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If a technician does not encounters any hardware problems, the time he requires to assemble a computer follows a normal distribution with a mean of $30$ minutes and a standard deviation of $3$ minutes. Let $T$ be the time in which he assembles a computer.

(a) Find the probability that it will take him more than $36$ minutes to assemble a computer given that he does not encounter hardware problems.

(b) When he encounters hardware problems the time to assemble a computer has a mean of 50 minutes and a standard deviation of $7$ minutes. Find the probability that it will take him more than $3$6 minutes to assemble a computer given that he encounters hardware problems.

(c) Suppose that he encounters a hardware problem $10\%$ of the time. If it took him more than $36$ minutes to assemble a computer, what is the probability that he encountered a hardware problem?

Let $Y$ be the event that the tech encounters a hardware problem.

For (a) I used $T\sim N(30,3)$ and I found $P(T>36\mid \overline Y) =1-P(T\leq36\mid \overline Y)=1-P(Z\leq2\mid \overline Y) = 1 - \Phi(2) = 0.0228$

Similarly for (b) I used $T\sim N(50,7)$ and found that $P(T>36\mid Y) = 0.9772$

Now for(c), I'm confuse as to how to set it up. Am I looking for a new random var?

I think is something like this:

How do I find $P(Y \mid T >36)$ How do I proceed?

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  • $\begingroup$ You have already computed $P(T>36|Y)$ and $P(T>36|Y^\complement)$ (where $Y^\complement$ means "not $Y$"). Do you see where you computed those things? Are you familiar with Bayes' Theorem, and if so, can you see how to apply it? $\endgroup$ – David K Jul 6 '15 at 22:22
  • $\begingroup$ @David K. We cover Bayes' rule in theory but never applied it on normal distributions. To be honest I don't see how to apply it to part (c) of this problem. $\endgroup$ – Jose soriano Jul 6 '15 at 22:31
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In part (a) you calculated $P(T > 36 \mid Y^\complement),$ where $Y^\complement$ is the event that the computer does not have a hardware problem and $T$ is the time taken to assemble the computer.

In part (b) you calculated $P(T > 36 \mid Y),$ where $Y$ is the event that the computer has a hardware problem and $T$ (as before) is the time taken to assemble the computer.

In part (c) you want to calculate $P(Y \mid T > 36)$. I suggest using Bayes' Theorem, which for events $A$ and $B$ says that

$$ P(A \mid B) = \frac{P(A) P(B \mid A)} {P(A) P(B \mid A) + P(A^\complement) P(B \mid A^\complement)}.$$

Now here is an important fact: Bayes' Theorem only cares about the values of the probabilities in the formula above. It doesn't care whether those probabilities were computed according to a normal distribution or handed to you by the Blue Fairy.

Since you have already calculated $P(T > 36 \mid Y)$ and $P(T > 36 \mid Y^\complement),$ that is suggestive that these should take the roles of $P(B \mid A)$ and $P(B \mid A^\complement)$ in the theorem. That is, we might try substituting $Y$ for $A$ and $T > 36$ for $B$ in the theorem's formula.

Try making these substitutions and see what you have on each side of the formula.

Also remember that you are given that $P(Y) = 0.1$ in part (c), from which you can easily find what $P(Y^\complement)$ is. Plugging in the values for all known probabilities, what probability can you compute using the theorem?

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$$P(Y\mid T>36)=\frac{P(Y\cap\{T>36\})}{P(T>36)}=\frac{P(T>36\mid Y)P(Y)}{P(T>36\cap Y)+P(T>36\cap \overline Y)}=$$$$=\frac{P(T>36\mid Y)0.1}{P(T>36\mid Y)0.1+P(T>36\mid \overline Y)0.9}$$

You know how to compute the conditional probabilities appearing in the derivation above.

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To begin, I checked your computations in (a) and (b), using R statistical software. They are correct:

 1 - pnorm(36, 30, 3)
 0.02275013  # part (a)
 1 - pnorm(36, 50, 7)
 0.9772499   # part (b)

If you know about the Empirical Rule you can get very close to the answer to (a) without tables or extra computation: The probability of a result more than $2\sigma = 6$ away from $\mu = 30$ (in either direction) is about 5%. You are looking only in the upper direction so it should be about half that or 2.5%, which is very near to 0.2275. You should be able to verify similarly that part (b) is about right.

For the rest, you need to find the 'overall' probability that it takes longer than 36 minutes. The 'law of total probability' gives (.9)(.0228) + (.1)(.9772). Do you see how I got that?

While I have been typing this I see someone else has shown how to finish. I hope you will take time to understand the problem and think the steps through for yourself. In particular, please take time to compare what you see here with the discussion of Bayes' Theorem in you text or lecture notes. (You won't have such eager help to give the away answer when you take an exam on this material.)

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