# Normal distribution probability question

A survey reveals that the waiting time for the bus during off-peak hours is approximated by a normal distribution, whose mean and standard deviation are respectively 7 and 2 minutes.

Assume that you have waited 5 minutes at the bus stop. What is the probability that you wait for a total of more than 8 minutes before boarding.

any tips? i tried to calculated by 13minutes but answer is wrong.

• It's $P(X>8 \mid X>5)=\frac{P(X>8 \text{ and } X>5)}{P(X>5)}=\frac{P(X>8)}{P(X>5)}$. (Note that they talk about a total of 8 minutes, so that is counting the 5 minutes you already waited.) – Ian Apr 2 '17 at 16:36

The mistake you made is not counting the 5 minutes that the person in the problem already waited, because the problem is asking about the probability to wait for a total of 8 minutes. Thus you are asked for $P(X>8 \mid X>5)=\frac{P(X>8 \text{ and } X>5)}{P(X>5)}=\frac{P(X>8)}{P(X>5)}$ where $X$ is normally distributed with the given parameters. Note that this is also not the same as $P(X>3)$, because the normal distribution is not memoryless (in contrast to the exponential distribution).
Yes, if you tried to calculated by total 13 minutes then you are wrong, as they asked the probability that you wait for a total of more than 8 minutes given that you have waited 5 minutes at the bus stop. So now it's clear that you have to find out $P(X>8\ |\ X>5)$ with mean $\mu=7$ and standard deviation $\sigma=2$. \begin{align*} P(X>8\ |\ X>5)&=\dfrac{P(X>8\cap X>5)}{P(X>5)}\\ &=\dfrac{P(X>8)}{P(X>5)}\\ &=\dfrac{Q\left(\frac{8-7}{2}\right)}{Q\left(\frac{5-7}{2}\right)}\\ &=\dfrac{Q(0.5)}{Q(-1)}\\ &=\dfrac{Q(0.5)}{1-Q(1)}\\ &\approx\dfrac{0.30854}{0.84134}\qquad \mathrm{(from\ table)} \\ &\approx0.367 \end{align*}