6. (10 pts.) A show is schoduled to start at 9: 00 A.M.9:30 A.M, and 10:00 A.M.

Question

6. (10 pts.) A show is schoduled to start at 9: 00 A.M.9:30 A.M, and 10:00 A.M. Once the show starts, the gate will be closed. A visitor will arrive at the gate at atime uniformly distributed between 8:30 A.M and 10:00 A.M. Determine the following (a) The distirbution function of the time (in minutes) betwwn arrival and 8:30 A.M. (b) The mean and variance of arrival time (in minmtes). (c) Probability that a visitor waits less than 15 minutes for a show. (d) Probability that a visitor waits more than 20 ninutes for a show

Explanation / Answer

for arrival time is between 8:30 am and 10 am

let us consider 8:30 am as t=0 minute ; therefore 10 am is equivalent to t=90 minute

a) therefore distribution function of time between arrival and 8:30 am is uniform with parameters a=0 and b=90.

b) mean =(a+b)/2 =(0+90)/2 =45

variance =(b-a)2/12 =(90-0)2/12 =675

c)

probability that a visitor waits less than 15 minutes =P(arrive from 8:45 am to 9:00 am or arrive from 9:15 am to 9:30 am or arrive from 9:45 am to 10:00 am)

=(15/90)+(15/90)+(15/90)=45/90=1/2=0.5

d)

probability that a visitor waits more than 20 minutes =1-P(waits less than 20 minutes)

=1-P(arrive from 8:40 am to 9:00 am or arrive from 9:10 am to 9:30 am or arrive from 9:40 am to 10:00 am)

=1-((20/90)+(20/90)+(20/90))=1-(60/90)=30/90=1/3 =0.3333

please revert for any clarification required

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