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The mean daily production of herd of cows is assumed t0 be normally distributed with mean of 35 liters and standard deviation of 2.3 litersWhat is the probability t...

Question

The mean daily production of herd of cows is assumed t0 be normally distributed with mean of 35 liters and standard deviation of 2.3 litersWhat is the probability that caily production is between 38.4 and 41 liters? Do not round until you get your your fina answer:Answer=Round your answer to decimal places.Warning: Do not Ue the 2 Norma Tables they may not be accurate enough since WAMAP may look for more ccuracy than comes from the table

The mean daily production of herd of cows is assumed t0 be normally distributed with mean of 35 liters and standard deviation of 2.3 liters What is the probability that caily production is between 38.4 and 41 liters? Do not round until you get your your fina answer: Answer= Round your answer to decimal places. Warning: Do not Ue the 2 Norma Tables they may not be accurate enough since WAMAP may look for more ccuracy than comes from the table



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Suppose the mean number of days to germination of a variety of seed is 22, with standard deviation 2.3 days. Find the probability that the mean germination time of a sample of 160 seeds will be within 0.5 day of the population mean.

For this exercise, we are given a continuous random variable that is uniformly distributed On the interval. 374- 380. And for part A were asked for the mean and standard deviation of the variable for a uniformly random variable, uniformly distributed random variable, the mean is equal to A plus B or two, where is the bottom of the interval and be at the top of the interval, And this is 377. And for the variance given by b minus A squared over 12, This is going to be six squared over 12 or 36/12, which is three or B. We are asked for the probability That X is less than 375. So let's first look at our density function for this random variable, it's equal to one divided by B -A. In other words, one of our six Forex between 374 and 380. So the probability is the area underneath the density curve, you're, the density curve is a rectangle because it's uniformly distributed. So the height is once 1/6, and the width is going to be 375 -374, that's the width of the area that we're trying to calculate. And so this comes out to 1/6, and next, for part C were asked for the value of X, for which 95% of the distribution exceeds it. So basically the probability that X is greater than what Is equal to 0.9 And actually that's not .9 but .95. So for what value of X does, 95% of the distribution lie above that value, so proceeding in a similar manner to part the the relevant area of the distribution that lies above X Is going to have a width of 380 minus x mm. And solving for X here Gives 374.3. So 95% of the distribution lies above the value of 374.3.

In this problem, we're assuming that cans of Coke are filled with a mean of 12 ounces of soda and a standard deviation of 0.11 ounces, and this problem comes in three parts. So part A is asking you to determine the probability that a single can has at least 12 0.19 ounces, so that means it's greater than 12.19 And technically, it's saying greater than or equal to and up to this point. You don't know how to solve problems with greater than or equal to, because it does require a continuity correction. So what we're going to do is we're going to solve it as if it said X is greater than 12.19 because that's going to give us a close enough answer. So what we're going to need to do is set up are bell shaped curve and the bell shaped curve is going to have the average in the center. So we're going to put our 12 in the center and we're trying to determine the area to the right of 12.19 and in order to do that, we are going to need to utilize these scores and to refresh your memory. Z is equal to X minus mu over Sigma. So we're going to calculate the Z score for 12.19 by doing 12.19 minus 12 over that standard deviation of 0.11 And in doing so, you're going to get a Z score of 1.73 so we could go back to the bell and we can put a 1.73 at the same location as 12.19 So when we are trying to determine the probability that the soda can has more than 12.19 it's no different than saying What's the probability that the Z score is greater than 1.73? And in order to solve that, we would use the standard normal table in the back of the textbook? Uh, it would be table a two in your textbook, and if you notice the standard normal table talks about the areas into the left tail of the bell rather than the right tail. So we're going to have to rewrite this problem as one minus the probability that Z is less than 1.73 And using that table, you find that the probability of Z being less than 1.73 is 0.9582 and then we would end up with an overall probability of 0418 So just recapping part a of this problem, it's asking us what's the probability that a single can that was selected is greater than or equal to or in the words they used in the problem, at least 12.19 is going to be close to 0.41 eight in Part B. We're now drawing a sample from the population of Coke cans, and we're going to use 36 cans, so our sample size is 36 and we want to find the probability that the mean of those 36 is at least 12.19 ounces. So again it's going to be greater than or equal to 12.19 And again, um, you would use continuity correction, but you haven't discussed that yet, so we're going to solve this as the probability that X is greater than 12.19 and it's going to give us a close enough probability now because we're talking sample means we're going to have to calculate the average of the sample mean and the standard deviation of the sample mean and the central limit theorem is going to help you do that. And the central limit theorem says the average of the sample means is going to be equivalent to the average of the population, and in this case it was 12 ounces. And the standard deviation of the sample means is going to be equals. Who? The standard deviation of the population divided by the square root of the sample size. So in this case, our standard deviation was 0.11 and our sample size was 36. We're going to structure part, be very similar to part A. We're going to draw that bell shaped curve, and we're going to put the average in the center and our average in the sample means was 12. And to accommodate the finding of a Z score, we're gonna have to modify our Z score formula a little bit. So this time we're going to use the formula Z equals X bar minus mu sub x bar over sigma sub X bar, and we're trying to find the probability that X is greater than 12.9. So we're going to need the Z score for 12.9. So we're going to do C equals 12.9, minus the news of X bar, which was 12 divided by the standard deviation, while the standard deviation is this a an expression. So we're going to use 0.11 divided by the square root of 36 here, and in doing so now, we're going to get a Z score equivalent to 10.36 And again, I always like to take our answer back onto the bell. So 12.9 as an ex score is comparable to 10.36 as a Z score. And we when we're solving the problem, asking what is the likelihood that um, our 36 cans have an average greater than 12.9 is no different than asking the question. What's the probability that the Z score is greater than 10.36? Again, the standard normal table in the back of the book is set up to determine areas or probabilities into the left tail, so we're going to have to rewrite this problem as one minus the probability that Z is less than 10.36 And when you go to the standard normal table again, table A to anything over 3.5 is going to be 0.9999 as a probability. And when you do one minus that, you're going to get an answer of 10.1 So again, to recap the probability that when you select 36 cans, their average of volume is at least 12.9 would be 0.1 and now we have part C to solve. And in doing part C, the question is saying, given the result from part B, is it reasonable to believe that the cans air actually filled with a mean, equal to 12 ounces? Well, we're seeing that there is a chance of having a mean of greater than 12.19 so we could say it appears that the cans contain an amount greater than 12 ounces. So is it reasonable to believe the cans are actually filled with a mean of 12? We're going to say it appears that it's more than and then the second part of part, C said. If the mean is not equal to 12 are consumers being cheated? And we could say since it's greater than 12 ounces. Therefore, since customers are getting a little more than 12 ounces, they're not being cheated because they thought they were buying 12 ounces. They're getting a little bit more for their money, so they're not being cheated.


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