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.
Best Answer
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.