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Question

# When making a connection at an airport, Jasmine arrives on a plane that is due to arrive at 2:15 P.M. However, the amount by which her plane arrives late has a normal distribution with a mean $\mu = 32$ minutes and a standard deviation $\sigma = 11$ minutes. Jasmine wants to transfer to a plane that is due to depart at 3:25 P.M., although the actual departure time is late by an amount that is normally distributed with a mean $\mu = 10$ minutes and a standard deviation $\sigma = 3$ minutes. If Jasmine needs 30 minutes at the airport to get from the arrival gate to the departure gate, what is the probability that she will be able to make her connection?

Solution

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Let $X$ represents the amount (in minutes) by which Jasmine's plane arrives late and $Y$ represents the amount (in minutes) of departure time late. It is given that

$X \sim N(32,11^2), Y \sim N(10,3^2).$

These times are independent. Using the general result about linear combinations of independent normal random variables, we get:

$X-Y \sim N(\mu, \sigma^2)\,,\text{ where } \mu=32-10=\boxed{22}\,,\sigma^2=11^2+3^2=\boxed{130}\,\,{\color{default}{(\star)}}$

Furthermore, let $A$ be the amount of time (in minutes) measured after 2:00 P.M. to arrival gate, and let $D$ be the amount of time (in minutes) measured after 2:00 P.M. to departure gate.

This is illustrated in Figure1, from where we can see that

$A=15+X \text{ and } D=85+Y.$

Notice that Jasmine will be able to make her connection iff

$A+30 \le D.$

Therefore, the requared probability is

\begin{align*} P(A+30 \le D)&=P(X-Y \le 40)\overset{{\color{default}{(\star)}}}{=}\Phi\left( \dfrac{40-22}{\sqrt{130}}\right)=\Phi(1.58)\\ &\overset{\text{Table I}}{=}\boxed{\bf{ 0.9429 }} \end{align*}

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