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Unformatted text preview: The 1/4 on the yaxis was found using the following equation: (x 2 x 1 ) * y = 1 so... (9  5) * y = 1 y = 1/4 5 9 1/4 =1  (POISSON(4,4.2,TRUE)) Probability Density Function Poisson with mean = 4.2 x P( X <= x ) 4 0.589827 HA 2201 Fall 2009 Homework III Solution Question 1. A. This is a uniform probability distribution because all possible times of arrival for each guest are equal between 5pm & 9pm. So X~U(5,9). B. P(X < 6.5) = 1.5*.25 (Since the interval from 5 to 6.5 has width 1.5) =0.375 or 37.5% C. P(X > 7.75)=1.25*0.25 (Since the interval from 7.75 to 9 has width 1.25) = 0.3125 or 31.25% D. P(5.75 < X < 8) = 2.25*0.25 (Since the interval from 5.75 to 8 has width 2.25) = 0.5625 or 56.25% E. ) You have 60 dining parties arrive in an evening. ASSUME each decides on its arrival time independent of all other dining parties. Let Y = the number out of 60 that decide to arrive between 5:45 pm but before 8:00 pm. i) Y is a binomial distribution where Y~B(60,0.5625) because there is an a priori cap of 60 guests that decide to arrive in an evening to dine between 5:45 pm and 8:00 pm that are independent of one another with either success or failure—guests arrive or do not arrive. ii) P(Y = 32) = 0.0927 or 9.27% = BINOMDIST(32, 60, 0.5625, FALSE Probability Density Function Binomial with n = 60 and p = 0.5625 x P( X = x ) 32 0.0927190 iii) P(Y < 32) = 0.37081 of 37.081% =BINOMDIST(32, 60, 0.5625, TRUE) Cumulative Distribution Function Binomial with n = 60 and p = 0.5625 x P( X <= x ) 32 0.370787 iv) P(Y < 32) = P(Y ≤ 32)P(Y = 32) = 37.081%  9.27% = 27.813% v) P(Y > 34) =1  P(Y ≤ 34) =1  0.5752 = 0.4248 or 42.48% 0.575= 0.425 or 42.5% =BINOMDIST(34, 60, 0.5625, TRUE) Cumulative Distribution Function Binomial with n = 60 and p = 0.5625 x P( X <= x ) 34 0.575208 Question 2. A. This is a Poisson distribution because there is no a priori maximum number of customers requesting tickets, with an average of 4.2 per hour. X~P(4.2). We must assume that the requests are independent of one another. In this case, we can assume that it’s VALID because there are a lot of tickets and shows in Vegas so the ticket requests likely have nothing to do with one another. however, the fact that friends may come up together to buy tickets cast some doubt on this assumption....
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 '06
 RLLOYD
 Normal Distribution, Probability theory, representative, Cumulative distribution function, business travelers

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