MyLab Operations Management with Pearson eText -- Access Card -- for Operations Management: Processes and Supply Chains
MyLab Operations Management with Pearson eText -- Access Card -- for Operations Management: Processes and Supply Chains
11th Edition
ISBN: 9780133885583
Author: Lee J. Krajewski, Manoj K. Malhotra, Larry P. Ritzman
Publisher: PEARSON
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Chapter B, Problem 1P
Summary Introduction

Interpretation: Probability of no customer arrival during a specific time and one and four customer arrival.

Concept Introduction:Probability of an event is how likely or how possibly an event take place. It shows an outcome of event which ranges between 0 and 1.

Expert Solution & Answer
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Answer to Problem 1P

The probability that during a specific minute no customer arrives is 0.1353. The probability that during a specific minute between one and four customer arrive is 0.8120

Explanation of Solution

Given information:

Number of customers arrive per minute = 2

In case of poisson distribution: The probability that n customers will arrive in a system in time period (T) , can be calculated using the following formula:

  Pn=(λT)nn!eλt

Where,

  Pn = probability of n arrivals in T time periods

Probability of no customer arrival (n=0) during a specific time (T = 1)

  Pn= (λT)nn!eλtPn= (2×1)00!e2Pn=0.1353

The probability that during a specific minute no customer arrives is 0.1353

The probability that during a specific minute (T = 1) between 1 and 4 customer arrives using the following formula:

  P0<n<5=P1+P2+P3+P4

  P0<n<5=(λT)11!eλt+(λT)22!eλt+(λT)33!eλt+(λT)44!eλt

  P0<n<5=(2×1)11!e2+(2×1)22!e2+(2×1)33!e2+(2×1)44!e2

  P0<n<5=0.8120

The probability that during a specific minute between one and four customer arrive is 0.8120

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