Question

The time required to finish a test is normally distributed (i.e. in a Gaussian distribution) with a mean of 60 minutes and a standard deviation of 10 minutes. What is the probability that a student will finish the test in less than 70 minutes?

Answer 1

About `84.13%` that a student finishes in less than `70` minutes.

---

Consider the Gaussian distribution function:

`f(t) = 1/sqrt(2pisigma^2) e^(-(t-t_"avg")^2//2sigma^2)`

You were given the average time and the standard deviation in the time:

• `t_"avg" = "60 minutes"`
• `sigma = "10 minutes"`

So now we have an explicit expression for `f(t)`:

`f(t) = 1/sqrt(200pi) e^(-(t-60)^2//200)`

This function, when integrated over a certain range, will give the area under the curve, and that gives the probability of one student finishing in the allotted time.

In other words,

`P = int_(0)^(t) f(t)dt`

is the probability that one student will finish the test within `t` minutes.

Thus, in order to finish the test in less than `70` minutes, we consider the range `t in [0,70)`:

`int_(0)^(70) f(t)dt = ???`

This is a non-elementary integral, so you can either use a calculator or Wolfram Alpha. And so:

`1/sqrt(200pi) int_(0)^(70) e^(-(t-60)^2//200)dt`

`= 0.841345`

So, there is about a `color(blue)(84.13%)` chance that a student finishes in less than `70` minutes.