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# Hi there: Have a great day. I understand this may be considered 3 questions.

Hi there: I've never done this before so please let me know how to ask a better question! Have a great day. I understand this may be considered 3 questions. There is a lot of background to this question including questions and answers from an assignment that lay the groundwork for the question I need help with.

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Skateboards are assembled in a particular manufacturing plant on an assembly line by three workers. Worker 1 performs the first three assembly steps, worker 2 performs the next three, and worker 3 performs the final three. Suppose there is no work-in-process inventory kept between workstations.

a) Suppose the time for worker 1 to perform each of his three assembly steps is a normally distributed random variable. The first step averages 30 seconds, the second step averages 20 seconds, and the third step 10 seconds. The standard deviations for each of the three steps are 6 seconds, 4 seconds, and 2 seconds respectively. What is the mean and standard deviation of the total time it takes for worker 1 to perform his three assembly steps?

**Let T = total time and let A, B, and C denote the times of his first, second, and third step respectively.**

**T = A + B + C, so T is the weighted sum of three normally distributed random variables where the weights are all +1. T is, therefore, normally distributed with **

**Mean = 30 + 20 + 10 = 60**

**Variance = 6****2 **** + 4****2**** + 2****2**** = 36 + 16 + 4 = 56**

**Standard deviation = sqrt(56) = 7.48 (approximately)**

b) The total time for worker 2 to perform her three assembly steps is normally distributed with an average of 70 seconds and a standard deviation of 4 seconds. Suppose worker 1 and worker 2 each start their part in the assembly operation (on different and successive skateboards) at the same time. What is the probability that worker 2 finishes first and is left waiting (assuming no work in process inventory between the two stations) with nothing to do until worker 1 finishes?

**Let T denote the time of the first worker and S denote the time of the second worker. Worker 2 finishes first if and only if S < T which is equivalent to S - T < 0.**

**Let W = S - T. W is a weighted sum of random variables where the weight on S is +1 and the weight on T is -1. W is, therefore, normally distributed with **

**Mean = 70 - 60 = 10**

**Variance = 16 + 56 = 72**

**Standard deviation = sqrt(72) = 8.49 (approximately)**

**We seek P{W <= 0} = normdist(0, 10, 8.49, 1) = 0.12 (approximately)**

c) Given the situation as stated in part b, what is the probability that worker 1 finishes first and is blocked (he can't start work on a new skateboard until he can hand the one he just finished off to worker 2)?

**Worker 1 finishes first if and only if worker two's time, S, is larger than worker one's time, T. In other words, if and only if S - T > 0. Thus, we seek P{W > 0}.**

**P{W > 0} = 1 - P{W <= 0} = 1 - 0.12 = 0.88.**

**The Question:**

Consider workers 1 and 2 as described in parts a, b, and c of the problem. Suppose workers 1, 2, and 3 each start their part in the assembly operation (on different and successive skateboards) at the same time. Let T = time of worker 1, S = time of worker 2, and R = time of worker 3.

a. What is the probability that worker 1 is idle for 10 seconds or less due to being blocked after finishing her skateboard?

b. What is the probability that worker 2 is idle for 5 seconds or less due to being starved after finishing his skateboard?