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Suppose that it is believed that on average, 387 million gallons of oil release into the sea each year; from natural and unintentional human causes. What is the pro...

Question

Suppose that it is believed that on average, 387 million gallons of oil release into the sea each year; from natural and unintentional human causes. What is the probability that a sample of 40 years collected by an environmental organization shows a sample average of more than 390 million gallons of oil leaked into the sea per year; if the sample standard deviation was 46 millions per year? Round all numbers to 2 decimal places and choose the value that is closest to your answer:Approximately 34

Suppose that it is believed that on average, 387 million gallons of oil release into the sea each year; from natural and unintentional human causes. What is the probability that a sample of 40 years collected by an environmental organization shows a sample average of more than 390 million gallons of oil leaked into the sea per year; if the sample standard deviation was 46 millions per year? Round all numbers to 2 decimal places and choose the value that is closest to your answer: Approximately 34% Approximately 56% Approximately 42% Approximately 26% Approximately 18%



Answers

Americans drank an average of 34 gallons of bottled water per capita in 2014. If the standard deviation is 2.7 gallons and the variable is normally distributed, find the probability that a randomly selected American drank more than 25 gallons of bottled water. What is the probability that the selected person drank between 28 and 30 gallons?

Let's first write down what we know. The population, meaning you is equal to 1 23 The population standard deviation Sigma is equal to 21 and the sample size and is equal to 15. Now we want to know the probability that the sample mean is in between 1 2126 In order to do so, we have to use the central limit zero, which says that the sampling distribution of the sample mean is approximately normal with the mean off new in the standard deviation of sigma over radical. So using that information, we can standardize thes two values and obtain to Z scores and find the corresponding probability values of those e scores and ultimately find the probability that the sample mean is in between these two values. So that was a lot, but let's take a step by step. First, let's find the Z score of 1 2126 to do that, we're gonna use this formula were the Z score is equal to the sample mean minus the population mean over the population standard deviation divided by radical and which is a sample size. So disease four for when the samples sample mean is 1 20 is 1 20 minus 1 23 over 21 divided by radical 15 and that is equal to negative point 55 in the Z score, when the sample mean is equal to 1 26 is equal to 1 26 minus 1 23 over 21 divided by radical 15 that is equal to point fry five. Now, if you use a Z score table, you'll realize that the probability value corresponding to negative 0.55 is equal to point to nine. We want to the probability value corresponding to the Z score 0.55 is equal 2.7088 Now you have to remember when you're reading the values given in a Z score table, it gives you the value left to that Z score. So what do we need by that? Let's first try to understand it visually, So our first see score is negative 0.55 which means that it's less than the average and it's corresponding probability value of 0.29 12 Is Ariel left of it? So it's this red area over here now, our second Z's war is 0.55 and it's a positive value, which means that it's above the average. Okay, it's a 0.553 here and it has a corresponding probability value of 0.7088 And again, this is the area left to this this see score, which is this green area shaded area over here. Now we want to know the probability that the sample mean is in between 1 2126 And we have to remember that the four first Z score is the stent re standardize the sample mean value of 1 20 to get this first see score, and we standardize the sample mean value of 1 26 to get the second Z score. So we have to realize that the probability that the sample mean is in between these two values is just the area in between these two z scores. This blue area that I shaded right over here. So in order to find the blue area, we just have to subtract thes two values, which is 0.7088 minus. That's a minus line 0.2912 and that is equal to point 4176 so we can conclude that the probability that the sample mean is in between 1 20

We have a normal distribution And we have a mean that we're assuming is 25 MPG And a standard deviation of one mile per gallon. And we are taking samples of size nine. And we're looking at what the sampling distribution would look like. So the mean, if I continue to take samples from this population, this population looks like this. If I have, this is for individual Individual sample. So if you just took the uh gas mileage for 11 sample, it would center at 25 and one standard deviation. Two standard deviations three standard deviations away would be up here at 28 but this is for individual just looking at 11 get tank of gas being checked and then 24 23 22. And so this would be our 68% of individuals would lie in between 24 26. But that's not what we have. We're taking nine Gas tanks, averaging them and then getting an X. Bar. And that distribution is having a mean, that would be centered at 25. And the standard deviation of that group would be the individual standard deviation divided by the square root event, which is going to be that one divided by the square root of nine or one third of a mile per gallon. And this is in miles per gallon, that picture is going to look like this, it's going to center. And so this is again, you want to make sure you think of it as you're taking a group of nine measures nine tanks of gas and finding out how many uh number of gallons you get per mile and finding that average, and we're doing that with nine. And if we are centered for that group at 25 MPG, and then one standard deviation away on each side is only a third of a unit. So that would be here at 24 2 3rd, and this would be up at 25 1 3rd, and that would include 68% of the means of nine samples, and then if we go out, two standard deviations go out, another one third were up here at 25 2 3rd as a high point, and we're down here at 24 1 3rd as a low point. And that interval from here To here would be 95 of your ex bars each coming from a sample size of nine. And then if we go out three standard deviations, Three standard deviations, what's going to be out here at 24 and op is high as 26 and that's going to include from here all the way to here, That would include 99.7 Of the X bars, each coming from a sample size of nine.

Were given a random variable X representing the amount of gasoline and gallons purchased by a randomly selected customer at a gas station. Yeah, we're told that the mean and standard deviation of X are 11.54 point zero. In part, they were asked to find in a sample of 50 randomly selected customers the approximate probability that the sample mean amount purchased is at least 12 gallons. So we have the X bar is the sample average amount of gas purchased by 50 customers. Now we know by the central Limit theorem, since 50 is greater than 30 supplies, we have that X bar is a approximately normal with expected value of X bar being the same as the average of the population which were given is 11.5 and the standard deviation of X bar is the standard deviation of the population. Sigma over the square it of the number of samples and and we're given that sigma is 4.0, and we have the end is 50 and so we have four over route 50 and therefore the probability that X bar is going to be at least 12. This is the same approximately as one minus Phi of 12 minus the mean 10.5 Sorry, 11.5 I mean standard deviation 4.0 over Route 50. This is approximately one minus five 0.88 And using a calculator or a table, we find that this is equal to approximately 0.1 94 0.1894 Next in Part B were given a sample of 50 randomly selected customers and rest to find the approximate probability that the total amount of gasoline purchased is at least 600 gallons. So now we're dealing with the total amount of gasoline purchased, so we'll let TBD total amount of gas purchased by 50 customers. Now we have that the customers total purchase is that least 600 gallons. Well, so T is when we greater than equal to 600. If and only if we have that customers, average purchase is going to be these at least 600 over the number of customers. 50. I'm sorry, not at least, but if their average purchase is 650 which is 12 gallons, but notice that finding the probability that T is greater for 600 this means this is finding the probability that X bar is greater than 12. So this is the same as part A. Therefore, we have that the probability that T is greater than or equal to 600 approximately equal to 0.1894 Finally, in part, C has to find the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers. Well, as from Part B, we're keeping the variable t and we have that by the Central Limit Theorem. Since we have at least 50 customers, T is approximately normally distributed, in particular with I mean, which is number of customers 50 times the mean cost, which is 11.5. This is equal to 575 gallons in a staring deviation which is square root of the number of customers, 50 times the standard deviation of the population, which is four. This is approximately 28.28 gallons. Um, using this information, you know, the 95th percentile of the standard normal distribution is about 1.645 follows the 95th percentile of tea. This is approximately while our mean of T 575 plus 1.645 times the standard deviation, which is approximately 28.28 So 95th percentile of tea is approximately 621.5 gallons.

In this problem, you're asked to determine the mean the variance and the standard deviation of a binomial distribution, and because they told you it was the binomial distribution, we're going to apply three different formulas to determine the mean you're going to do mean equals end times p To determine the variance you're going to use the formula n times, P Times Q. And to find the standard deviation, you're going to calculate the square root of N Times. P Times Q. So now we need to read the problem and determine the values of each of these variables. This problem is about life on Mars, and it tells you that 31% of adults think that life existed on Mars. At some point in time, we continue to read, and it tells us the random variable represents the number of adults who does think, um, that the life existed on Mars, so therefore, the 31% would represent our probability of success, or RPI value. You're randomly selecting six adults, so there's our value for N, and our Q value is the probability of failure. And if 31% believe that life existed on Mars, That means there are 69% of adults out there that do not believe there was life on Mars. So we're going to take each of these values, and we're going to substitute them into each formula. So to determine our mean, we're going to multiply six times 0.31 and you get a value of 1.86 are variants will be calculated, calculated by doing six times 60.31 multiplied by 0.69 So you're variance is one point 28 34 And to find your standard deviation since we've already calculated end times P. Times Q. To be 1.2834 we just need to take the square root of that value. So our standard deviation will be one point 13 to 9, and Part D asks you to interpret the results. So what do these three numbers mean? On average, 1.86 adults out of every six think life existed on Mars at some point in time, and since the standard deviation is about 1.13 than most, samples of six adults would differ from the mean by at most 1.13 adults


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-- 0.027634--