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 33. A random sample of 2 years is selected. Find the probability that the mean amount of greenhouse gases for the sample is (b) between 6000 and 6500 MMT CO2 eq. Compare your answers with those in Exercise 33.

Accepted Answer

Welcome back, everyone. The average annual water consumption in a city since 2000 is normally distributed with a mean of 4200 million gallons and a standard deviation of 210 million gallons. If a random sample of 4 years is chosen, what is the probability that the sample mean water consumption is between 4350 and 4500 million gallons? A says it's about 0.0745. B about 0.0894. C about 0.1028, and D about 0.1229. Now how can we figure out the probability that our sample mean water consumption is between these two values? Well if we think about it. Then we know that the average annual water consumption is normally distributed as pointed out in our problem. And according to the central limit theorem, if random samples of size n are drawn from a population that is normally distributed, then the sampling distribution of sample means is normally distributed for any sample size in. In other words, since we are drawing or we're trying to find a sample mean water consumption, whatever works with normal distributions for our population will work for the sample mean water consumption for this sampling distribution. So applying what we know, then the mean of the sample means mu x bar will be equal to the population mean of 4200. And the standard deviation of our sample will be equal to the standard deviation of the population divided by the square root of N, where N is equal to 4, OK, because we choose a random sample of 4 years. So that's gonna be 210 divided by the square root of 4. The square root of 4 is 2, and 210 divided by 2 equals 105. So we can use this information. To solve, OK, for the probability that our sample mean water consumption X bar is between 4350 and 4500 million gallons. Now how are we going to do that? Well, since it follows a normal distribution, first we can convert our values, 4350 and 4500 to the corresponding Z score. So, for our first value X bar 1 which is 4350, then its corresponding Z score would be equal to that volume minus or sample mean divided by or sample standard deviation. Which is 4350 minus 4200, OK. 4200 divided by 105. And that would be approximately equal to 1.429. On the other hand, for XY 2 E equals 4500, then its corresponding Z score would be 4500 minus 4200 divided by 105, which would be approximately 2.857. Now why is this important? Well, if we think about our normal distribution of the sample means, OK, 4 or mean of 4200. Then we're trying to find a probability. Let me put this in red. The probability that our value chosen is going to be between X1 and X2 X 1 and X by 2. Which for a normal distribution would correspond to Z1 and Z2, OK? So if we can figure out, let me write Z2 a bit closer. If we can figure out our probabilities here, which we just have, then we should be able to figure out. Or a probability that's in between these z squares because now we know that the probability. That our sample mean is between XB 1 and XB 2. Would be equivalent to the probability that our Z score is between Z1 and Z2. Which is going to be equal to the probability that Z is less than Z2 minus the probability that Z is less than Z1. No, for the Z scores that we have, we could go to our standard normal distribution table, but the problem is these values are not explicitly noted in the table. So that means we'll have to interpolate between the closest values. So from our table from the table. By interpolation, for example. Starting with the probability that Z is less than Z1 or 1.429. Then by interpolation, the probability Z is less than 1.429. Minus its closest value, the probability Z is going to be less than 1.42. Divided by, OK, the probability that Z is less than 1.43 minus the probability that Z is less than 1.42 is going to be equal to 1.429 minus 1.42. Divided by 1.43 minus 1.42. Now, if we do some solving here, OK. Then remember, we don't know if we, we can solve here for the probability Z is less than 1.429. Now, that is gonna be equal to this entire expression, which when we solve, it's gonna be equal to 0.9. Multiplied by the probability Z is less than 1.43, which is 0.9236 minus the probability Z is less than 1.42, which is 0.9222. OK. Plus, can we bring this value over? The probability Z is less than 1.42, which is 0.9222. OK? I know when we solve here, we should get this to be 0.9234. So, by interpolation, this is the probability Z is less than 1.429. OK? I put a P right there by accident, so let me just get rid of it. Now that we have this value, we can do the same. Let me put this in its own little box here, OK. Now that we have this value, we can do the same for uh the probability Z is less than. 2.857, OK. If I make another box here just to show that it's a separate probability value, OK. I think this is a this is a good size box. OK. Then no. OK, we can say the oops, my apologies. Now we can see the probability. Z is less than 2.857, using the same reasoning that we have here in the second line is going to be equal to. 2.857 minus 2.85 divided by 2.86 minus 2.85, the closest values multiplied by the probability. That Z is less than 2.86 A minus the probability that Z is less than 2.85. I know if we go ahead and solve here in the interest of time, I'll allow you to figure out what those probabilities are, but you should get this. To be equal to approximately 0.9978714286, OK? Know that we have both of our probabilities that Z is less than 1.429 and Z is less than 2.857, we can plug them into our formula to solve for the probability that it's between Z1 and Z2 because no, this would imply then. Substituting what we know about Z1 and Z2, the probability Z is between 1.429 and the 2.857 is going to be equal to the probability it's less than 2.857. Minus the probability it's less than 1.429. Which is gonna be equal to 0.9978714286 minus 0.9234. That would be approximately equal to 0.0745. Thus, since this corresponds to the probability, it's between our sample our 1st and 2nd sample mean 4350 and 4500. This tells us that for a random sample of 4 years, the probability the sample mean water consumption is between 4350 and 4500 million gallons is about 0.0745. If we look back at our answer choices, that means A 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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