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 (c) more than 11.1 picomoles per liter. Compare your answers with those in Exercise 34.

Accepted Answer

Welcome back, everyone. In this problem, 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 randomly chosen, what is the probability that the sample mean rainfall is greater than 22.8 millimeters? It says it's 0.0051, B is 0.0104, C 0.0483, and the D 0.0837. Now how can we figure out the probability that the sample mean rainfall is greater than 22.8 millimeters? Well, note here that we are told the daily rainfall in the region 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 a daily rainfall, then the sampling distribution of sample means is normally distributed for any sample size n. In other words, we can use whatever we know about normal distributions to help us figure out this probability for the sample mean. Now, in that case then we can note for by the central limit theorem the mean of the sample population, or sorry, the mean of the sample will be equal to the mean of the population, which is 18.2. 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 N. Which is 6.3 divided by the square root of 10 or approximately 1.99. Now we can use this to help us solve for the probability that, for the probability that our sample mean is greater than 22.8. Let's try to picture it on a normal distribution. Now here on our normal distribution, if we take our mean, or if we put our mean of 18.2 on the distribution, and we're trying to find a values greater than 22.8 millimeters, OK, which we could say that is right here. Then what we're looking for is this shaded region on our normal distribution. Now, this shaded region is going to correspond with a Z value. So if we can figure out the Z value, then we can be able to solve. So let's start by trying to find a Z value using the formula for Z scores. Z We put it in black actually. Our corresponding Z value is going to be equal to. Or sample mean minus or, or value minus or sample mean divided by or a sample standard deviation. Which is going to be 22.8 minus 18.2, divided by 6.3 divided by the square root of 10. And that value is going to be approximately equal to 2.309, OK? So now, we're trying to find the probability, OK? In other words, the probability that our sample mean is greater than 22.8 will correspond to the probability that our Z value is greater than 2.309. And using what we know about probabilities, that's gonna be equal to 1 minus the probability that Z is less than 2.309. Now why do we write it this way? Well, our standard normal distribution table can help us to find the probabilities that Z is less than a certain value. However, in our table, we don't have exactly Z equals 2.309. So that means we'll need to figure out that probability by interpolation with its closest values. So from the table. From the table by interpolation. OK Then The probability that Z is less than 2.309 minus the probability that Z is less than 2.30. OK. Minus all divided by the probability that Z is less than 2.31 minus the probability that Z is less than 2.30 is going to be equal to 2.309 minus 2.30. Divided by. 2.31. Minus 2.30. OK. In other words, we can use this to solve. Now, if we transpose here to get the probability Z is less than 2.309 and substitute all of these probabilities that we can get from our table, then we're going to have the probability Z is less than 2.309 being equal to. 0.9, the result of our expression here, OK. Multiplied, OK, by the probability Z is less than 2.31, which is 0.9896 minus the probability that Z is less than 2.3, which is 0.9893, OK? Plus our probability Z is less than 2.3, which we said is 0.9893. I know when we go ahead and solve here, we should get this to be approximately equal to 0.9896. So what does that mean? Well, no, to find the probability, so let's go back here, to find the probability that Z is greater than 2.309. It's going to be equal to 1 minus the probability is less than 2.309, which we found to be 0.9896, or I'll write the exact value, 0.9895689392. OK? Remember, this is just an approximation, or this is just the exact value of our approximation. And when we go ahead and solve here, we should get this to be approximately equal to 0.0104. Since this is the probability for or Z score, it corresponds with the probability for or X value, which means that which is our sample mean, which means that for a sample of 10 days, the probability that the sample mean rainfall is greater than 22.8 millimeters is about 0.0104. Therefore, B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Sampling Distribution

Central Limit Theorem (CLT)

Hypothesis Testing

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