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 (a) less than 5500 MMT CO2 eq. Compare your answers with those in Exercise 33.

Accepted Answer

Hello everyone. Let's take a look at this question together. 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 less than 4000 million gallons? Is it answer choice A about 0.0225, answer choice B about 0.0284, answer choice C about 0.0734, or answer choice D about 0.0799. So in order to solve this question, we have to recall how to determine the probability, so that we can determine the probability of the sample mean water consumption being less than 4000 million gallons, if 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. And we can recall that 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 N. Therefore, in In this case, the mean of the sample means is 4200 million gallons, and the standard deviation of the sample means is calculated using the following formula, where we have the standard deviation of the sample mean is equal to the standard deviation divided by the square root of our sample size N. Which when substituting the values that we are given, we have 210 divided by the square root of 4. Which is equal to 210 divided by 2, which is 105. And so using the values for the mean of the sample means and the standard deviation of the sample means, we know we want to find the probability that X is less than 4000. Of which we know that the Z score that corresponds to 4000 is calculated using the formula Z is equal to the sample mean subtracted by the mean of the sample means divided by the standard deviation of the sample means, so substituting the values we have 4000 subtracted by 4200 divided by 105. Which is equal to -40 divided by 21, so that is approximately 1.905. So we know the probability that X is less than 4000 is equal to the probability of a Z score less than -1.905. So then using the standard normal distribution table, we need to find the probability that Z is less than -1.905. However, we know that we do not have exactly Z equals -1.905 in the standard normal distribution table, and since -1.905 is between -1.90 and -1.91, we can interpolate. And here we have our standard normal distribution table, and by interpolation, we have the following formula, which we can use. This formula to rearrange the values and solve for the probability of Z being less than -1.905, which, when simplified, our formula gives us 0.0284. Therefore, for a random sample of 4 years, the probability that the sample mean water consumption is less than 4000 million gallons is about 0.0284. Which looking at our answer choices, we can identify is answer choice B, so answer choice B is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Sampling Distribution

Central Limit Theorem (CLT)

Z-Scores and Normal Distribution

Watch next