A movie theatre scheduled three sessions for a movie at: 5:00pm, 7:00pm and 9:00

Question

A movie theatre scheduled three sessions for a movie at: 5:00pm, 7:00pm and 9:00pm. Once the movie starts, the gate will be closed. A visitor will arrive at the movie theatre at a time uniformly distributed between 4:00pm and 9:00pm. (i.e., t~U/[4: 00pin, 9: 00pm]) Determine the cumulative distribution function of the arrival time in minutes Use the cumulative distribution function to determine the following: The probability that the visitor attends at least one movie session The probability that the visitor waits more than 20min for a movie session The probability that the visitor waits less than 10min for a movie session The probability that the visitor attends the second show knowing that he missed the first show.

Explanation / Answer

from uniform distribution function parameter a =0 minute at 4:00 pm and b=60*5=300 minute at 9:00 pm

hence cumulative distribution function P(X<x) =(x-0)/(300-0 )=x/300

1) probabilty that vistor attends at least one seesion =Probabilty arrives before 9:00 pm =(300-

2)probabilty wait more then 20 minute =P(arrive before 4:40 or after 5:00 and before 6:40 or after 7:00 before 8:40)=(40-0)/300 +(160-60)/300+(280-180)/300 =240/300=0.80

3)probbailty that waits less then 10 min =P(visit b/w 4:50 and 5:00 or visit b/w 6:50 and 7:00 pm or visit b/w 8:50 and 9:00 pm) =(60-50)/300+(180-170)/300+(300-290)/300=0.1

4)probabilty attanding second show given missed first show =P(X<180|X>60) =P(60<X<180)/P(X>60)

=(180-60)/300/(300-60)/300 ==0.5

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