Try Magic Notes and save time.Try it free
Try Magic Notes and save timeCrush your year with the magic of personalized studying.Try it free
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 μ=32\mu = 32 minutes and a standard deviation σ=11\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 μ=10\mu = 10 minutes and a standard deviation σ=3\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

Verified
Step 1
1 of 3

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

XN(32,112),YN(10,32).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:

XYN(μ,σ2), where μ=3210=22,σ2=112+32=130()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 AA be the amount of time (in minutes) measured after 2:00 P.M. to arrival gate, and let DD 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 and D=85+Y.A=15+X \text{ and } D=85+Y.

Notice that Jasmine will be able to make her connection iff

A+30D.A+30 \le D.

Therefore, the requared probability is

P(A+30D)=P(XY40)=()Φ(4022130)=Φ(1.58)=Table I0.9429\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*}

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Introduction to Probability and Statistics for Engineers and Scientists 5th Edition by Sheldon Ross

Introduction to Probability and Statistics for Engineers and Scientists

5th EditionISBN: 9780123948113Sheldon Ross
575 solutions
Probability and Statistics for Engineers and Scientists 9th Edition by Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers

Probability and Statistics for Engineers and Scientists

9th EditionISBN: 9780321629111 (9 more)Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers
1,204 solutions
Probability and Statistics for Engineers and Scientists 4th Edition by Anthony J. Hayter

Probability and Statistics for Engineers and Scientists

4th EditionISBN: 9781111827045 (2 more)Anthony J. Hayter
1,341 solutions
Applied Statistics and Probability for Engineers 6th Edition by Douglas C. Montgomery, George C. Runger

Applied Statistics and Probability for Engineers

6th EditionISBN: 9781118539712 (1 more)Douglas C. Montgomery, George C. Runger
2,026 solutions

More related questions

1/4

1/7