Calculating probabilities in the normal distribution, example problems. Let's take a look at our first example. Suppose the weights of pound cakes produced by your bakery are normally distributed with a mean of 16.08 ounces and a standard deviation of 0.05 ounces. What percent of the pound cakes will way less than 16.00 ounces? Please pause the video and attempt to solve the problem. When you have your answer resume the video and I will go over the solution. We begin by drawing a normal curve, center at the mean of 16.08 ounces. We want the probability of being less than 16.00 on this curve, so mark 16.00 somewhere to the left of 16.08. It doesn't matter that you get in the perfect spot, and then shade that area in. Notice that you will have shaded less than half the curve, so you should expect your final answer to be less than 0.5. We next calculate our z score, this is our x of 16.00 minus our mu of 16.08 all divided by sigma of 0.05. We obtain -1.6, We then go to our table and look up z of -1.60 by going down to -1.6.0 and over to 0.00. This is 0.0548, this means that the area to the left of -1.6 on the standard normal curve is 0.0548. And that means the probability of being less than 16.0 ounces in our example is also 0.0548. This problem can also be solved in Excel. When using Excel, remember that you don't need to calculate your z. You can just enter the x, mean, standard deviation, into the formula and then we will always use the word true as the last argument in this syntax. Notice here that we obtain the same solution. Let's do one more example. Suppose now we want to know the probability of being greater than 16.121 ounces in the same scenario. How does that change the problem? Go ahead and pause the video and see if you can work out this problem. We again draw a normal curve centered at 16.08, but now we mark 16.121 on the right of 16.08, since 16.121 is greater than 16.08. And we shade the area to the right of 16.121, since we're interested in the probability of being greater than 16.121. Notice that your shaded area is again less than one-half of the curve, so you expect your final answer to be less than one-half. We again calculate our z, and this time obtain z=0.82. We again look up our z value in the table, this time going down to 0.8, and over to 0.02, since 2 is in the hundredths place of our z. The value is 0.7939, but remember this is the area to the left of 0.82, and this time we want the area to the right. So we must subtract our value from 1. Our correct answer is 1-0.7939 which is 0.2061 or 20.61%. Remember from when we do the curve, that we expected an answer less than one-half. Taking that extra step can keep us from making mistakes, and that if we had tried to stop at 0.07939. We would hopefully realize our answer didn't make sense since it was too large. This can save you from a lot of mistakes. And again, this problem could have been solved in Excel instead. Just as when solving by hand, we need to subtract our answer from Excel from 1 to get the area to the right of the x value. We obtain the same answer as when solving by hand.