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The problem statement said:

The servicing of a machine requires two separate steps, with the time needed for the 1st step being an exponential random variable with mean 10 minutes and the time for the second step being an independent exponential random variable with mean 15 minutes. If a repairperson has 20 machines to service, compute the aproximate the probability that all the work can be completed in 8 hours.

My work:

*) Let Xi be the time to service the ith machine, then

 E[Xi]=E[X1]+E[X2]=10+15=25min (Thanks Andre!)

 V[Xi]=V[X1]+V[X2]=100+225=325

**) Therefore, Xi is modeled by a normal r.v. with parameters

 E[X]=20*E[Xi]=500
 V[X]=20*V[Xi]=6500  then std=80.622

***) Last Part

Now, I know that X is the sum of 20 independent and identict distributives r.v. x1, x2 ….x20 where each xi is the time for the repair of the ith machine.

P(X<8*60min)=P(z<(480-500)/80.622)=P(z<-0.248071)= 2.74855E-10

Question: Is may procedure correct? Looks very small probability, it is right? Thanks so much for your help.

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    $\begingroup$ Note that $10+15=25$, not $35$. $\endgroup$ Jan 30, 2016 at 1:55
  • $\begingroup$ Thanks, I fix my mistake but I'm still concern about that the probability is pretty low. $\endgroup$
    – DarioC
    Jan 30, 2016 at 2:06

1 Answer 1

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Everything looks right, except for the last value. You made a mistake using your calculator or reading values from the table. I got $$ P\left(Z<\frac{480-500}{\sqrt{6500}}\right) = P(Z<-0.2480695) = 0.4020403$$ Notice that this value is "close" to $50\%$. That makes sense since the standardized value is very small (-0.248), close to the mean.

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  • $\begingroup$ Thanks! Yes know the probability looks logicaly correct :-) $\endgroup$
    – DarioC
    Jan 30, 2016 at 2:17

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