Susan commutes daily from her home to her office. The average time for a one-way trip is $24$ minutes with a standard deviation of $3.8$ minutes. Assume that the trip time follows a normal distribution.

(f) A trip to a client's office from her home takes $30$ more minutes than twice the time to her own office. Let $W$ be the time for a trip to the client's office.

Find the probability that a trip to the client's office takes more than $1$ hour but less than $1.5$ hours.

There are seven parts to this question, but I'm confused on just this part.

What I have tried so far:

So I know that $E(Y) = 24$, $V(Y) = 3.8^2$, and $SD(Y) = 3.8$

I also know that $W = 2Y + 30$

I've also calculated:

$$E(W) = 2E(Y) + 30 = 78$$

$$V(W) = 2^2V(Y) \approx 57.76$$

But at this point, I'm not sure how to proceed. I suppose I'd be looking for $P(60 < W < 90)$, but I'm not sure how to go about calculating this.

  • 1
    $\begingroup$ Are you allowed to be using a graphing calculator? $\endgroup$ – Karn Watcharasupat Dec 14 '17 at 15:20
  • 1
    $\begingroup$ oh wow...ok this is gonna be interesting... Let me think of a way to bypass some calculations... $\endgroup$ – Karn Watcharasupat Dec 14 '17 at 15:23
  • 2
    $\begingroup$ Your problem statement says that the standard deviation of the (to work) trip length is 3.8 minutes; but then you use 3.8 for the variance. $\endgroup$ – paw88789 Dec 14 '17 at 15:25
  • 2
    $\begingroup$ You say you can't use a graphing calculator. Do you have a table of values (areas) for a (standard) normal distribution? $\endgroup$ – paw88789 Dec 14 '17 at 15:28
  • 2
    $\begingroup$ Oh if you have the table, then just normalize the RV and break the probability up into two subtracting CDF's. $\endgroup$ – Karn Watcharasupat Dec 14 '17 at 15:35

If $\mu$ and $\sigma^2$ denote mean and variance of $W$ then $U:=\frac{W-\mu}{\sigma}$ has standard normal distribution.

So: $$\mathsf P(60<W<90)=\mathsf P(60<\mu+\sigma U<90)=\mathsf P\left(\frac{60-\mu}{\sigma}<U<\frac{90-\mu}{\sigma}\right)$$

This can be found by means of a table.

  • $\begingroup$ Wonderful! Thank you for the help, I really appreciate it. $\endgroup$ – Doadle Dec 14 '17 at 15:40
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Dec 14 '17 at 15:40
  • $\begingroup$ I don't see how the suggestion to use a table qualifies. How was the table tabulated? With a slide rule? What's the difference between using a calculator and a table? $\endgroup$ – wolfies Dec 14 '17 at 15:42
  • $\begingroup$ @wolfies the OP said in a comment that a table was provided, and had not the possibility to use a calculator. $\endgroup$ – drhab Dec 14 '17 at 15:45
  • 1
    $\begingroup$ @wolfies Yes it is. But that's the OP's situation so we solve it according to that. Usually taking away calculator is meant to prevent some complex calculations being done without the understanding of its mechanics. Making life slightly more difficult with simple questions such as this is just a side effect that to me is somewhat acceptable. $\endgroup$ – Karn Watcharasupat Dec 15 '17 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.