 0.1423 . b) What is the probability that at least 260 of the sample members favor the tax increase? P( X 260 ) = P( X > 259.5 ) = P( Z > 1.99 ) = 0.0233 . c) What is the probability that at most 235 of the sample members favor the tax increase? P( X 235 ) = P( X < 235.5 ) = P( Z < – 0.46 ) = 0.3228 . d) What is the probability that between 225 and 245 (both inclusive) of the sample members favor the tax increase? P( 225 X 245 ) = P( 224.5 < X < 245.5 ) = P( – 1.58 < Z < 0.56 ) = 0.6552 . e) What is the probability that exactly 234 of the sample members favor the tax increase? Compare the answer with the exact probability. P( X = 234 ) = P( 233.5 < X < 234.5 ) = P( – 0.66 < Z < – 0.56 ) = 0.0331 . Binomial: P( X = 234 ) = ( ) ( ) 166 234 40 . 0 60 . 0 234 400 = 0.033565 .
3. Suppose the number of e-mail messages received by Mike follows Poisson distribution with the average rage of 42 messages per week. Use Normal approximation to find the following: Poisson, λ = 42. μ = 42, σ = 6.48. a) What is the probability that Mike would receive over 50 e-mail messages in one week? P( X > 50 ) = P( X > 50.5 ) = P( Z > 1.31 ) = 0.0951 . b) What is the probability that Mike would receive exactly 50 e-mail messages in one week? Compare the answer with the exact probability. P( X = 50 ) = P( 49.5 < X < 50.5 ) = P( 1.16 < Z < 1.31 ) = 0.0279 . Poisson: P( X = 50 ) = ! 50 42 42 50 - e = 0.02748 . c) What is the probability that Mike would receive at most 45 e-mail messages in one week?

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