 We have some data and we plotted a histogram and have decided that the data is approximately normally distributed. We're going to compare the actual occurrence of data points with the probabilities that you get by using the normal distribution. To do that, I first set up some tables. I got the count of the data point 636 using the count function, the average, which gives me the mean, the average function gives me the mean, and the standard deviation. This is a sample, so we use ST.S for the range, and those are the values. I first set up a table, and I'm going to get the actuals, and I need the count of the actuals data points that are less than 5.33. And we get that using a count ifs function. We put in the range, and then the argument enclosed in quotation marks is greater than or equal to 5.33. So that gets every data point that is greater than 5, equal to 5.33 or greater. And then I calculated the proportion, which is just the count divided by the total count, which is .0189. And then I did a similar thing for the counts less than 4.85, and I used the count ifs again, same range, but changed the argument again inside quotation marks to less than or equal 4.85 to make sure we got everything including 4.85. The probability of that is again, just the count divided by the total count, which gives you about .503. And remember when we're doing probabilities, you just take the thing you're interested in, the number of ways that you can be successful, and divide it by the total number of occurrences to get the probability. So these percentages, these decimals are the same thing as probabilities. Looking down at our table of normal probabilities, we wanted first of all the probability that it's greater than 5.33. And I should point out that, again, for a number of probabilities that is actually calculating greater than or equal to 5.33. And to get that, we use 1 minus the norm dot dysfunction. The arguments for that are the x value we're interested in, 5.33, the mean up here, d2, and the standard deviation, d3. And true means we want the cumulative probability from the left infinity to that point, which in this case would be from here to here. But because we want the greater than the right tail, we should track that from 1 to get that probability of .0130. For the less than, we just use the norm dot dist with the same x, the mean, standard deviation, and true because we want that left tail. And here we get a probability of .0787. You notice they're different. They're not extremely different, but you would expect that. Remember the histogram is the actual counts. And if you remember back in our histogram, we had some outliers, some points that were sticking out to the right side of our curve. So it wasn't perfectly normal. It was a normal approximation. So that's the reason these are close, but not exact.