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Johnny B works 5 days a week for a food delivery app. On averagehe makes 100 deliveries/day. Assume a σ = 20 deliveries. A randomsample of 50 days is taken. a. W...

Question

Johnny B works 5 days a week for a food delivery app. On averagehe makes 100 deliveries/day. Assume a σ = 20 deliveries. A randomsample of 50 days is taken. a. What is the probability that thesample mean is more than 105? b. What is the probability that thesample mean is less than 95? c. What is the probability that thesample mean is between 95 and 105?

Johnny B works 5 days a week for a food delivery app. On average he makes 100 deliveries/day. Assume a σ = 20 deliveries. A random sample of 50 days is taken. a. What is the probability that the sample mean is more than 105? b. What is the probability that the sample mean is less than 95? c. What is the probability that the sample mean is between 95 and 105?



Answers

Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron's, February $18,2008 ) .$ Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study.
a. Show the sampling distribution of $\overline{x},$ the sample mean average for a sample of 50 unemployed individuals.
b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean?
c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1$/ 2$ week of the population mean?

So recall that the formula for confidence interval around mean is expire, plus or minus the critical value set off over two times the standard error, which is given by Sigma over the square root of n. So for a we just directly plug into the formula. So we're going to have 1.52 closer minus or critical value times Sigma, which is going to be the square root of sigma squared since we're given the variants in the question. So the square root of 2.25 divided by the square root of the sample size button. So now we need to figure out the critical value. So since we have a 95% confidence interval, we have Alfa being equal. Teoh 0.5 So are critical. Value is going to be found as zed 0.25 which is 1.96 when we use our table. So now we can just directly plug the center a calculator and are lower limit is going to be given by 1.5 to minus 1.96 times the square root of 2.25 divided by the square root of 20 which gives us roughly zero point 863 And the upper limit is gonna be 1.52 plus 1.96 times the square root of 2.25 divided by the square root of 20 and that gives us roughly 2.177 So that's for a for B. We're going to increase the sample size to 32 so we're going to keep everything else constant, which means we're gonna have 1.52 plus or minus 1.96 times the square root of our variants divided by the square root of our new sample size, which is 32. So now we conclude that directly into our calculator, and the lower limit is given by 1.5 to minus 1.96 times the square root of 2.25 divided by the square root of 32. And that gives us approximately one. The upper limit is 1.52 plus 1.96 times square root of 2.25 divided by the square root of 32 which gives us approximately 2.0, for okay And for C were asked what effect we have when we increase the sample size. What effect does that have on our confidence interval? So let's just draw quick number line. So I'm gonna draw the first interval first. So let's say this is 0.863 here and this is 2.177 here. So that's where my first confidence interval and for my second confidence interval gonna actually be somewhere inside that So we can see that when we increase the sample size from 20 to 32 we actually decreased the size of our confidence interval or we made it more narrow.

You know that someone takes the bus five days per week to their job. We're told that the waiting times until they can board the bus or a random sample from a uniformed distribution on the interval from 0 to 10 minutes in part a were asked to find the probability density function and then the expected value of the largest of the five waiting times. So to find the probability density function first we have that because X, which is the waiting time, is uniformly distributed on the interval from 0 to 10 minutes, we have ffx is equal to 1/10 minus zero, which is 10 so 1/10 and therefore cumulative distribution function for X capital F of X is X over 10 begin to roll this and therefore we have that g five of y. This is the largest of the five waiting times by a formula from this section is equal to five times why over 10 so ever. Why, to the five minus one is fourth power times lower case F of y, which is simply 1/10 which in turn is equal to five times wider, the fourth over 10 to the fifth. Yeah, and this is only for y between zero and 10. This is the probability density function now to find the expected value. The expected value of y five largest of the times is the integral over the possible values of why so from 0 to 10 of why times the probability density function of why? Which is five times wire the fourth over 10 to the fifth de y And evaluating this integral, we get 50 over six, which is approximately 8.33 and the unit is in minutes. Expected waiting time. Largest waiting time is 8.33 minutes. Next in part B were asked to find the expected value of the difference between the largest and smallest times. Well, again from part A. We have that the expected value of why one could be found by finding G one of why, and this could be found using a formula from this section. And so, if you are to carry out this computation, we find that the expected value of Y one is 10 6 or approximately one point 66 minutes, and therefore the expected value of range y five minus y one this is a linear combinations. This is the same as the difference of expected values. Affected value of y five. I asked the expected value of Y one and using the previous exercise we have 56 minus 16th. So we get 46 which is approximately 6.67 And this is also in minutes. This is the expected value of the range in Part C were asked to find the expected value of the sample Median waiting time. Well, because we have five samples, the median waiting time, the five samples is going to be. Why three. And we have again from a formula in this section that the pdf of white three is G three of y, which is given by five factorial over two factorial times. One factorial tends to factorial times capital f of why the CDF. This is going to be squared times a pdf f of why to the first power times one minus the CDF of y. You mean second power which putting in calculations from part A. This gives us 30 times why squared times 10 minus y squared all over 10 to the fifth. And this is also for why lying between zero and 10 And so it follows that using integration or by symmetry, we find it. The expected value of Y three is going to be five course. You could also do this by direct integration in the unit. Here, again is minutes. Finally in part D were asked to find the standard deviation off the largest time. We're looking for the standard deviation of Y five. Well, first, it's find the second moment of Y five. This is the integral from 0 to 10 of why squared times the pdf of y five, which, when there was five times why did the force over 10 to the fifth de y and evaluating this integral we get 500 over seven, therefore by the shortcut formula for variance the variance of y five. Well, this is going to be the second moment of y five minus the first moment of y five squared. And so we have. This is 500 over seven minus. And from the first part, this is 50/6 squared and calculating the obtained 125 over 63 which is approximately 1.984 from which we obtain, but the standard deviation of why five is the square root of this and is equal to approximately 1.409 minutes.

Alright in this problem we're looking at a probability distribution about um the number of days in the summer months that they can't work because of weather. Um a construction group. So Um on part they were finding the probability that no more than 10 days um which is going to be No more than 10 days is the same as less than or equal to 10. And then that's going to be all of the possibilities that are less than Or equal to 10. So we are going to include 10 there, that is still no more, it's not more than 10. So we will include that. Um which if we add up all those possibilities We get 65 Alright. Part be found the probability That from 8 to 12 days. So that's just between eight and 12 Which is going to be 15 plus 20 Plus 19 plus 16 plus 10. Just probably going to be our most likely. Yeah, so about 80%, chance that it's going to be somewhere between um 8 to 12 days lost, for part C. The probability that X is equal to zero. If you look at your probability distribution, that is not an option. And just to check things out here, I'm gonna add up all of our probabilities and make sure That they all total up to one. So there is no option of there being no days loss which we know that's not really going to happen, whether it's always going to play a part, um and then last are mean and standard deviation, I'm going to uh enter that into our calculator, it's a whole lot faster than doing the formulas by hand. So we enter that in do first variable stats And our frequency list is our probabilities. So too, and we get a mean of 9.79 and a standard deviation of 1.86. Now, the interpretation for the mean is that you would expect. It is the expected value of days missed. You would expect to miss about 979 days. So somewhere between nine and 10 days is what you would expect the construction crew to miss over the course of the summer.

Questions or, um, a B C. So for part A, they are asking us what kind of distribution it is and with being uniformed distribution, Uh, each one is equally likely. So that means it's uniformity even say it's uniform. So we go from 24 26 with a mean of 25 and they want the standard deviation. The verse standard deviation were using the formula being my most a one of next year over 12. So here I have, uh, 26 minus 24 water. I swear, over 12. And it's where word of that so that becomes to square, which is four over 12 ends. Where of that. So it's really this where we're 1/3 which equals by 77 or so My standard deviation. Wait, I 774 in r B. They want us to find it in a normal distribution for a normal distribution. I would have. I mean, you know, 25 they mean but my standard deviation would be my standard deviation divided by the square root of and so my standard deviation with 0.5774 divided by this, we're route of 100 which I know to be 10 so 0.5774 divided by 10 gives me a week zero by seven cabin. So for part B, I have a normal distribution. 25 is my name. Wait, 05 77 is my standard deviation. And for part C, we're looking at a probability. Find the probability, which means normal. See? Yeah, I'm going to go lower. My lower bound is negative Infinity and I usually just put in the year 99999 You put in native one e 99 It just represents negative infinity. And in here we have our upper bound at 24. Weight now are mean and our standard deviation. And when we do all that, you should that zero for one five.


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