1. Data Introduction

We are going to use the laws of probability and properties of random variables to plan a trip. We will warm up by considering how conditional probabilities differ from marginal probabilities.

2. Analysis Goal

Ultimately, the goal of this analysis is the estimate the cost to fly several of us across the globe for a statistics conference. We will do this by using random variables to model the airline ticket prices of two different destinations.

Click this link to view the worksheet for experiments with random variables.

3. Summary of Results

\[E(W) = E(X) + E(Y), \quad\text{and } Var(W) = Var(X) + Var(Y).\]





Solutions to Part 3

4. Define two random variables and use them to express the total amount of funding we will need to request to fly to both of these statistics conferences.



One possible solution is to let \(X\) represent the random cost of an airline ticket from Philadelphia to Toronto, and let \(Y\) represent the random cost of an airline ticket from Philadelphia to Chengdu:

\[X = \text{ cost of a flight from Philadelphia to Toronto} \\ Y = \text{ cost of a flight from Philadelphia to Chendu}\]

Then, according to the information above, we have that

\[E(X) = 563, \quad Var(X) = 70^2 \\ E(Y) = 5400, \quad Var(Y) = 750^2.\]

We can model the total cost of airfare to and from these conferences as a function of these two random variables: \(2(3X + 5Y)\).



5. Find the expected value and standard deviation of the cost of the one-way-trip to get all students to their destinations. Do we need to make any assumptions in calculating these means or standard deviations?

\[E(3X + 5Y) = E(3X) + E(5Y) = 3 E(X) + 5 E(Y) = 3(563) + 5(5400) = 28,689 \text{ dollars}\] \[Var(3X + 5Y) = Var(3X) + Var(5Y) = 3^2 Var(X) + 5^2 Var(Y) = 9(70^2) + 25(750^2) = 14,106,600 \text{ dollars}^2\] \[sd(3X+5Y) = \sqrt{14,106,600} = 4,755.88 \text{ dollars}\] Variance formula assumes \(Cov(X,Y)=0\).