# Calculate probability from normal distribution WITHOUT calculator

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.

• Are you allowed to be using a graphing calculator? – Karn Watcharasupat Dec 14 '17 at 15:20
• oh wow...ok this is gonna be interesting... Let me think of a way to bypass some calculations... – Karn Watcharasupat Dec 14 '17 at 15:23
• 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. – paw88789 Dec 14 '17 at 15:25
• You say you can't use a graphing calculator. Do you have a table of values (areas) for a (standard) normal distribution? – paw88789 Dec 14 '17 at 15:28
• Oh if you have the table, then just normalize the RV and break the probability up into two subtracting CDF's. – 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: $$\mathsf P(60<W<90)=\mathsf P(60<\mu+\sigma U<90)=\mathsf P\left(\frac{60-\mu}{\sigma}<U<\frac{90-\mu}{\sigma}\right)$$