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 (c) greater than 5900 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 greater than 4500 million gallons? A says it's about 0.0021. B about 0.0084. C about 0.9916, and D about 0.9979. Now, If we're going to find the probability that the sample mean water consumption is greater than 4500 million gallons, let's first think about our average annual water consumption. Notice our problem told us that it's normally distributed, and according to the central limit theorem, if random samples of size N are drawn from a population that is normally distributed or annual water consumption. Then the sampling distribution of sample means is normally distributed for any sample size N. In other words, we can use what we know about our normal distribution to help us find a probability for the sample mean water consumption. Now if we think about it here, if we were to represent our sampling distribution as a normal distribution, OK, and where we're told that our the mean of the sample means mu x will be the mean of the population by the central limit theorem which is 4200, then if we put that mean on our normal distribution. And we want to find the probability the sample mean is greater than 4500 million gallons. Then here, essentially, we're looking for this probability shaded in red on our normal distribution, OK? So let's say that's 4500. This would represent the probability that X is X bar or sample mean is greater than 4500. Now to find this probability, we can use what we know about normal distributions. So, let's first try to find the Z score that corresponds to 4500. Now, to do that, we'll need the standard deviation of our sample mean, Sigma X bar, which is going to be our equal to our standard deviation be divided by the square root of the sample size, and that is gonna be 210 divided by the square root of 5, sorry, the square root of 4. It's a random sample of 4 years. And that is equal to 210 divided by 2, which is 105. So now, Solving for our corresponding Z score, we know it will be equal to our sample mean, 4500 minus. Sorry, our value 4500 minus our sample mean 4200. Divided by 105. Which equals. 300 divided by 105, which is approximately, let me put that here, approximately 2.857, OK. Then we're trying to find the probability that Z is greater than 2.857, OK? In other words, the probability X is greater than 4500. Corresponds to the probability that Z is greater than 2.857. And using what we know about probabilities, that's going to be the probability, or sorry, that's going to be 1 minus the probability that Z is less than 2.857. Now, we write it this way because we should be able to figure out the probability that Z is less than 2.857 using the standard normal distribution table. But the problem is that we don't have exactly Z equals 2.857 in our table. However, we know that 2.857 is between 2.85 and 2.86, so we can interpolate to solve for that probability. In other words, from the table. By interpolation. OK. This means That the probability that Z is less than 2.857, OK, minus the probability that Z is less than 2.85. OK. Divided by the probability that here. Z is less than 2.86 minus the probability that Z is less than 2.85. Is going to be equal to 2.857 minus 2.85 divided by 2.6. 2.86, sorry. -2.85. If we do some transposing here, that means I'll just move it around just to make some space. Then we can multiply, or if we multiply both sides by the denominator on the left hand side. OK. Then we'll get rid of this probability and actually we'll, you know what, let me write it out here. OK. This means then that the probability Z is less than 2.857, OK? It's gonna be equal to this expression. Let me paste it here. OK. Multiplied by our denominator on the left hand side. Probability Z less than 2.86 minus the probability Z is less than 2.85, OK. 2.85, my apologies. OK. And that result will subtract, that product will subtract the probability that Z is less than 2.85. Now, I'll allow you to substitute these probabilities here, but when you do that, OK, then you should get it to be equal to 0.9978714286 or approximately 0.9979. Thus if we use what we know, that means then. That the probability Z is greater than 2.857 is going to be equal to 1 minus this probability, which is 0.9978714286. And when we subtract, we should get this to be approximately 0.0021. Since this probability for a Z2 corresponds to the probability for the value, this tells us that for a random sample of 4 years, the probability that the sample mean water consumption is about 0.0021. A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Sampling Distribution

Central Limit Theorem (CLT)

Z-Score

Watch next