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Suppose that visitors at a theme park spend an average of $100on souvenirs. Assume that the money spent is normally distributedwith a standard deviation of$10.a)Plo...

Question

Suppose that visitors at a theme park spend an average of $100on souvenirs. Assume that the money spent is normally distributedwith a standard deviation of$10.a)Plot the PDF of thisdistribution covering the three standard deviations on either sideof the mean. b)What are the chances that randomly selected visitorswill spend more than $120? c)What is the chance that a randomlyselected visitor will spend between $80 and$90(inclusive)?d)What are the chances of spending within onestandard deviation,

Suppose that visitors at a theme park spend an average of $100 on souvenirs. Assume that the money spent is normally distributed with a standard deviation of$10.a)Plot the PDF of this distribution covering the three standard deviations on either side of the mean. b)What are the chances that randomly selected visitors will spend more than $120? c)What is the chance that a randomly selected visitor will spend between $80 and $90(inclusive)?d)What are the chances of spending within one standard deviation, two standard deviations, and three standard deviations, respectively? e)Betweenwhattwo values will the middle 90% of the money spent fall? f)If the theme park gives a free T-shirt for the top 5% of the spenders, what will be the minimum amount you have to spend to get the free T-shirt? g)Showa plotfor10,000 visitors using the above distribution.



Answers

A box of tickets has an average of $100,$ and an SD of $20 .$ Four hundred draws will be made at random with replacement from this box.
(a) Estimate the chance that the average of the draws will be in the range 80 to $120 .$
(b) Estimate the chance that the average of the draws will be in the range 99 to $101 .$

So then problem number 15. There are really two questions, and both of those questions revolve around the same given information. So the given information is that the average cost of an overseas trip is 2708 with a standard deviation of 405 and the other given information is that we're going to assume normal distribution. So because of the nature of this problem, we know that the average always goes in the center of our bell, and we're ready to ask our questions. So the question here is what is the probability that the cost said the probability that the cost of a randomly selected trip So it's one trip, one piece of data, So we're going to say X is more than which is greater than $3000. So 3000 on the bell would end up being to the right of 2708 and we're talking about being greater than so. We're looking for the area of the curve to the right of 2000 or of 3000. In order to solve this, we're going to need to switch the 3000 into its standard score or it Z score and you have a formula for Z score as X minus new, divided by sigma. So in this case, it would be 3000 minus 2708 all over 405 which is equivalent 2.72 So on our bow we can put a 0.72 at the same line as the 3000. So instead of us, thinking of this problem is being greater than 3000, we can also talk about it as being the probability where Z is greater than 0.72 When you look in the back of the book at the standard normal table, you're going to find that the table always goes to the left, the area to the left. So if we look up 0.7 to the area to the left of this line is 0.76 four to. So if we want the area to the right, we're gonna have to do one. Take away the area to the left with a probability that sees to the left of 10.72 because the entire curve represents an area of one. So, in essence, are problems gonna read one minus 10.7642 which would result in 0.23 five eight. So in summary, the probability that a randomly selected trip costs more than $3000 is 0.2358 Now, we're ready to tackle the second part of this problem. So I'm just gonna scoot up a little bit and we're going to discuss what was given are given was still that the average cost of a trip overseas was 2708 and our standard deviation was still 405. They want you to realize that this is information about the population, all trips overseas. And in the second part of this problem, where you're gonna be talking about a sample from the population and the sample from the population is going to be of size 30 it says if we select a random sample of 30 overseas trips, we want to ultimately find the probability that the mean of those trip costs is greater than 3000. Ultimately, that's what we're trying to solve Now. In order to solve this type of problem, we will need the central limit theorem, So we will need to find the average of the sample means and we will need to find and let me fix that one second here. The average of the sample means would be new X bar, and we will need to find the standard deviation of the sample means and the central limit. Theorem says that the average of the sample means is equivalent toothy average of the population, which in this case is 2708. And the standard deviation, or of the sample means or the standard error of the mean is equivalent to the standard deviation of the population divided by the square root of the sample size. So in this case, it would be 405 divided by the square root of 30 again, we're going to use our bell shaped curve, so we're gonna draw at our bell shaped curve and at the center of our bell shaped curve, we're gonna put our average of the sample means 72,708 and we're talking about when the average is greater than 3000. So we're going to want to put 3000 on our curva swell, and we're talking about being greater, So just like the first part of this problem, we needed to switch the 3000 into its standard equivalent. So we're going to also find a Z score. But this time the Z score is going to look slightly different because instead of a single data, we're talking about an average of 30 pieces of data, so it's gonna be X minus. The average of the sample means divided by the standard deviation of sample means. And we know that the X bar is 3000 minus. The average of the sample means 2708 and the standard deviation of the sample means is 400 over the square root of 30. And if you pick your calculator up and calculate that you end up with approximately a 3.95 so up here on our bell, we can put 3.95 And when we talk about to the right of 3000, it's no different than saying that we're to the right of 3.95 so we can rewrite our problem to read. The probability the Z is greater than 3.95 and greater than 3.95 is the same as one, minus the probability that Z is less than 3.95 And then when we go to the standard normal chart in the back of your textbook, the 3.95 is nowhere to be found. Your chart goes up to almost 3.5. Um, and if you look in that lower right hand corner, you've got a an area or a probability of 0.9998 So 3.95 is larger than what's there, so we know that the probability is going to be something greater than 0.9998 so we can make an assumption that it's it's really, really, really close toe one. So what we're gonna do is we're just going to use something that's really close toe one. We're gonna say one minus 10.99999 and we're going to get an answer of 0.1 or somewhere close. So in summary, our best bet is to just say that the probability that the average of those 30 trips is greater than $3000 is less than 0.1

Here we are told that 30% of all students want to buy a new text for a certain course, and we define that as the success. And so the 70% of students want a used copy, and we have a random selection of 25 purchasers. So we have an equals 25. We have probability of success. Is it what is 0.3? So if we define the random variable X as the number of successes that is the number of students or people who want to buy the new book, it follows a binomial distribution based on the size of 25 and a probability of success of 0.3. Now for part A were asked for the mean value and the standard deviation on the number who want the new copy. So that's the mean and standard deviation on the number of successes. So, for a binomial random variable, the mean is given by 10 times p just 25 times 0.3, which comes out his 7.5, and the standard deviation on X is given by and that comes out to you about 2.29 now for part B were asked, What is the probability that a number of successes will be more than two standard deviations away from the mean value? So I mean value of 7.5 and standard deviation is about 2.3. So basically, the question is asking, you know, what is the probability that that number of successes is? Well, I'll give the range first. So 7.5 is the mean closer minus two standard deviations of two times 2.29 So that's 5.21 or 9.79 So we're really looking for the probability of being more than two standard deviations above 7.5 or more than two standard deviations below 7.5. So we can say we're looking for the probability at the number of successes is less than or equal to now successes. It's a discreet distribution, so we can only have entered your numbers of successes. So if we have five or less successes, we're more than two standard deviations below 7.5. But we can also have at least 10 successes, which would put us more than two standard deviations above 7.5 this could be written as probability of at most five successes, plus one minus the probability of that most nine successes. So this is equal to zero point 193 plus one minus zero point 811 and that comes out to zero point 383 So now for Apart, See told that a bookstore has 15 new copies and 15 years copies and that 25 people are coming to purchase the text. So what is the probability that all of them we'll get the book that they want? So since there are 15 copies of the new text available, we want the number of Successes x, where the number of people looking to buy the new book to be at most 15. If it's any higher than 15 then somebody won't be satisfied. But we also want the people who are looking for the used book to be satisfied so out of 25 and we only have 15 used books as well. So out of the 25 people we want, at most 15 looking for the used book, which means that we want at least 10 people that's 25 minus 15. We want at least 10 people who wants the new book. So what we're really looking for is probability of the number of successes being between 10 and 15 inclusive, which is equal to the probability of at most 15 successes, minus the probability how that most nine successes. And so this comes out to approximately one minus zero point 811 gives us 0.189 and finally for party. We're told that new copies cost $100 used copies or $70 and we're told that the bookstore has 50 new and 50 used copies, so everybody will be satisfied. So what is the expected value of the total revenue from the sale of the next 25 copies of the book? So revenue will be given by a function of the number of successes, which is $100 times the number of successes. So for each new book sold, we get $100 revenue plus $70 times the number of used books sold, which will be the difference of 25 the number of successes. So this is our than your equation for revenue, and we can simplify this too. This So if we want to find the expected revenue, you take your expectation of both sides. And because of the linearity of expectation, this comes out to so the expected value of a constant is just that number. So this comes out to 17 50 plus 30 times the expected value expected value of X, which we already know from previously. That's 7.5, So this comes out to $1975. That's the expected revenue for the next 25 customers.

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.

This question, We're told that he normally distributed population has a mean of 57,800 and a standard deviation of 7:15. And were asked for us for the probability, since this is a normally distributed population probability that a single randomly selected element is between 58 1,057 1000. So we convert this to our standard normal random variable And we when we do that conversion, we get c between .27 and -1.27 And we look for these two values in our table, so we get .6064 -14-3, Which gives us .4641. We're party part B find the mean and standard deviation for samples of size hundreds, So are mean Is basically our population mean, which is 57,000 800. Our standard deviation is basically 750 over squared of 100 which is 75 Part C. Find the probability that the mean of a sample of 600 Between 58,057,000, and we're going to use, our answer is calculated in part B to convert our X to the standard normal random variable, so we can see between 2.67 and -10.67. That's just probably dizzy, less than 2.67, Which is .9962.


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