Find what percent of data lies between two numbers

Question

I have a set of data relating to customer waiting times for an order at a restauraunt

Mean$ = 4.308$

Standard Deviation$ = 4.102$

Min$ = 0.1$

Max$ = 16.3$

I'm asked to find what percentage of customers had waiting times between $0.9$ and $9.6$ minutes. What sort of formula would I use to do this?

Answer 1

Have you studied exponential distributions recently? They are often used to model waiting times.

One property of an exponential population is that its mean and standard deviation are the same. I notice that for your data $\bar X = 4.308 \approx S = 4.102,$ so the sample mean and SD are nearly equal. Also, an exponential distribution is strongly right skewed so it would not be surprising for the mean to be much closer to the minimum than to the maximum.

So it seems reasonable that the sample might be from an exponential population with $\mu \approx 4.308.$ In that case, the CDF is $$F_X(t) = 1 - e^{-t/4.308},$$ for $t > 0.$ There is strong circumstantial evidence that this is nearly the correct distribution; it is a good guess, but maybe not correct. (It would be nice to see the data that you summarize in your Question.)

Then you can find $P(.9 < X \le 9.6) = F_X(9.6) - F(0.9).$ This would be only an approximate answer because we know the population mean only approximately, and there is no guarantee that the sample will closely imitate the population. (You do not give the sample size, but the approximation would be better for large $n.)$

The exponential rate $\lambda = 1/\mu$. R statistical software uses the rate to parameterize the exponential distribution. So in R, the computation of the desired probability would about 70%, computed as shown below, and you can check to see that the formula above gives the same result.

diff(pexp(c(.9, 9.6), 1/4.308))
[1] 0.703766

In the figure below, about 70% of the area under the density curve of this exponential distribution lies between the vertical broken lines.