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A tricky question from my textbook: A garage expects to fix a mean of 32 cars per week (assuming 6 day week). Each car takes up two days of a mechanic's time. If a car is brought in during any given week, it is allocated to a mechanic for completion during the following week. If the car is not fixed in the following week the customer will take his car somewhere else. Calculate the amount of mechanics needed so the garage can state that 75% of all cars will be repaired during the second week.

I'm quite certain this is Poisson distribution but I'm unsure of how to start it

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  • $\begingroup$ I'm not sure I understand the objective and setup here. I assume the number of customers is a Poisson random variable with mean $32$. You could choose a number of mechanics so that the probability of satisfying all customers is at least 0.75; this is a well-defined question with a finite answer. You could also choose a number of mechanics so that the probability of satisfying at least $75\%$ of the customers is $1$. This is a well-defined question, but (under the Poisson assumption) the answer is $\infty$, which is bad. $\endgroup$
    – Ian
    Commented Mar 6, 2016 at 17:56
  • $\begingroup$ That is one of the odd things about this question. I think that's what it is trying to get at. I think it must be trying to work out the probability of satisfying all customers, but as you say the wording is misleading. Would there be another way of working it out if the objective is to satisfy at least 75% of customers, without poisson? $\endgroup$ Commented Mar 6, 2016 at 17:58

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Here is how I believe the problem was supposed to be solved.

Since one mechanic fixes a car in two days, three mechanics can fix three cars in one week.

Suppose that the number of cars $X$ that the garage has brought in in one week is Poisson distributed $\text{Pois}(32)$ with mean $32$. We need to find the number $x$, such that $$P(X \leq x) \geq 0.75.$$ According to Wolfram Alpha, we have $P(X\leq 35)\approx 0.74$, and $P(X\leq 36)\approx 0.79$, so we choose $x=36$.

We need $36/3 = 12$ mechanics to make sure that we can fix $36$ cars a week. With this number of mechanics, in at least $75\ \%$ of all weeks, we have enough mechanics to fix the incoming cars the next week.

Notice however that since the garage fixes all cars in weeks with at most $36$ incoming cars, and since in weeks where the number of incoming cars exceed $36$, the garage still fixed $36$ of them, the probability of fixing some incoming car will be larger than $75\ \%$. As the problem is stated, this is alright though. The garage can state that for an arbitrary car, it will be fixed with probability $75\ \%$.

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  • $\begingroup$ I agree, I think this was the intended working. A poorly worded question it seems. Thank you for your help :) $\endgroup$ Commented Mar 6, 2016 at 18:29

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