Exponential distribution as model for purchasing airline tickets

Question

I, again, have a statistics problem that I am totally clueless about how to solve it.

On average, flight passengers purchase airplane tickets 15 days in advance. Assuming an exponential distribution:

1. What is the probability that a passenger buys her ticket ten days or less before departure?
2. How many days until departure do half of the passengers wait?

My problem in about part 2). I do have the supposed correct result here (which is 10.3972) but I have no idea how to approach it and Google doesn't show me any similar questions.

I know you guys always want to see how we ourself can solve it, but I really have no idea :(

Thanks a lot!

Answer 1

The probability density function of the exponential distribution is:

$$f(x)=\left\{\begin{array}{rl}\lambda e^{-\lambda x},&x\ge0\\0,&x<0\end{array}\right.$$

To solve for the median number of days waited, which is what question 2 is after, we need to find $t$ in:

$$\begin{align}\int_0^tf(x)\;\text dx&=\frac 12\\\int_0^t\lambda e^{-\lambda x}\;\text dx&=\\-\left[e^{-\lambda x}\right]_0^t&=\\e^{-\lambda t}&=\frac 12\\-\lambda t&=-\ln 2\\t&=\frac{\ln 2}\lambda\end{align}$$

The expected value of the pdf is $\lambda^{-1}$ (which you can verify by evaluating $\displaystyle \int_{x\in X}x\ f(x)\;\text dx$). From the given information, $\lambda=\frac{1}{15}$. Substituting this in, we get the result $t=15\ln 2$.

Answer 2

Graphical comment: Here is a graph of the density function of the distribution $\mathsf{Exp}(\mu = 15).$ The total area beneath the curve is $1.$ The area to the left of the vertical dotted red line is the probability someone buys a ticket 15 days of less ahead of time. The vertical broken blue line is at the median; half of the area beneath the curve is on either side of this line. So the figure illustrate both the mean (red) and median (blue) of this distribution. In such a right-skewed distribution, it is typical for the mean to be larger than the median.