In Exercises 21–24, find the mean and median for each of the t...

Question

In Exercises 21–24, find the mean and median for each of the two samples, then compare the two sets of results.

It’s a Small Wait After All Listed below are the wait times (minutes) at 10 AM for the rides “It’s a Small World” and “Avatar Flight of Passage.” These data are found in Data Set 33 “Disney World Wait Times.” Does a comparison between the means and medians reveal that there is a difference between the two sets of data?

Data table showing wait times in minutes for

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says, listed below are the drive-through wait times in seconds at 12 p.m. for two different fast food chains. Find the mean and median for each set of data, then compare the two sets of results. Does a comparison between the means and median reveal that there is a difference between the two sets of data? And below the problem we're given a table that contains the wait times for chain A in the first column and the wait times for chain B in the second column. In the first column for chain A, we have the values of 120, 135, 140, 125, 130, 145, 150, and 155. For chain B, we had the values of 310, 290, 320. 275, 330, 295, 280, and 310. We're also we have 4 possible choices as our answers. For choice A, we have the wait times for chain B are significantly longer than for chain A as shown by both the mean and median. For choice B, the wait times for chain A are longer than chain B, based on the mean, but not the median. And for choice C, we have the means suggests that both fast food chains have identical wait times, and for choice D, the medians indicate that chain A has a higher wait time than chain B. So, the first thing we need to do in order to solve this problem is to calculate the mean and median or both data sets. So, recall your definition for the mean, that is X bar is equal to the sum. Of X of I. Divided by N. Where X by here is our individual data points and N is the number of data points. So we're gonna calculate this first for chain A. So we'll have X bar A. And this is going to be equal to, and the numerator, we'll just need to add up all of our wait times, so we'll have the quantity of 120. Plus 135. Plus 140. Plus 125 plus. 130 Plus 145. Plus 150 plus 155 in quantity, and all of that is divided by the number of data points, which is 8. And when we evaluate this expression, this is going to be equal to 137.5 seconds. So that is the first average that we need to calculate. Now we need to do something similar for chain B. So, we need to calculate X bar. Or B And this is going to be equal to in numerator, we will have the quantity of 310 plus 290. Plus 320. Plus 275 plus 330 plus. 295 plus 280. Plus 310 in quantity, and all that is divided by the number of data points, which again, in this case is 8. And evaluating this expression, this is going to be equal to. 301. 25 seconds. And so that is the 2nd average that we need to calculate. Next, we need to calculate the median for both chains. And recall that the median is the value that shows up in the middle of our data set. So, the first thing we need to do is to order our data from the smallest to largest for each chain. So, we're gonna start off with chain A. And when we order our data from the smallest to the largest, we will have 120. Then 125. Then 130 Then 135. 140. Then 145. Then 150. And then 155. Now, since we have a data points that's an even value, so the value that would show up in the middle of our data set actually falls between 135 and 140. So we need to find that midpoint. So that means that the median. 4 chain A. is going to be equal to. Now to find the midpoint we'll just add those two values that appear in the middle together and divide by 2. We'll have the quantity of 135. Plus 140 in quantity, divided by 2. And that is going to be equal to. 137.5 seconds. Now, we're going to have to calculate the medium for chain B, so we'll need to first order the data points for chain B. In doing so, we will have 275. Then 280. Then 290. Then 295. Then 310. Then 310 again. Then 320. Then finally, 330. And so now we just need to find the value that shows up in the middle of our data set. And so the median. For chain B. is going to be equal to, and here again, our median value is going to fall between two values, since we have an even number of data points, and so it's going to fall between 295 and 310. So to calculate that midpoint, we'll have the quantity of 295. Plus 310 in quantity, divided by 2. And that is going to be equal to. 302 0.5 seconds. So that is the 2nd median that we needed to calculate. And so now we just need to compare our means and medians, and see if we can draw any conclusions based upon those. Now, the mean and median for chain A are exactly the same, 137.5 seconds. So this suggests a symmetrical distribution. If we look at the mean and median for chain B, they're also very close together. Which that suggests a balanced distribution. Now, since the mean and median or chain B um are substantially higher than those of chain A, we can conclude that chain B generally has longer wait times. So, based upon that conclusion, we see that the correct answer choice for this answer is going to be answer choice A. Which says that the wait times for chain B are significantly longer than for chain A, as shown by both the mean and median. So I hope that this explanation has been useful, and I'll see you in the next video.

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