Title: Using the standard normal distribution to make descriptive summary statements
Course: UGQuants (SSPS) > Section: Probability

Using the standard normal distribution to make descriptive summary statements

Let us look briefly at how we can use the standard normal distribution to make summary statements about groups of cases that are approximately normally distributed.

1 | For a normal distribution, and with any standard deviation, we can convert the actual value for any observation (or case) into a z score in a standard normal distribution with a mean of zero and standard deviation of 1.
2 | For any standard unit or z score distance from the mean, we can calculate the corresponding area under the curve, or (in what is the same thing), the proportion of cases lying within that area.
3 | For any proportion of cases above or below the mean (and thus proportion of the area under the normal curve), we can calculate the corresponding z score.

How do we standardise a normal distribution? Let us say we have a group of 200 people and we know that their ages are normally distributed with a mean age of 40 and a standard deviation of 10. Because the distribution is normal, even without standardizing it, we would know that 137 of them will be between the ages of 30 and 50 - that is within one standard deviation above or below the mean.)

How do we know this? Recall that we know that the proportion of the area under the curve which lies between the mean and a point one deviation above it is 0.03413. Thus we will find a proportion of 0.3413 within one standard deviation above the mean and 0.3413 below it.

The proportion of people that will have ages within one standard deviation of the mean thus equals (0.3413 + 0.3413 =) 0.6826

The number of people thus equals (0.6826 * 200 = ) 136.52, which we can round to 137.

Using the same logic, we would also know that 9 people (2 * 0.023 * 200 = 9.2) would have ages below twenty and above 60 - that is beyond two standard deviations from the mean).

However, if we wished to calculate the answers to more specific questions (e.g. how many are aged under 35 or what is the span of ages of the middle 25% of this group) we need to standardize and then refer to z scores.

We standardize the value for any observation in the distribution by subtracting the mean value from it and dividing the result by the standard deviation. This converts that value into a z-score, by converting all the values for the original distribution into a distribution with a mean of 0 and standard deviation of one.

Once we have the z score, we can use statistical tables to find the corresponding proportion of cases. A z score can be defined as follows:

Let’s run through some examples. Remember, we have a group of 200 people and we know that their ages are normally distributed with a mean age of 40 and a standard deviation of 10.

Example 1 – How many people are aged between 40 and 50?

Given our mean of 40 and a standard deviation of 10, those aged 50 lie one standard deviation above the mean, so here z = 1:

The area under the curve corresponding to the distance on the X axis from the mean to z = 1 is 0.3413.

Therefore (0.3413 * 200 =) 68 people will be aged between 40 and 50

Example 2 – how many people are aged between 34 and 40?

The z score for age 34 would be (34-40)/10 = -0.6

The area under the curve corresponding to the distance on the X axis from the mean to z = -0.6 is 0.2257

Again we find this value in the standard normal tables.

Therefore (0.2257 * 200 = ) 45 people will be aged between 34 and 40

Example 3 - How many people would be aged over 64?

Z = (64-40) /10 = 2.4

The area under the curve corresponding to the distance on the X axis from the mean to z = 2.4 is 0.492 (from standard normal tables).

Therefore (0.492 * 200 =  ) 98 people will be aged between 40 and 64.

Since we know that half the group (i.e. 100 people) will be older than the mean, the (100 – 98 = ) 2 people will be aged over 64.

Example 4 - What is the age range of the youngest 5% of respondents?

Remember here that 50% of all people fall under the curve below the mean. So we need to find the z value for a point on the x axis that would correspond to 45% of the area below the mean. That will allow us to work out what ages remain even further below the mean – i.e. the youngest 5%.

From statistical tables, the z score corresponding to 45% of the area under the curve from the mean to a point below it is -1.645.

If Z equals -1.645, and our original standard deviation was 10, to convert the z score into our original units we multiply it by the standard deviation (= -16.45) and then add this to the mean: 40 + (-16.45) = 24.5.

Therefore the youngest 5% will be aged 24 or less.

When we undertake these calculations, we are always interested in knowing the distance relative to the mean and locating the precise position on the normal curve of the area we wish to measure. That is to say, if we have two points either side of the mean, we need to calculate the distance to the mean from each point and add the two corresponding areas. If we have two points on the same side (either above or below) of the mean, we need to subtract the area corresponding to the distance to the mean of the nearest point, from the area corresponding to the distance to the mean of the further point.