Items produced by a manufacturing process, in a single file, have a final check
ID: 3219462 • Letter: I
Question
Items produced by a manufacturing process, in a single file, have a final check before the items are packed into boxes. On average every three minutes a defective item is noticed and removed from the line. These inter-arrival times are to be modelled by an exponential distribution. Use R as much as possible to answer the following questions. (a) Taking an event as the location of a defective item explain in context why this stochastic process is a counting process (Refer p117 Course Notes). (b) Find the probability that five defective items are noticed and removed in a fifteen minute period. (c) Find the probability that no more than four defective items are noticed and removed in 10 minutes. (d) What is the probability that the 6^th defective item is noticed and removed 20 minutes after the start-up of the line? (e) Find the probability that the time elapsed between noticing and removing four successive defective items is at least 13.5 minutes. Assume inter-arrival times are independent and identically distributed. (f) Suppose the line speed was set at 15 cans per second. Estimate the ppm (parts per million) of defective items. (g) The cost of repairing each defective item is a random variable with a mean of $7.50 per item and a standard deviation of $1.95 per item. The dollar amount of repairing the defective items in a 24 hour period is approximately normally distributed. Find the mean and standard deviation of these repair costs based on a 24 hour period. (h) Referring to part (g), find the probability that the dollar amount of repairs in 24 hours is between $3500 and $3800.Explanation / Answer
Part (a)
Given on an average every 3 minutes one defective item is produced, suppose time taken for the first 10 defective item to be produced are 1, 4, 3, 2, 2, 4, 5, 2, 4, 3 minutes respectively. Here the event is the time taken for the event to occur. The very same phenomenon can be described as ‘10 defective items are produced in 30 minutes’ Here the event is the number of times the defective items occur. Thus, an inter-event time event is equivalent to a counting process of number of times the event occurs in a common unit of time. The given situation is: average time for a defective to be produced is 3 minutes which in a counting process would mean that on an average, number of defective units produced per minute is 1/3.
In particular,
if inter-arrival time is Exp(), then number of events per unit time is Poisson(), where = 1/. ………………(1)
Part (b)
For this part and the next 2 parts, we employ the above connection between Exponential and Poisson distributions and also the property of Poisson described below.
If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(), then Y ~ Poisson (k) …............................................................………….. (2)
Combining (1) and (2) above, if Y = Number of defectives produced in 15 minutes, then
Y ~ Poisson(5) [5 = 15x(1/3)]
Probability that 5 defectives are produced in 15 minutes = P(Y = 5) = 0.1755 ANSWER
Part (c)
Here, Y = Number of defectives produced in 10 minutes, and hence Y ~ Poisson(10/3).
Probability that no more than 4 defectives are produced in 10 minutes
= P(Y 4) = 0.7565 ANSWER
Part (d)
Here, Y = Number of defectives produced in 20 minutes, and hence Y ~ Poisson(20/3).
Probability that 6th defective is produced in 20 minutes = P(Y = 6) = 0.1552 ANSWER
NOTE: All the above probabilities are obtained using Excel Function.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.