# Combining two standard normal distributions

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?

## 1 Answer

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 $$P(Z>61.5)$$.

• I see, I misinterpreted the $\mathbb EZ$... Thank you sir – novo Oct 5 '18 at 15:45
• You are welcome. – drhab Oct 5 '18 at 15:49
• 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? – novo Oct 5 '18 at 16:12
• 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 $\overline X$ – drhab Oct 5 '18 at 16:38
• 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! – novo Oct 5 '18 at 16:54