d. Suppose that the person has already been waiting for 1.2 minutes. Find the probability that the person's total waiting time will be between 2 and 3.2 minutes e. 86% of all customers wait at least how long for the train? minutes.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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D and E please

A bus comes by every 7 minutes. The times from when a person arives at the busstop until the bus
arrives follows a Uniform distribution from 0 to 7 minutes. A person arrives at the bus stop at a
randomly selected time. Round to 4 decimal places where possible.
a. The mean of this distribution is 3.5
b. The standard deviation is 2.0207
c. The probability that the person will wait more than 2 minutes is 0.7143
d. Suppose that the person has already been waiting for 1.2 minutes. Find the probability that
the person's total waiting time will be between 2 and 3.2 minutes
e. 86% of all customers wait at least how long for the train?
minutes.
Transcribed Image Text:A bus comes by every 7 minutes. The times from when a person arives at the busstop until the bus arrives follows a Uniform distribution from 0 to 7 minutes. A person arrives at the bus stop at a randomly selected time. Round to 4 decimal places where possible. a. The mean of this distribution is 3.5 b. The standard deviation is 2.0207 c. The probability that the person will wait more than 2 minutes is 0.7143 d. Suppose that the person has already been waiting for 1.2 minutes. Find the probability that the person's total waiting time will be between 2 and 3.2 minutes e. 86% of all customers wait at least how long for the train? minutes.
Expert Solution
Step 1

Solution for part d:

Given a=0 and b=7

Let X be the number of minutes that a person waits for the bus.

Then, X~0,7

The probability density function is:

fx=17-0=17

Since it is given that the person is already waiting for 1.2 minutes the probability density function changes to:

fx=17-1.2=15.8 

for 1.2<x<7.

Note that Pa<x<b|x>c=b-a1b-c.

Therefore,

P2<x<3.2|x>1.2=3.2-215.8=1.25.8=0.2069

Therefore, the probability that the person's total waiting time will be between 2 and 3.2 minutes is 0.2069.

 

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