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Assume that the time used for someone driving to work has a normal distribution with expected value $E(X) = 27$ and standard deviation $\sigma = 2.5$. Driving from work, we have the same distribution, but now with $E(X) = 31.5$ and $\sigma = 2.5$.

What is the probability that a given individual will use more than a total of $61.5$ minutes going to and from work on a given day?

(Time used to and from work on a given day and time used to and from work on different days are indepedent).

I'm confused about how I'm supposed to combine these two different distributions. I tried using $E(X) = 27-31.5 = -4.5$ and $\sigma = \sqrt{2.5^2+2.5^2} = 3.5355$ but this doesn't make sense when I want to find $P(X>61.5$).

Any ideas?

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Let $X$ denote the time driving to work and $Y$ the time going back home.

Then $Z:=X+Y$ is the total, and has normal distribution with expectation: $$\mathbb EZ=\mathbb E(X+Y)=\mathbb EX+\mathbb EY=27+31.5$$

If moreover $X$ and $Y$ are independent then:$$\mathsf{Var}(Z)=\mathsf{Var}(X)+\mathsf{Var}(Y)=2.5^2+2.5^2$$

The distribution of $Z$ is determined by expectation and variance, so this together enables you to find $P(Z>61.5)$.

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  • $\begingroup$ I see, I misinterpreted the $\mathbb EZ$... Thank you sir $\endgroup$ – novo Oct 5 '18 at 15:45
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Oct 5 '18 at 15:49
  • $\begingroup$ If I'm going to find the probability that an individual on average over the course of $4$ days uses between $25$ og $29$ minutes to work, how do I do that? I know that $P(25<X<29) = 0.5763$ given $E(X) = 27$ and $\sigma = 2.5$ But I don't see how to relate this to average over 4 days... Do you also have a hint on this one? $\endgroup$ – novo Oct 5 '18 at 16:12
  • $\begingroup$ If there are $4$ days and you must look at $\overline X=\frac14(X_1+X_2+X_3+X_4)$ where the $X_i$ are iid and distributed as $X$. Again the distribution is normal and you can find expectation and variance of $\overline X$ $\endgroup$ – drhab Oct 5 '18 at 16:38
  • $\begingroup$ I see, I think I got it right! Thank you. For the last part, if you wouldn't mind giving one last hint; What is the probability of someone using less time driving from work than to work on a given day? I appreciate the help! $\endgroup$ – novo Oct 5 '18 at 16:54

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