Section 5.5 Lecture Problems

# Section 5.5 Lecture Problems - ASW ESBE(5e Chapter 5...

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ASW ESBE (5e) Chapter 5 Solutions to Selected Problems Section 5.5 for the Poisson distribution. Problems 40, 41, 43, 44, 45. 40. What kind of problem is this? “Phone calls arrive at the rate of 48 per hour.” In this problem, counting the number of call that arrive during an interval of time. This sentence has all of the key words of a Poisson problem: “arrive,” “rate,” and “per.” So this is a Poisson problem. Need μ must refer to the same length of interval during which x number of calls is counted. Given: μ = 48 per hour (first sentence). Let X be the r.v. that counts the number of calls received during a given time interval during which μ calls are expected. The values of x are x = 0 (none), 1, 2, …. Probabilities calculated by with the μ for the same length of time interval during which x is counted (the interval specified in the parts). Per minute, expect μ = (48 call/hour) × (1 hour/60 min) = 0.8 calls/min a. Three calls arrive means x = 3. Want prob ( x = 3 in a 5 minute interval) In 5 minutes , expect μ = (0.8 calls/min) × 5 min = 4 calls/5-min interval. So want Prob ( x = 3 | μ = 4) Some Section 5.4 Problem Solutions 1 ( 29 ! x e | x X Prob x μ = μ = μ -

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( ) ... . ... . ! e | x prob 1953668 0 = 6 64 × 0183156 0 = 3 4 × = 4 = 3 = 3 4 - μ So prob ( x =3| μ =4) = 0.1954. b. Ten calls means x = 10. Want prob ( x = 10 in a 15 minute interval) In 10 minutes , expect μ = (0.8 calls/min) × 15 min = 12 calls/10-min interval. So want Prob ( x = 10 | μ = 12) So prob ( x =10| μ =12) = 0.1048. c. No calls means x = 0. Want prob ( x = 0 in a 5 minute interval) In 5 minutes , expect μ = (0.8 calls/min) × 5 min = 4 calls/5-min interval, as before. So want Prob ( x = 0 | μ = 4) ( ) ... 0183156 0 = 1 1 × = ! 0 4 × = 4 = | 0 = 4 0 4 . e e x prob - μ Some Section 5.4 Problem Solutions 2 ( ) ( ) ... 0.10483725 = 3628800 4 6191736422 × ...10 6.14421235 = ! 10 12 × = 12 = | 10 = 6 - 10 12 - e x prob μ
So prob ( x =3| μ =4) = 0.0183. d. No calls means x = 0. Want prob ( x = 0 in a 3 minute interval) In 3 minutes , expect μ = (0.8 calls/min) × 3 min = 2.4 calls/3-min interval. So want Prob ( x = 0 | μ = 2.4) ( 29 ... e ! . e . | x prob . . 0.0907179 1 1 0 4 2 4 2 0 4 2 0 4 2 = × = × = = μ = - - So prob ( x =0| μ =2.4) = 0.0907. Additional questions: 1. What is the chance of as many calls as expected arrive during 5 minutes? In 5 minutes , expect μ = (0.8 calls/min) × 5 min = 4 calls/5-min interval. “as many as expected” means x = E ( x ), which is 4, so x = 4. So want Prob ( x = 4| μ = 4) ( ) 1954 0 = 24 256 × 0183156 0 = 4 4 × = 4 = 4 = 4 4 . ... . ! e | x prob - μ So prob ( x =3| μ =4) = 0.1954. 2. What is the chance that half as many calls as expected arrive during 5 minutes? Some Section 5.4 Problem Solutions 3

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In 5 minutes , expect μ = (0.8 calls/min) × 5 min = 4 calls/5-min interval. “half as many as expected” means x = ½× E ( x ), which is ½× 4, so x = 2. So want Prob ( x = 2| μ = 4) ( ) 1465 0 = 2 16 × 0183156 0 = 2 4 × = 4 = 2 = 2 4 . ... . ! e | x prob - μ So prob ( x =2| μ =4) = 0.1465. Some Section 5.4 Problem Solutions 4
41. Similar to problem 40. Let X be the r.v. that counts the number of calls received during a given time interval during which μ calls are expected. The values of x are x = 0 (none), 1, 2, … . This is a Poisson situation, because calls are received during specified intervals of time. (Note the key words which all indicate the Poisson distribution.) Probabilities will be calculated with the Poisson probability distribution, given by ( 29 ! x e | x prob x μ = μ μ - with the appropriate μ for the time interval of interest in the question.

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