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The assembly of a machine requires two stages, that proceed consecutively and independently of one another. The first stage takes a mean time of 20 minutes with a standard deviation of 8 minutes. The second stage takes a mean of 30 minutes with a standard deviation of 12 minutes. The next 100 machines assembled will be sent in a shipment to Portland.

And the question:

What is the probability that the total time spent assembling 100 machines is less than 80 hours?

I am not 100% sure on how to go about solving this problem as an example does not exist within my professors notes, however my intuition is telling me to create a normal variable with the mean being the sum of both stages means, and the variance being the sum of squared SDs and then using a normal cumulative distribution function to solve for the probability p(X<80)… Am I on the right track or am I completely off?

Thank you in advance for any help provided!

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  • $\begingroup$ I think your intuition for the first part id correct, however the seond part should be $80$ hours divided by $100$, which gives $48$. $\endgroup$ – StAKmod Apr 9 at 19:31
  • $\begingroup$ @StAKmod Would it also work if multiplied the means and variances by 100 before adding the two stages together and then using a normalCDF? $\endgroup$ – Chase Allen Apr 9 at 19:41
  • $\begingroup$ I believe that will give the same answer, but I am not sure if that answer is correct. $\endgroup$ – StAKmod Apr 9 at 19:43
  • $\begingroup$ @ChaseAllen: Let me make a suggestion that will improve your ability to answer questions. Eliminate all irrelevant data. Do you think it is relevant that the machines are being shipped to Portland? Really? Would the problem change if they were shipped to Paris? Of course not. So delete that. $\endgroup$ – David G. Stork Apr 9 at 22:02
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You are on the right track.

Let $X$ be the time to assemble one machine. You should be able to compute $E[X]$ and $\sigma^2_X$.

Now let $Y= X_1 + X_2 + ... +X_{100}$ be the time needed to build 100 machines. (I'm assuming here that the machines are build in series, one starts after the previous finishes - this should be made more explicit). Then, $Y$ will approach a normal distribution with $E[Y]=100 E[X]$ and $\sigma^2_Y=100 \sigma^2_X$

Then to estimate $P(Y\le 80)$ you need to evaluate the normal cumulative distribution.

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  • $\begingroup$ (+1) for being first! $\endgroup$ – BruceET Apr 9 at 22:43
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Assuming that the total assembly time for one machine will be the sum of two normal distributions (as implied by your title), the mean and variance of the normal total will be as you surmise. Specifically, assuming normality, $T + X + Y,$ so that $X \sim \mathsf{Norm}(\mu = 20, \sigma = 8)$ and $Y \sim \mathsf{Norm}(\mu = 30, \sigma = 12),$ we have $$T \sim \mathsf{Norm}(\mu = 20+39 = 50, \sigma = \sqrt{8^2 + 12^2} = \sqrt{208} = 15.42).$$

However, your question has to do with $S = \sum_{i=1}^{100} T_i.$ This $S$ is again a normal random variable. (That's assuming machines are assembled sequentially.) Can you find its mean and SD? And then the probability $P(S < 4800\, \text{min})?$

Notice that $E(X) = 100E(T) = 5000$ min. So the result will surely be less than half?

In order to avoid working this problem completely, I'm showing a simulation for a million such shipments, each consisting of 100 sequentially manufactured machines. That will give a fairly accurate answer (about 8.2%) for you to check against.

set.seed(409)
s = replicate(10^6, sum(rnorm(100, 50, sqrt(208))))
mean(s < 80*60)
[1] 0.08288

enter image description here


Addendum: As @leonbloy ha pointed out, the question is ambiguous. If the 100 machines are assembled simultaneously, and we're wondering whether even the last one to be finished will be ready within 80 hrs, that is a much more difficult analytic problem. I think it isn't likely that the max was intended. But just for fun, a simulation shows it's pretty sure all 100 machines will be ready within 4800 minutes. (120 min should be plenty of time). Notice that the maximum of IID normal distributions is not normal.

set.seed(2019)
b = replicate(10^6, max(rnorm(100, 50, sqrt(208))))
mean(b < 4800)
[1] 1

enter image description here

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  • $\begingroup$ We don't know (as far as I see) that the individual times are normally distributed. And we don't need that assumption, what matters is that the sum will (approximately) be. $\endgroup$ – leonbloy Apr 9 at 22:55
  • $\begingroup$ Right. The question is sufficiently vague for several interpretations. // Of course the CLT should work fine with $n=100,$ unless the individual assembly times are very strangely distributed. But the title does seem to imply normality. $\endgroup$ – BruceET Apr 9 at 23:14

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