9.3.5 Normal Probabilities

It follows from Activity 5 that if we could evaluate the integral

exactly, we could write down a formula for `Ftext[(]t text[)]`. However, `f text[(] s text[)] = e^(-s^2text[/]2)` is one of those functions that does not have an elementary antiderivative, that is, it does not have an antiderivative that can be described using algebraic combinations of the functions that we know. For the time being, if we want to know a value of `F`, we have to approximate an integral.

Example 1

Assume you have a data set with the standard normal distribution. Find the probability that a random data value from this distribution lies between `0` and `1`.

Solution   This probability is

Using a computer or calculator (with `c = 0.3989`), we find the value to be approximately `0.3413`.

Example 2

Assume you have a data set with the standard normal distribution. Find the probability that a random data value lies between `-1` and `0`.

Solution   This probability is

Since the integrand is an even function, by symmetry this is the same as the integral we estimated in Example 1: `0.3413`.

Example 3

Use your computational tool to find the probability that a random data value from a set with the standard normal distribution is greater than `1.47`.

Solution   You will find that the probability of a value less than `1.47` is approximately `0.9292`. Then the probability of a value greater than `1.47` is obtained by subtracting the "less than" probability from `1`. In this case the value is approximately `0.0708`.

Note 1 – Tools

Checkpoint 6

Example 4

Suppose we have a normally distributed data set with mean `m = 2.3` and standard deviation `sd = 1.63`. Find the probability that a random data value from this data set lies between `1` and `3`.

Solution   We use the fact that the standardized data set obtained by subtracting the mean and dividing by the standard deviation has the standard normal distribution. Suppose `v` is a random value from the original data set. Then

exactly when

which is true exactly when

which is true exactly when

and that is true approximately when

Now the quantity `w=(v - 2.3)/1.63` is a random value from a data set that has the standard normal distribution. Thus our original probability question has the same answer as the question of the probability that a random value from a standard normal data set lies between `-0.80` and `0.43`. Using our computational tool, we see that the probability is `0.4545`.

Checkpoint 7

Checkpoint 8