In Exercises 55–60, find the indicated probabilities and inter...

Question

In Exercises 55–60, find the indicated probabilities and interpret the results.

Refer to Exercise 34. A random sample of six days is selected. Find the probability that the mean surface concentration of carbonyl sulfide for the sample is (a) between 5.1 and 15.7 picomoles per liter. Compare your answers with those in Exercise 34.

Accepted Answer

Welcome back, everyone. The mean daily rainfall in a rainforest region is normally distributed with a mean of 18.2 millimeters and a standard deviation of 6.3 millimeters. If a sample of 10 days is chosen, what is the probability that the sample mean rainfall is between 14.4 and 21.7 millimeters? A says about 0.1634, B about 0.8213. C about 0.9323, and D, about 0.9671. Now how are we going to figure out the probability that the sample mean rainfall is between 14.4 and 21.7 millimeters? Well, note that in our problem we were told that the mean daily rainfall is normally distributed, and according to the central limit theorem, if random samples of size n are drawn from a population that is normally distributed in this case or daily rainfall, then the sampling distribution of sample means is normally distributed for any sample size n. So in other words, we can apply what we know about normal distributions to help us figure out the probability for the sample mean rainfall. Now, let's make a note here, OK? Because in this case, the mean of the sample means mu x bar will be equal to the population mean of 18.2 millimeters, OK? And the standard deviation of the sample means sigma X bar will be equal to the standard deviation of the population divided by the square root of the sample size, which is 6.3 divided by root 10, which is going to be approximately 1.99. So we can use this information to help us solve for the probability. That X bar, or sample mean is between 14.4 and 21.7 millimeters, OK? Considering X bar 1 to be 14.4 and X bar 2 to be 21.7. Now since we can use what we know about normal distributions to help us, then if we were to put this on a normal distribution. Essentially OK, let me draw that curve a little better. Essentially what we're trying to find here if we have our sample mean at 18.2. Then what we're really trying to find is the probability that our the value we choose is going to be X 1 14.4 and X 2. In other words, the probability that it will be in this region. And because we're working with a normal distribution, it would be a good idea to find the corresponding Z scores for both of those X bar values. So, starting with Z1 is gonna be equal to X 1 minus mu x bar divided by sigma X bar. And that would be equal to 14.4 minus 18.2. Divided by or or approximate value of 1.99 and that would be approximately equal to negative 1.907. Next, for our second Z value, OK, it's gonna be equal to X2 minus mu x bar divided by sigma X bar. And that is gonna be equal to 21.7. -18.2. Divided by 1.99, which again is gonna be approximately equal to 1.757. So now that we have our corresponding Z squares here. Then we need to figure out what's the probability or value lies in this region. Well, if we think about it, the probability that Xpi lies between X1 and X2. And It's gonna be equivalent to the probability that Z lies between Z1 and Z2. And we know we can calculate that probability by subtracting the probability, or sorry, by using the idea that we can. Say the probability that Z is less than Z2 minus the probability that Z is less than Z1, OK? So we need to figure out what these probabilities are. Now notice for both of our results -1.907 and 1.757, we could use the standard normal distribution table, but it does not have those exact Z values. So we'll have to interpolate with the nearest or the closest possible values that are in the table. So let's start with the probability so from the table. By interpolation to get those probabilities. Let's start with the probability. That C is less than -1.097. Now by interpolation, we'll have this probability minus the probability that Z is less than -1.91, OK, divided by the probability that Z is less than -1.9 minus the probability that Z is less than -1.91. OK. And that is gonna be equal to. That is going to be equal to -1.907 minus -1.91 divided by -1.90 minus -1.91. So what is that gonna be equal to? Well, here, if we do some transposition, then we should find that we get the probability that Z is less than -1.907 is going to be equal to. 0.3, OK. Multiplied by our probabilities that were in our denominator here. So the probability Z is less than -1.9, it's gonna be 0.0287, while the probability Z is less than -1.91 is gonna be 0.0281, OK. Minus the probability Z is less than -1.91, which we said is 0.0281. No, when we go ahead and solve here, OK, or rather this is gonna be plus, my apologies, plus that value. And when we go ahead and solve, we get that to be 0.0283 or approximately 0.0283. So this will be the probability Z is less than -1.907. Now we can apply the same idea here to figure out our second probability, which is the probability Z is less than 1.57. Or 757, sorry. Because by the same idea, if we take this line that we just figured out here using this idea, OK. Then this is going to be. Here, if we do some, some, some solving here, this is gonna be 1.1.757, OK, minus 1.75 divided by 1.76 minus 1.75 because 1.757 is between 1.75 and 1.76. OK. Multiplied by the probability that Z is less than. It's less than 1.76 here minus the probability that Z is less than 1.75. OK. Plus, the probability, I'm just trying to make some space here, my apologies, plus the probability that Z is less than 1.75. Let me just fix that here in the way I wrote it. In this line, OK? I know when we go ahead and solve, I'll leave that up to you, but you should find that the probability here is going to be approximately 0.9323. I'm sorry, 0.9605. I don't know where I got that from. But here this is gonna be our probability. Now that we have both of these, OK. Then this would imply. Substituting what we know for Z1 and Z2, the probability that Z is between -1.907. And 1.757 is going to be equal to the probability Z is less than -1, less than 1.757, which is 0.9605 minus the probability that Z is less than -1.907 or 0.0283. And when we go ahead and solve remembering these are approximate values, then we should end up with approximately 0.9323. Since this corresponds to the probability that it's between those same X the corresponding X values, then for a random sample of 10 days, the probability that the sample mean rainfall is between 14.4 and 21.7 millimeters is about 0.9323. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Sampling Distribution

Central Limit Theorem (CLT)

Probability Calculation

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