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### hw8sol

Course: SMAM 351, Fall 2009
School: RIT
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Word Count: 318

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351 Homework SMAM 8 Solution 1. A. Using the lack of memory property of the exponential distribution P[X &lt; 0.7) = 0 .5e.5x dx = e.5x B. The number of customers that arrive each minute is a Poisson distribution with mean 5. The number that arrive in 10 minutes is Poisson with mean 5 minutes. P[N &gt; 4) = 1 P[N 4] = 1 .440 = .560 Alternatively P[N(10) &gt; 4] = P[N(10) 5] = P[T5 10] 10 1 = x 4e...

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351 Homework SMAM 8 Solution 1. A. Using the lack of memory property of the exponential distribution P[X < 0.7) = 0 .5e.5x dx = e.5x B. The number of customers that arrive each minute is a Poisson distribution with mean 5. The number that arrive in 10 minutes is Poisson with mean 5 minutes. P[N > 4) = 1 P[N 4] = 1 .440 = .560 Alternatively P[N(10) > 4] = P[N(10) 5] = P[T5 10] 10 1 = x 4e x/2 dx = .559 5 0 2 (5) The above integral may be done with a graphing calculator, computer algebra system or integration by parts four times. C. The time until one customer calls in has a mean of 2 minutes and a variance of 4 minutes. The time until 5 coustomers call in has a mean of 10 minutes and a variance of 20 minutes.Alternatively the gamma distribution in B has parameters =5 =2. The result then follows using the formulae for a mean and variance of a Gamma distribution. A number of you had trouble with the above problem. Please study the solution carefully. 2. A. Using the formulae on page 175 together with the properties of the Gamma function 3 1 = 900( 2 ) = 900( 1 ) ( 2 ) = 450 = 797.6 2 0.7 0.7 0 = 1 e .35 = .295 = 2 (900)2{ (2) 4 } = 173827.5 B. Using 4.12 P(T < 800) = 1 e (800/900) = .546 2 C. P(T > 797.6) = e (797.6/900) = .456 2 Computer software is not needed to do the problem if you have a good calculator. However 1 A, B and 2 B,C may be done using Minitab. Here is the solution Worksheet size: 100000 cells 1A. MTB > CDF .7; SUBC> Exponential 2. Cumulative Distribution Function Exponential with mean = 2.00000 x 0.7000 B. MTB > cdf 4 c1; SUBC> poisson 5. MTB > let c2=1-c1 MTB > print c2 Data Display C2 0.559507 Alternatively MTB > CDF 10; SUBC> Gamma 5 2. Cumulative Distribution Function Gamma with a = 5.00000 and b = 2.00000 x 10.0000 2B MTB &...

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