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Have three errands to take care of in the Administration Building Let X; the time that it takes for the ith errand (i = 1, 2, 3), and let X4 the total time in minut...

Question

Have three errands to take care of in the Administration Building Let X; the time that it takes for the ith errand (i = 1, 2, 3), and let X4 the total time in minutes that spend walking to and from the building and between each errand_ Suppose the X s are independent; and normally distributed_ with the following means and standard deviations: U1 13, 01 4, 2 7, 02 1, U3 10, G3 = 2, 44 11, plan to leave my office at precisely 10:00 A.M: and wish to post note on my door that reads; "I will ret

have three errands to take care of in the Administration Building Let X; the time that it takes for the ith errand (i = 1, 2, 3), and let X4 the total time in minutes that spend walking to and from the building and between each errand_ Suppose the X s are independent; and normally distributed_ with the following means and standard deviations: U1 13, 01 4, 2 7, 02 1, U3 10, G3 = 2, 44 11, plan to leave my office at precisely 10:00 A.M: and wish to post note on my door that reads; "I will return by t A.M: How long should estimate my trip will take if want the probability of the trip taking longer than my estimate to be 0.01? (Round your answer to two decima places_ 52.76 min Need Help? Read It Watch It Talk to Tutor



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A professor has three errands to take care of in the Administration Building. Let $X_{i}=$ the time
that it takes for the $i$ th errand $(i=1,2,3),$ and let $X_{4}=$ the in minutes that spends walking to and from the building and between each errand. Suppose the $X_{i}$ s are independent, normally distributed, with the following means and standard deviations: $\mu_{1}=15, \sigma_{1}=4$
$\mu_{2}=5, \sigma_{2}=1, \mu_{3}=8, \sigma_{3}=2, \mu_{4}=12, \sigma_{4}=3 .$ She plans to leave her office at precisely $10 : 00$ a.m. and wishes to post a note on her door that reads, "I will return by $t$ a.m." What time $t$ should she write down if she wants the probability of her arriving after $t$ to be .01$?$

In this exercise we are told that the arrival times between taxis arriving Is exponentially distributed with the mean of 10 minutes. For part A. Were asked for the probability that we wait for longer than one hour. This is the probability that X is greater than 60 minutes. This also means that lambda which is one over the main Is 0.1. This is equal to one minus the CDF evaluated at 60 And this comes out 2.0025. So it's a very small probability that we will wait more than an hour for part B. We're told that we are already waiting for an hour. What is the probability that we We'll have a taxi within the next 10 minutes? This is the probability that X is less than 70, given that we've already waited an hour which because of the memory list feature of the exponential random variable Is equal to the probability of waiting less than 10 minutes. And this gives us a probability of .632. 1 part C asks for what time is the probability that we wait longer than that time Equal? The probability that we wait longer than certain amount of time Is equal 2.10 means that either the negative lambda to x Is Equal 2.10. Take the natural algorithm of both sides and this gives us X equals 23 point 03 minutes. Now we want to find the time such that the probability that we wait less than that time Is 0.9. This is equal to 1- the probability that we wait more than that time which is equal to one minus now. We've already solved this in part C. So let's just get this straight so that's Equal 2.9. Which means that probability that X is greater than X is equal to 0.1. That's just subtracting .9 from both sides and then adding the probability that X rated X to both sides. And so we've already made this calculation part C. So we know that the answer is 23 point 03 minutes. And then for e we're asked to find the time such that the probability that we wait Less than that time is .5. Take the natural log of both sides would get -101. Is equal to the natural log of .5. And this gives us X equals 6.93. So there's a 50% chance that the first arrival will occur within 6.93 minutes.

We're told that waiting time for a bus in the morning is uniformly distributed on the interval 08 and waiting time in the evening is uniformly distributed on the Inter Bull 0 10 independent of the morning Waiting time In part a Suppose that we take the bus each morning and evening for a week were asked to find our total expected waiting time. So first, let's define some random variables So let x one through x five These fiber and and variables be the morning times during this week and x six the rex 10 be the evening times we have that we want to find the total expected waiting time. This is the expected value of the some of X one through x 10 because this is a linear combination. This is the same as some of the expected values. So expected value of X one up to expected value of X 10. Now we have that all the morning wait Times are the same. So they all have the value expected value of X ones. We have five times expected value of X one and all the evening times they're the same. They have the same expected value. So we have five times the expected value of x six the evening times and forgiven. The expected values for the morning times is four so five times four and the expected value for the evening time is five. So we have five times five. So you have five times four plus five times five, which is 45. So they expected total Wait. Time is 10 Mm in part B were asked to find the variance of the total waiting time. Well, we have the variance of X one summed through x 10. This is the same as be some of the variances x one up through the variance of x 10 and then because we have the evening times, their independence, the morning waiting times, then this is simply what the variance is. And so this could be simplified since the more in times of the same variance to five times the variance of X one in the evening. Times also have the same variance plus five times the variance of x six. And we're given that the variances for X one next six thes are is calculated by squaring the length of the interval so we have five times 64 and dividing by 12. And then we have squaring the other length. This is 10 squared or 100 began, divided by 12 and so we get 820 over 12 which is equal to 68.33 This is our variance in part C, whereas to find the expected value and variants of the difference between morning and evening waiting times on a given day. So we have the expected value of, say, more in Thai next one, minus the evening time X six for a given day. This is the same as the expected value of X one minus expected value of back six and we have that expect the value of the morning times is the midpoint of the interval, just four and then the expected value of the evening times is the midpoint of the interval, which is five. So we get negative one and likewise the variance of X one minus x six. This is going to be variance of X one plus and here is a plus because we have plus negative one squared which is just plus one times the variance of x six and again because x one x six or independent. This is all we have the variance and calculate the variance, as we did in part B, squaring the length of the interval. So 64 God by 12 And then adding to that variants of X six square the length of the interval 100 over 12. So we get 1 64 over 12 which is approximately equal to 13.67 finally in part D, whereas to find the expected value and variance of the difference between total morning waiting time in total the evening waiting time for a particular week. Now we're finding the expected value in variance. Uh x one plus x two plus x three plus x four plus x five. This is a total morning time. My institute for evening time, which is x six, added up through x 10 and this is equal to since this is a linear combination, this is the linear combination of the expectations, So expected value of X one. Plus we have to the expected value of X five minus expected value of x six all the way up to expected value of X 10 and we have that expected value of X one directs five of the same. This is five expected value of X one and expected values of X six through X 10 are the same. So we have minus five tons. Expected value of X six expected value. This uniform distribution is found by taking the midpoint of the intervals. So we have five times four minus five times five or negative five for the expected value on the variance variants of X one. Added up through x five minus x six added up through X 10. Once again, this is a linear combination and so this is going to be well. We have the coefficients squared times, the variances. So we have one squared times variance of x one all the way up through one squared times the variance of x five and then plus negative one squared which is one times the variance of X six all the way up through negative one squared which is again is one times the variance of x 10. And because x one through extender independent, it follows that this is the end of the variance and we have that the variance for X one X five for the same. So we have five times the variants next one plus five times well, the variance of X six and extender the same. So the variance of x six And this is the same variants as we found in a previous problem. It's the same as the variance this time in part me kids, He 8.33

Okay, So we're given that the amount of time it takes for class one to end is normally distributed. The amount of time it takes for the class to to start is normally distributed. And the amount of time it takes to get from class one to class two is also normally distributed. And they have these normal distributions, I have been down here. X one is the time that it takes to get to The time it takes for the first class to end. X two is the time it takes for the second class to begin. An X three is the time It takes to get from Class 1 to class too. So what I'm gonna do is I'm gonna let t equal X one plus X three and X one plus X three is the time it takes for the first class to end, added to the time it takes to get from class one to class too. So this will be equal to This will be equal to do one of these lines is equal to a normal distribution where we add the main values and we add the variances and this is allowed because all three of these are normally distributed and they're all independent of each other. We can do that when they're independent of each other. So if we add 902-6 minutes we get 908 and if we add one squared to 1.5 squared 1.5 squared is 2 to 5. So this is 3 to 5. And now that we have this normal distribution of the time it takes to get to class two added to the time it takes for the first class to end. We can now set this up With this 3rd distribution. So what we're trying to find is we want this time this normal distribution to be less than at the time it takes for the class to to start. So this normal distribution, so we want t To be less than X two. And so what we're gonna do now is I'm gonna rewrite this by just minus ng ti from both sides. So you get zero Is less than X two -T. And now we can set this up as another normal distribution by minus ng the two means and then we still add the variances. So this would be normal Where we have 908. We're sorry, we have 9 10 minus 98 just to two minutes and we have 3.25 Added to one squared up here. So that is 4.25 as our variants. And now what we can do is just set up a Z score. So right now we want to find the probability that X two minus T is greater than zero, but I'm going to change this. So we've found the probability of Z is greater than 0-. Are you divided by our variance? I'm sorry, our standard deviation and this will let us use then the standard normal distribution table to find our probability. So if we let zero minus you, if we set up this, so we actually find zero minus, you will use to so we know zero minus you is negative two and the standard deviation, It's just the square root of our variance, which is 4-5. So we're trying to find the probability that Z is greater than -2 over the square to 4.25. And we can do this by setting this up a different way, we can say that this is equal to one minus the probability that Z is actually less than or equal to -2 over the square to 4.25. And this will make it. So we can actually look at our distribution table And find the value that we need to, which is the value at -2 divided by the square to 4.25. So if we simplify this further, We'll see that -2 divided by the square Of 4-5, goes to about -97. And now if we go and look at our Normal distribution table, we'll see that the value at negative .97 Is just about .16 60 And so now all we have to do is take one and subtract this value of .1660. And we'll get a resulting value of .834, which is then our probability.

91 Which problem? 20 Which in a, um, you nothing is equal to. And one times me one plus one times Mewtwo, which is 118.6 Sigma Squared, which is 298.28 and the Central Division the squared off this value, which is 17.2, seven or eight for coaching be, um for A is equal to 1.5 and be equal to 2.75 So we will make the same calculations here instead. One, we will use 1.5 and here is 2.75 So, um is equal to 20 to 8 to 91.25 and the standard the variance is 1898.53 and the standard deviation squared off. This value is equal to 43.57 to 1 using the same concept but using different uh, so we use A is equal to 50 and B is equal to 1.5. So mu l is 15 plus 1.5 times you won, which is 92.45 This is a linear function that in your function is a plus. plus the X. Okay, so this is the mean and the variance is 1.5 square seen my one square, which is 151.29 And the standard deviation is the square root off this value which is 20.3.


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