Question

if we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot.

Answer 1

To find this probability we have to take integral (corresponding to normal distribution)

$$1 / (\mathrm {s s q r t} (2 \mathrm {p i})) \exp (- (\mathrm {x - m}) ^ {\wedge} 2 / (2 \mathrm {s} ^ {\wedge} 2))$$

where $s = 1$ is standard deviation and $m = 3.5$ is mean and $x$ is values of length in limits from 2 to 4.5 we can do it numerically only. So, the value of this integral is

0.7745

That is the probability we were looking for.