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PLEASE SHOW WORK TO GET FULL CREDITProblem 5) This problem has two independent parts a) and b) a) The time between calls to a research laboratory is exponentially d...

Question

PLEASE SHOW WORK TO GET FULL CREDITProblem 5) This problem has two independent parts a) and b) a) The time between calls to a research laboratory is exponentially distributed with a mean of 10 minutes. i) What is the probability that there are no calls within 30 minutes? (6 points) (Round your answer to four decimal places)ii) Determine X such that the probability that there are no calls within X minutes is 0.01. points) (Round your answer to three decimal places)

PLEASE SHOW WORK TO GET FULL CREDIT Problem 5) This problem has two independent parts a) and b) a) The time between calls to a research laboratory is exponentially distributed with a mean of 10 minutes. i) What is the probability that there are no calls within 30 minutes? (6 points) (Round your answer to four decimal places) ii) Determine X such that the probability that there are no calls within X minutes is 0.01. points) (Round your answer to three decimal places)



Answers

The probability that your call to a service line is answered in less than 30 seconds is $0.75 .$ Assume that your calls are independent. (a) What is the probability that you must call six times in order for two of your calls to be answered in less than 30 seconds? (b) What is the mean number of calls to obtain two answers in less than 30 seconds?

In this problem, it is given that the probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are independent. It is asked what is the probability that you must call four times to obtain the first answer? In less than 30 seconds. Let x denote the number of calls until we are answered in less than 30 seconds. So X is a geometric random variable with P is equal to 0.75. We know that if x is a geometry random variable, then probability of X equal to X is one minus P raised two x minus one in to pay. We are X ranges from one to up to infinity. So this is equal to 1 -0.75 is 0.25. So this is equal to 0.25 Raised two x -1 Into 0.75, we have to find the probability that we must call four times. So we have to find the probability of X equal to four, Probability of x equal to four is equal to 0.25, raised to X -1. But here X is four, so 4 -1, that is three Into 0.75. 0.25, rest 2 3 In 20.75. This is equal to zero point 01 56 25, 0.015, Into 0.75. This is equal to zero point 01 17, 0.0117. So the probability that we must call four times to obtain the first answer in less than 30 seconds is 0.0117. In the next part it is asked what is the mean number of calls until you are answered in less than 30 seconds. You know Dad for a geometry random variable X. I mean that is expected X is equal to one divided by p Mean is equal to one divided by P. Here P is equal to 0.75. So this is equal to one divided by zero point 75. Multiplying and dividing by 100. We get this is equal to 100 Divided by 75. So this is equal to we here 25 into 425 into three is 75. So this is equal to four divided by three. That is the mean number of calls until we are answered in less than 30 seconds is 4x3.

In this question, we're going to use exponential distribution, which is defined as the probability distribution that describes the time between events which occurred continuously and in divinity at an average time. The probability density function for this distribution is if flicks equal Lambda E Power Negative Mondex or zero for longer able negative fix some banks its range from X bigger than or equal zero and zero with the zero elsewhere. Net Capital X Let X has exponential distribution with promise to London, then e X, or equal to one over Lambda and evolve actually equal to one over Lambda Square. We have one call every two minutes so we can say that average time for one called what equal to two minutes The time between colt is exponential distributed so we can say the athletics be fine between two calls. X also has exponential distribution with average too. X has also exponential distribution, with average time to on Do we need to find the parameter London so we can get it as follows to equal E X equals one over London. From this, we can say that Lambda would equal to have 1/2. So ex approximately the exponential, huh? Now we need to find the average time for five calls you need have time four five, cause cause we know that for one call we need two minutes when we finish it for a second. We also need, on average, two minutes. So for two calls, we need time. What? Equal to to by two. Equal Four minute now for five calls. We need 552 What? It could do 10 minutes for the other question be we have to find the probability then Steve function four x f of x equals toe half by e power Negative health X for X Bigger than equal 00 elsewhere. Now we need to find the probability that we need more three minutes to take the next call. So the probability the probability four more three minutes toe take the next cold will be probability of X probability for X bigger than three. We're equal integration from three to infinity. A few clicks the X for equal to integration from three to infinity for a question half e power negative hub X, the T. So, by substitution, you can see that you equal to have X so do you? We're able to help. It will be equal integration Form one and half to infinity e forward Negative. You d you After integration, we get negative. Evil or negative? You from 1.5 to infinity will equal to negative zero minus e power. Negative 1.5 Well equal to oh, point 22 3. So we can say that the probability that next call will come more than three minutes is all 0.223 Moving to the next question number C in question. See, we need to find the nineties percentile of this distribution of time between two calls. It means that with probability, a 0.9 which happened X calls what X is greater than the Europe. So you have that probability 4.9 Well, happen ex calls where X is greater than zero. If X is, If X is listened, then zero. Then we have the probability X capital less than a small what equal to integration from negative Infinity two x small off. Zero d x will equal to zero so we can have the following 0.9 equal be off X capitalist than a small well equal integration from negative infinity. Two weeks a 50 DT would equal to integration from negative infinity to zero zero deity plus integration from zero to x have e power negative 4.5 xdd By substitution, we put you equal 4.5 t and D you will be 0, 00.5 DT so it would equal to zero plus integration from Europe toe a 0.5 x for e power Negative you d you would equal to negative e forward negative you from the you to a 0.5 x well equal to one minus e power Negative 4.5 x We can minus one from opened nine. So we get ive power 0.5 x equal to 4.1 Take limb for the both sides We get that 4.5 x equal then for over 81 so X would equal to 4.6. And that's the final result for question number C moving to point D in pointy. We have that two minutes paused from last coal. Now we need to find the probability that the next call will come in next the probability Welcome. And next third minute it means that we need to find the probability that coal will happen between the second and the third minute. So probability off time. Listen, 30 minutes or x more than two minutes were equal to probability off to this, then X. This is a three over B X, more than two well equal to integration from 2 to 3 a few weeks d x an integration from two to infinity for every weeks the X for that intervention from two to ST 4.5 e power negative 0.5 X DT over division from two to infinity over and five eating it Power negative over and five x the X Why substitution you would equal to 4.5 x. Do you recall too? All 0.5 you the X. So it's equal to integration from one to 1.5 e por negative You you over integration from Wantagh infinity E power Negative! You d you After we evaluate this integration, we get that negative e power negative You from 1 to 1 toe 0.5 over negative e for negative. You, from one to infinity would equal toe negative boy e power negative 1.5 minus e for negative one over E power. Negative one equals two. All 0.393 And the final result for question number D moving to the next question e We have that one call in two minutes so we can define the variable capital. Why? Which describes the number off course per hour is the number off course per hour. Then why has postal distribution then? Why has bought so distribution? We have 60 minute in one hour. So we have 60/2 equal toe certain 30 calls per hour on average. So why I approximately b p off 30. We need to find the probability that there is less than 20 calls per hour. So probability for the there is less calls per hour. Why less than 20? It was equal to some mission from oh equals zero 19 for probability that why equal I so it would equal to some mission. I equals zero to 19 for e power Negative 30 boy, 30 I over factorial oil would equal to all point all too and that the final result Thank you

In this problem, it is given that the probability that you're called to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are independent. Let X denote the number of times the Call is answered in less than 30 seconds. So X is a binomial random variable with probability of success, P is equal to 0.75. We know that for a binomial random variable, X probability of X equal to X is n C X into P race to X into one minus P raised two and minus X. The x ranges from 01, two up to end. So this is equal to n C X In two, 0.75, French two, X 1 -4 is 1 -0.75. So into 0.25, 1 -0.75 is 0.25 raised to end minus X. In the first part it is asked if you call 10 times, what is the probability that exactly nine of your calls are answered within 30 seconds. We call 10 times. So here and is equal to 10 probability of exactly nine calls is probability of X equal to nine. That is we have to find probability of X equal to nine. This is equal to then C nine. Yeah, X is nine and 10 is 10. So 10 C9 In two, 0.75, raised to nine into zero point 25 Raised to 10 -9. That is one 10 C9 into 0.75 raised to nine into 0.25. Race to one, 10 C nine is 10. So 10 into zero point seven, faith raised to nine In two, 0.25 raise to 1 0.25. So 10 into 0.75. Race to nine into 0.25. This is equal to zero but it 77. So If you call 10 times the probability that exactly nine of your calls are answered within 30 seconds is 0.1877 In the next part. Disaster. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds. If you call 20 times gives us and is equal to 20 Probability of at least 16 calls. That is we have to find probability of greater than or equal to 16. This is equal to Probability of X greater than or equal to 16 is equal to Probability of x equal to 16 Plus probability of x equal to 17 Plus. The probability of x equal to 18 Plus. The probability of x equal to 19 Yeah. Plus the probability of x equal to 20. The binomial random variable ranges up to end here is 20 so we will find up to probability of X equal to 20. This is equal to 20 C 16 for X equal to 16. This is equal to 20 C 16 In 20.75 raised to 16 In two, 0.25, Raised to 20-16. That is four Plus now we will find for x equal to 17. So it is 20 C17 In two, 0.75 reach to 17 In two, 0.25, Raised to 20-17. That is three, 2017 plus 0.75, raised to 17 In two, 0.25. race to three plus 20 C 18. Yeah. Into 0.75, raised to 18. Into 0.25, raised to 20-18. That is raised to two 20, satiating into 0.75 raise to 18. 0.25 race too. Yes. Now for x equal to 19, this is 20 C19 In 20.75 raised to 19 In two, 0.25 Raised to 20-19. That is well, Now for x equal to 20, this will be 20 C20. Yeah. Into 0.75 raised to 20. Mhm In two, Raised to 20-20. That is 0.25. Race 20. This is equal to zero point one egg 97 plus zero point but three. Mhm. Today nine. Yes. Zero point 06 69. Yeah plus zero point right. 02 Yeah. 11 plus zero point 00 32. We have to add all these probabilities. So this is equal to zero 41 48. 0.4148. So if you call 20 times the probability that at least 16 calls are answered in less than 30 seconds is 0.41 for it. In the last part it is asked if you call 20 times what is the mean number of calls that were answered in less than 30 seconds? So again we have N is equal to 20. We have to find the mean number of calls mu It is our mean that is expected X for a binomial random variable X expected X is equal to and be. So this is equal to We have an is equal to 20 And P is 0.75. P is given to be 0.75, 20 in two, This is equal to 15. So if you call 20 times the mean number of calls that are uncertain less than 30s against is 15.

In this problem, it is given that the probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are independent. Let x denote the number of times the call is answered in less than 30 seconds. So x follows a binomial distribution with P is equal to 0.75. We know that for a binomial random variable X, probability of x equal to X is n C. X into periods two X into one minus P. Raised two and minus X. They're x ranges from 01 up to N. So this is equal to n C X. In 20.75 raise two x. P is 0.75. So in 2 0.75 raised two X Into 1 -4. That is 1 -0.75 which is equal to 0.25 Raised two and -X. In the first part he's asked if you call 10 times what is the probability that exactly nine of your calls are answered within 30 seconds. We call 10 times so and is equal to 10. We have to find the probability that exactly nine of the costs are answered within 30 seconds. That is we have to find probability of X equal to nine. This is equal to 10 C nine and is still an X. S name. So 10 39 Into 0.75, raised to nine Into 0.25, Raised to 10 -9. That is one 10 C nine into 0.75. Raised to nine into 0.25. Raised to one. This is equal to 10. is 10 Into 0.75. Raised to nine. 0.75 raised to nine in 2 0.25 raise to one is 0.25. So 10 into 0.75 raised to nine into 0.25. This is equal to zero point 18 77. So The probability that exactly nine of the calls are answered within 30 seconds. 0.1877. In the next part it is asked If you call 20 times So here and is equal to 20. What is the probability that at least 16 calls are answered in less than 30 seconds. So we have to find probability of X greater than or equal to 16. This is equal to Probability of x equal to 16 Plus probability of x. equal to 17 Plus probability of x equal to 18 Plus probability of x equal to 19 Plus. The probability of x equal to 20. The binomial random variable eggs ranges up to end Here and is equal to 20. So we will compute the probabilities up to 20. This is equal to Rain. d. c. 16 For X. equal to 16. This is equal to 2016 in 2 0.75 raised to 16 Into 0.25 Raised to 20-16. That is four Plus. Now we will find for x equal to 17. So this is 2017 In two, 0.75 raised to 17 Into 0.25, Raised to 20-17. That is three now for X equal to waiting. This is equal to 20 satiated. Yeah. Into 0.75, raised to 18. Yeah. Into 0.25, raised to 20-18. That is to less Now for x equal to 19. This is 2019 Into 0.75. Raised to 19 Into 0.25, Raised to 20-19. That is one Now for x equal to 20. This is 2020 into 0.75, raised to 20 Into 0.25 Raised to 20-20, that is zero. So now this is equal to zero point 18 97. Bless zero point 13 39 less. Zero point 06 69 plus zero point 02 11 plus zero point 00 32. We have to add all these probabilities to get the required probability this is equal to zero point 41 four, 0.4148. So If we call 20 times the probability that at least 16 culture answered in less than 30 seconds is 0.41. For it. In the next part it is asked if you call 20 times What is the mean number of calls that are answered in less than 30 seconds. So here also is equal to 20. We have to find the mean number of calls that is, we have to find expected X. We know that for a binomial random variable X expected X is equal to N. P. Here and is equal to 20, And P is given to be 0.75. So expected X is equal to 20 into 0.75. This is equal to 15. So If we call 20 times The mean number of cars that are inserting less than 30 seconds is 15.


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