How to find percentile of exponential distribution with ti-84

Question

Hello I am taking an online statistics class as part of a nursing program that I am in and we are using a free online textbook from lumenlearning. I am currently learning about uniform and exponential distributions and understand everything so far but on the exponential distributions I am having a lot of trouble with how to find a percentile, please look at the exact resource I am using here, I am specifically on the third box down on the page: https://courses.lumenlearning.com/introstats1/chapter/the-exponential-distribution/ I am seeing how they got these probabilities easily with my TI-84 Plus but once they start with trying to find the 50th percentile of this exponential distribution where the mean is the amount of time spent with a customer, and the average amount of time spent with each customer is four minutes. This is what they have for how they found the probability:

$P(x < k) = 0.50$, $k = 2.8$ minutes (calculator or computer)

Half of all customers are finished within $2.8$ minutes.

You can also do the calculation as follows:

$P(x < k) = 0.50$ and $P(x < k) = 1 – e^{–0.25k}$

Therefore, $0.50 = 1 − e^{−0.25k}$ and $e^{−0.25k} = 1 − 0.50 = 0.5$

Take natural logs: $ln(e^{–0.25k}) = ln(0.50)$. So, $–0.25k = ln(0.50)$

Solve for $k$: $k=\frac{\ln(0.50)}{−0.25}=0.25=2.8$ minutes

And I am not understanding that at all, what exactly is the formula for finding the percentile and what should I be entering in my calculator? I see that if I enter $LN(0.5)/-0.25$ I get $2.77$ which rounds up to $2.8$, but I don't understand how they got there? Also the next box down has a very similar problem where they want you to find the 50th percentile with an average time of 15 days in an exponential distribution, but I am not seeing how they are getting an answer of $10.4$ days for that one?

Any help is very much appreciated I have not taken any math classes for several years so am having to look up what a lot of these things mean but when it comes to this I am just clueless. Thanks.

Answer 1

Comments interjected below.

You can also do the calculation as follows:

$P(x < k) = 0.50$ and $P(x < k) = 1 – e^{–0.25k}$

The left equation is what we want; we're hunting for a value $k$ that marks the median of the distribution -- i.e. a we want the $x$-coordinate where half the distribution lies to its left and the other half to its right.

The right equation uses the CDF (cumulative distribution function) of $X$, which they mention, but don't seem to prove, up above:

$\color{blue}{\text{The cumulative distribution function (CDF) gives the area to the left.}}$

$\color{blue}{P(X < x) = 1 – e^{–mx}}$

On that linked page, this fact sort of just seems to appear, but it's completely crucial to everything that follows. If you actually wanted to show why that's true, you'd need calculus (which I'm guessing you're not familiar with). But this CDF is exactly what we need; it's a function that takes in an input ($x$) and outputs the probability of being at or below $x$. In other words, we are hunting for exactly the place where that CDF reaches the height of $0.5$.

Therefore, $0.50 = 1 − e^{−0.25k}$

Here, they've just written down an equation summarizing my previous comment.

and $e^{−0.25k} = 1 − 0.50 = 0.5$

On this step, they just did some algebra; namely, they subtracted $e^{-0.25k}$ and $0.5$ from each side, then simplified. Now, they're trying to solve for $k$, but it's locked up in the exponent of something, so they'll need a way to knock it down. The best tool to do this is the natural log $(\ln)$, because this function is defined as the inverse of $e^x$, so it cancels $e$ when it is found as a base of other stuff. That is, $\ln(e^x) = x$, so $\ln(e^{\text{whatever}}) = \text{whatever}$.

Take natural logs: $ln(e^{–0.25k}) = ln(0.50)$.

They just took a natural log of both sides of the equation; remember that you can do whatever you want to both sides of an equation so long as you do the same thing to both sides. Next, they'll use that cancellation property I mentioned:

So, $–0.25k = ln(0.50)$

Finally, they divide both sides by $-0.25$.

Solve for $k$: $k=\frac{\ln(0.50)}{−0.25}=0.25=2.8$ minutes

The steps for the other problem are the same; the crux is to solve the equation $0.5 = 1 - e^{-1/15 x}$, where we use $-1/15$ in the exponent because it's the reciprocal of the mean, as they describe in the first example box.