Ask Your Teacher Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter μ = 20 (suggested in the article "Dynamic Ride Sharing: Theory and Practice"†). (Round your answer to three decimal places.) (a) What is the probability that the number of drivers will be at most 13? (b) What is the probability that the number of drivers will exceed 25? (c) What is the probability that the number of drivers will be between 13 and 25, inclusive? What is the probability that the number of drivers will be strictly between 13 and 25? (d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

a) 0.00613 b) 0.113 c) 0.821, 0.804 d) 0.99997

Step-by-step explanation:

Cumulative Poisson probability=∑(e^-μ) (μ^x) / k!

μ: mean=20

x: max number of successes

k: number of success from 1 to x

e= 2.71828

In this question use poisson distribtuion calculator

a) x≤13

(e^-20) (20^13) / 1! + (e^-20) (20^13) / 2! + + (e^-20) (20^13) / 13!

probability= 0.00613

b) x>25

probability= 1- P(X≤25,20)

                = 0.113

c) probability( 13≤x≤25)= P(X≤25,20)-P(X≤13,20)

                                           0.88728 - 0.06613

                                       = 0.821

probability( 13<x<25)= P(X<25,20)- P(X<13,20)

                                    = 0.84323-0.03901

                                        = 0.804

d) In poisson distribution, variance = mean

So, here x=40

P(X<40) = 0.99997

77

step-by-step explanation:

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

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