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Numerade Educator

# The mean precipitation for Miami in August is 8.9 inches. Assume that the standard deviation is 1.6 inches and the variable is normally distributed.a. Find the probability that a randomly selected August month will have precipitation of less than 8.2 inches. (This month is selected from August months over the last 10 years.)b. Find the probability that a sample of 10 August months will have a mean of less than 8.2 inches.c. Does it seem reasonable that a randomly selected August month will have less than 8.2 inches of rain?d. Does it seem reasonable that a sample of 10 months will have a mean of less than 8.2 months?

## a. 0.3300 b. 0.0838c. yes d. yes, but not as likely.

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### Video Transcript

right. The precipitation for Miami in August. It averages at 8.9 inches and the standard deviation is about 1.6 inches and this represents population data. We're talking about the entire, um, set of data for Miami. It does say that the variable is normally distributed, So therefore, we know we're working off of the bell shaped curve when solving this problem. So for part A in this problem, it's asking us to find the probability that a randomly selected month will have a precipitation of less than 8.2. So we're gonna go over to our bell curve, and we're gonna put 8.9 in the center and 8.2. It would be somewhere right in here. So what we're gonna need to do is find the Z score or standard score associated with a point to, um so Z equals X minus mu over. Sigma is the formula you would utilize, so we're gonna have eight point to minus 8.9, divided by 1.6, and we're going to get a Z score of approximately negative 0.44 So when we're asking for the probability that the randomly selected August month will be less than 8.2. We're also saying, What's the probability that the Z score is less than negative 0.44? And at that point we would just go to our standard normal table, and we will find the answer in that table to be 0.3300 So summarizing the probability that a randomly selected August month in Miami has a rainfall of less than 8.2 inches is 0.3300 Now let's complete part B in Part B. We are going to switch to a sample of 10 August months. So now we're going to have to do talk in terms of the average of the sample means and the standard deviation of the sample means. So again the sample size was 10 and the average of the sample means is going to be equivalent to the average of the population, which in this case is 8.9, and that's due to the central limit. Their, um, and the standard deviation of the sample means or the standard error of the mean is going to be equivalent to the standard deviation of the population divided by the square root of end because of that central limit theorem again, and in this context it would be 1.6 divided by the square root of 10. So if we were to construct a new bell to answer this question again, we're gonna have our 8.9 in the center. And the question in Part V says, find the probability that the average or the mean of these 10 months is less than 8.2. So now it's going to say X Bar is less than 8.2. So on our bell, 8.2 again would be somewhere here and again. We're going to need R Z score, but ours thescore is going to be modified to account for the fact that we're dealing with an average rather than a single piece of data. So our Z score will be Z equals X bar, minus the average of the sample means divided by the standard deviation of the sample means. So in this instance, it's going to be each point to minus 8.9, divided by 1.6 over the square root of 10 and the Z score ends up being approximately negative one point 38 So when we are asking the question, what's the probability that X is less than 8.2 inches? It's the same thing is asking the question. What's the probability that your Z score is less than negative? 1.38? And again, you would look in the back of your book at your standard normal table, and you will find that this probability is equal 2.838 So summarizing part B. The probability when we select 10 random August months that their average precipitation will be less than 8.2 is going to be 0.838 in part C of this problem, it's asking, does it seem reasonable that a randomly selected August month will have less than 8.2 inches of rain? So we found the probability that a single August month having less than 8.2 inches of rain we found it to be 0.3300 That was our part. A. So we're basically saying there's a 33% chance of that happening, So yes, that is a reasonable statement. And in part D, we're asking, Does it seem reasonable that a sample of 10 months will have a mean So if we go back to Part B, which was the probability that the average is less than 8.2 was 0.838 Well, that is comparable to approximately 8.38% of the time. So, yes, it's reasonable. But it's less likely than selecting one month, because when you compare the two values that we have for Part C and D, this is less likely to happen. So it's reasonable and usually are threshold for what is usual. Versus unusual is the 5% mark, so it is reasonable at 8%. But it is less likely to happen than selecting 1 August month and having a rainfall, er a precipitation of less than 8.2 inches.

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