Operations Management: Processes and Supply Chains, Student Value Edition Plus MyLab Operations Management with Pearson eText -- Access Card Package (12th Edition)
Operations Management: Processes and Supply Chains, Student Value Edition Plus MyLab Operations Management with Pearson eText -- Access Card Package (12th Edition)
12th Edition
ISBN: 9780134855424
Author: Lee J. Krajewski, Manoj K. Malhotra, Larry P. Ritzman
Publisher: PEARSON
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Chapter B, Problem 1P
Summary Introduction

Interpretation: The probability that during a particular time, no customer arrives. Further, customers arrive between one and four is to be determined.

Concept Introduction:

The probability between the arrival of one and four customers can be calculated by the Poisson distribution.

Expert Solution & Answer
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Explanation of Solution

The formula of the Poisson distribution is as follows,

  Arrival rate (λ) = 2/minutes Thus,Probability that a specific time, no customers arrivesBy the formula,P(0) = λke K!where e is teh Euler's numberP(0)=e 2 ×201= 0.1353

Probability of customer during one time period

  Arrival rate (λ) = 2/minutes Thus,Probability when 1 customer arrives By the formula,P(0) = λke K!where e is teh Euler's numberP(0)=e 2 ×211!= 0.2706

The probability when 4 customer arrives

  Arrival rate (λ) = 2/minutes Thus,Probability when 4 customer arrives By the formula,P(0) = λke -λK!where e is teh Euler's numberP(0)=e 2 ×414!= 0.0902

Probability when 1, 2, and 3 customers arrive at any specific time.

1= 0.2706

2=0.2706

3=0.1804

4=0.0902

Thus, the probability that between 1 and 4 customer arrives.

  Sum of the probabilities0.2706+0.2706+0.1804+0.0902= 0.8118

Thus,

The probability that no customer arrives is 0.1353

The probability that between 1 and 4 customer arrives is 0.8118

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Chapter B Solutions

Operations Management: Processes and Supply Chains, Student Value Edition Plus MyLab Operations Management with Pearson eText -- Access Card Package (12th Edition)

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