6. Consider a coffee shop that uses a single-server queue. The inter-arrival time isexponentially distributed with a mean of 10 minutes and the service time is alsoexponentially distributed with a mean of 8 minutes. Calculate the:(i)mean wait in the queue (2)(ii) mean number in the queue (2)(iii)the mean wait in the system (2)(iv)mean number in the system (2)(v)proportion of time the server is idle. (2)[10]7. The arrival of customers at a bank follows a Poisson distribution with a mean arrivalrate of 10 customers per hour. The service time at the bank follows an exponentialdistribution with a mean service rate of 6 minutes per customer. Calculate theaverage number of customers in the system (including those being served andwaiting) and the average time a customer spends in the system. How is the systemperforming?[8]

The objective of the question is to calculate various parameters of a single-server queue system in…

Answered: 6. Consider a coffee shop that uses a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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18. On average, a banana will last 6.3 days from the time it is purchased in the store to the time it is too rotten to eat. Is the mean time to spoil greater if the banana is hung from the ceiling? The data show results of an experiment with 15 bananas that are hung from the ceiling. Assume that that distribution of the population is normal. 5.4, 8.2, 7.8, 6.8, 6.4, 8.3, 5, 7.5, 7, 5.3, 6.9, 8.8, 8.9, 9, 8.5 What can be concluded at the the αα = 0.05 level of significance level of significance? For this study, we should use T- Test, Z test The null and alternative hypotheses would be: H0: P or U, <>=, H1: P or U, <>=, The test statistic T,Z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is <> αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population…
Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed with a mean of 8 million dollars and a standard deviation of 1.3 million dollars. One health care information systems company considers a quarter a "fallure" if its sales level that quarter is in the bottom 10% of all quarterly sales levels. Determine the sales level (in millions of dollars) that is the cutoff between quarters that are considered "failures" by that company and quarters that are not. Carry your Intermediate computations to at least four decimal places. Round your answer to one decimal place. million dollars X S Save For Later Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Terms of Use 4 O * Check tab esc ! 1 ← 9. a @ 2 → W S # N = 3 C e $ 4 d r % 5 f acer Oll A t 6 g O. & 7 y h * u 8 j ( 1 9 k m 0 O P 1 M Submit Assignment Privacy Center Accessibility 10:12 backspace { 1 8民口團 11 D ?
3. The table below shows the characteristics of cases and controls from a case-control study into stressful life events and breast cancer in women. Identify the number items on the table whether they are: Nominal, Ordinal, Interval and Ratio Breast cancer group (n = 106) Control group (n=226) Variable P value 1 Age 61.6 (10.9) 51.0 (8.5) 0.000* -Social class* (%): I 10 (10) 20 (9) II 38 (36) 82 (36) III non-manual 28 (26) 72 (32) 0.094¹ III manual 13 (12) 24 (11) IV 11 (10) 21 (9) V 3 (3) 2 (1) VI 3 (3) 4 (2) No of children (%): 0 15 (14) 31 (14) 1 16 (15) 31 (13.7) 0.97 2 42 (40) 84 (37) >3 32 (31) 80 (35) 21.3 (5.6) 20.5 (4.3) 0.500* 12.8 (1.4) 13.0 (1.6) 0.200* Age at birth of first child Age at menarche Menopausal state (%): Premenopausal Perimenopausal 14 (13) 66 (29) 9 (9) 43 (19) 0.000 117 (52) Postmenopausal Age at menopause 83 (78) 47.7 (4.5) 38 0.001* 45.6 (5.2) 61 Lifetime use of oral contraceptives (%) 0.000¹ 3.0 (5.4) 4.2 (5.0) 0.065 No of months breastfeeding (n=90) (n =…
Suppose that the average salary of health systems professionals is normally distributed with µ=KES 40,000 and σ = KES 10,000. What proportion of professionals will earn KES 24,800 or less? What proportion of professionals will earn KES 53,500 or more? What proportion of professionals will earn between KES 45,000 and KES 57,000? Calculate the 80th percentile. Calculate the 27th percentile
6. A kinesiologist wanted to examine the effects of temperature and humidity on exercise performance. Each participant was asked to walk on a treadmill at a normal pace for 60 minutes, and the kinesiologist recorded the distance walked (miles). She assigned participants to one of 4 conditions that varied in temperature (normal or high) and humidity (normal or high). Mean distance walked by the four groups are shown in the table below. Assume a = .05 and Tukey HSD = 0.39. Normal temperature High temperature Normal humidity M = 3.00 M = 2.80 High humidity M = 2.80 M = 2.00 Complete the following conclusion: The factorial ANOVA revealed significant main effects of temperature [F (1, 24) = 25.00, p<.05] and humidity [F (1, 24) = 19.65, p<.05] and a significant Temperature x Humidity interaction [F (1, 24) = 9.25, p < .05].
b. The compensation received by the highest-paid executives of 12 international companies (in $'000) is as follows: 801 820 2215 803 850 856 899 846 902 856 911 947 i. Find the mean and the median compensation of the executives. Comment on the skewness of the distribution of the executives' compensations. 2
4. Researchers working for a sports drink company would like to know if a new blend if its popular sports drink can help improve marathon runners' final running time. Marathon times, Y, are known to follow a Normal distribution. Suppose that runners using the old blend sports drink have a mean finishing time of 300 minutes (H = 300 minutes). A random sample of 20 runners was selected to use the new blend of sports drink in a marathon, finishing with an average time of 270 minutes and a standard deviation of 20 minutes (y = 270 min, s = 20 min). Can it be claimed that the new blend lowers finishing times? Or is it possible that random chance alone can explain the discrepancy? Carry out a hypothesis test of the true mean u using the 4- step procedure. Set your alpha-level to 0.05. a. Which is the correct null hypothesis? O i. Ho: µ = 300 min O ii: Ho: µ = 270 min %3D O ii. Ho: µ < 300 min iv: Ho: µ < 270 min
12. x: 471.7 659.3 740.5 790.5 781.3 1047.3 969.5 1206.1 y: 446.3 680.2 725.2 765.3 740.4 1005.6 936.3 1156.8 Using the data find the y mean 219.60 833.28 205.42 807.01
1. Kindly help me to answer this problem with complete solution. Thank you’
. A process that cuts lengths of metal with high precision has historically produced metal rods with length that is normally distributed and with a mean of 800 millimetres (mm). Management at the factory is concerned that after an overhaul of the machine the mean length of the rods being produced could differ from the required 800 mm. If it is found that the rods are not being produced to specifications then the machine will have to be stopped for readjustment, with a consequent loss of production. The production line manager tasked with investigating the issue finds that a (random) sample of 25 rods yielded a mean length of 806 mm and a sample standard deviation of 10 mm. i. The mean of the sample is clearly not the required 800 mm. Say why this is not enough information in itself to say if the machine is operating correctly or not. ii. State the null and alternative hypotheses for this situation.
23. A TV station claims to its advertisers that the mean number of minutes people view the station is 30. The station conducted a survey of 500 of their viewers who commute. The sample statistics are shown below. 29.11 Sx 20.718 A simulation was run 1,000 times based on the results of the survey. The results of the simulation appear below. 40 samples = 1000 mean = 29.101 SD = 0.934 30- 10 27.0 28.0 29.0 30.0 31.0 32.0 Time Based on the simulation results, is the claim that viewers watch the station for an average of 30 minutes plausible? Explain your response, including an interval containing the middle 95% of the data, rounded to the nearest hundredth. Frequency
An industrial oil heater is tested by its ability to heat up a large room on a cold day and in particular, the rate at which it reaches a desired average temperature. The size of the room and the room temperature prior to switching on the heater were identified as two important factors of the rate at which the average room temperature increases. The experiment was replicated 3 times for each combination. The time (in minutes) till an average temperature increase of 10°C was measured is tabulated below: Factor B Large Classroom (200m2) Large Auditorium (400m2) (1) Treatment Combination (2) Factor A 7°C (1) 8,9,8 22,18,21 12°C (2) 9,10,12 19,29,25 Let factor A be the initial temperature and factor B be size of the room.

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