Standard Normal Distribution in Math Problems

The graph of a standard normal distribution showing the location of z on the bell curve
Standard normal deviation.

Courtesy of C.K.Taylor (author)

The standard normal distribution, which is more commonly known as the bell curve, shows up in a variety of places. Several different sources of data are normally distributed. As a result of this fact, our knowledge about the standard normal distribution can be used in a number of applications. But we do not need to work with a different normal distribution for every application. Instead, we work with a normal distribution with a mean of 0 and a standard deviation of 1. We will look at a few applications of this distribution that are all tied to one particular problem.


Suppose that we are told that the heights of adult males in a particular region of the world are normally distributed with a mean of 70 inches and a standard deviation of 2 inches.

  1. Approximately what proportion of adult males are taller than 73 inches?
  2. What proportion of adult males are between 72 and 73 inches?
  3. What height corresponds to the point where 20% of all adult males are greater than this height?
  4. What height corresponds to the point where 20% of all adult males are less than this height?


Before continuing on, be sure to stop and go over your work. A detailed explanation of each of these problems follows below:

  1. We use our z-score formula to convert 73 to a standardized score. Here we calculate (73 – 70) / 2 = 1.5. So the question becomes: what is the area under the standard normal distribution for z greater than 1.5? Consulting our table of z-scores shows us that 0.933 = 93.3% of the distribution of data is less than z = 1.5. Therefore 100% - 93.3% = 6.7% of adult males are taller than 73 inches.
  2. Here we convert our heights to a standardized z-score. We have seen that 73 has a z score of 1.5. The z-score of 72 is (72 – 70) / 2 = 1. Thus we are looking for the area under the normal distribution for 1<z < 1.5. A quick check of the normal distribution table shows that this proportion is 0.933 – 0.841 = 0.092 = 9.2%
  3. Here the question is reversed from what we have already considered. Now we look up in our table to find a z-score Z* that corresponds to an area of 0.200 above. For use in our table, we note that this is where 0.800 is below. When we look at the table, we see that z* = 0.84. We must now convert this z-score to a height. Since 0.84 = (x – 70) / 2, this means that x = 71.68 inches.
  4. We can use the symmetry of the normal distribution and save ourselves the trouble of looking up the value z*. Instead of z* =0.84, we have -0.84 = (x – 70)/2. Thus x = 68.32 inches.

The area of the shaded region to the left of z in the diagram above demonstrates these problems. These equations represent probabilities and have numerous applications in statistics and probability.