Question

The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?

Short answer

The value of the number of days half of the travelers wait is $10.40$days.

Step-by-step solution

Given information

The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days.

Solution

It is shown that the number of days is exponentially allocated with mean $15$days. It is known that the decay rate $\left(m\right)$is reciprocal of mean $\left(\mu \right)$for the exponential distribution. So,

$m=\frac{1}{\mu }$

$=\frac{1}{15}$

$=0.066667$

Thus the number of days is exponentially distributed as below:

$X~\mathrm{Exp}\left(0.066667\right)$

Here,

$X=\text{Random variable for the number of days}$

Solution

Here it is clear that cumulative distribution function of exponential distribution is:

$P(X<x)=1-e^{-mx}$

So, the required probability can be calculated:

$P(x<10)=1-e^{-0.066667\times 10}$

$=1-0.513416947$

$\approx 0.4866$

Thus the value of $P(x<10)$is $0.4866$.

Now you must compute the number of days that half of the travellers must wait, which is the same as the data's ${50}^{{}^{th}}$ percentile. So, here's how you figure out what the ${50}^{th}$ percentile is:

$P(x<k)=1-e^{-0.066667k}$

$0.50=1-e^{-0.066667k}$

$e^{-0.066667k}=1-0.50$

$e^{-0.066667k}=0.50$

Take logarithmic on both the sides. Then,

$\ln \left(e^{-0.066667k}\right)=\ln 0.50$

$-0.066667k=-0.6931$

$k=\frac{0.6931}{0.066667}$

$k\approx 10.40$