A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of 9.5 vehicles per day (randomly). The mechanic crew can service an average of 13 vehicles per day with a repair time distribution that approximates an exponential distribution.

What is the utilization for this service system?

A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of 9.5 vehicles per day (randomly). The mechanic crew can service an average of 13 vehicles per day with a repair time distribution that approximates an exponential distribution.

What is the average time (in days) before the facility can return a breakdown to service from vehicle arrival until service complete?

A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of 9.5 vehicles per day (randomly). The mechanic crew can service an average of 13 vehicles per day with a repair time distribution that approximates an exponential distribution. Of the time before the facility can return a breakdown to service, how much of that time (in days) on average is spent waiting for service to begin?

Part 2. (10.5 points) Identify the distribution. The main purpose of probability density function is to model a random variable. We use the model to find probabilities and expectations for new values. The data collected in the file DA2_Cont RV_Data.csv represents 1000 measurements for three continuous random variables. Use this data in association with the R script, DA2_Cont_RV.R, to visualize and calculate the expectation and variance for each random variable. For each random variable: a. (1.5 points each) Based on the shape of the data (paste your plots from R), you need to identify which common probability density function best fits the data. Uniform, Exponential or Normal distribution. Explain your reasoning. b. (1 point each) From the mean and variance, calculate specific parameter values that describe the distribution. Show work! Round to nearest whole number *Exponential round to two decimal places. c. (1 point each) Based on the parameters, state the unique probability density function. Include the range.Random Variable 1 Random Variable 2 Random Variable 3 500 150 400 300 100 Frequency Frequency 200 8 8 - 100 O 0 10 20 30 40 50 1000 2000 3000 300 400 500 600 Continuous Units Continuous Units Continuous Units# Select and run lines 11 t # Expectation and Variance, > mean (cont RVs$rvl) [1] 6.264699 > var (cont RVs$rvl) [1] 35. 09125 > # Expectation and Variance, >mean (cont RVs$rv2) [1] 2012.003 > var (cont RVs$rv2) [1] 156673.2 > # Expectation and Variance, > mean (cont RVs$rv3) [1] 450.9865 > var (cont RVs$rv3) [1] 7428. 629 >STAPLE OR PAPERCLIP YOUR WORK, PLEASE PROBLEM 2 [20 POINTS = 5+5+5+5] TWO RANDOM VARIABLES, X AND Y, ARE INDEPENDENT. THEIR MARGINAL DISTRIBUTIONS ARE SPECIFIED AS FOLLOWS. X -5 0 10 PROBABILITY 40% 35% 25% Y -4 6 PROBABILITY 40% 60% 1. FIND EXPECTATION FOR EACH VARIABLE. E [XI = E [YI = 2. FIND THE VARIANCE FOR EACH VARIABLE. VAR [X] = VAR [YI = 3. WHAT IS THE EXPECTATION OF THE NEW RANDOM VARIABLE, Z =3X-2Y? E [ZI = 4. FIND THE VARIANCE OF Z (=3X-2Y). VAR [Z] =