8.055 Empirical rule with z-scores

As we discovered previously, the shape or spread of a normal distribution is affected by the standard deviation, which varies depending on the data set. To directly compare multiple normally distributed data sets, we need a common unit of measurement. In statistics involving the normal distribution, we use the number of standard deviations away from the mean as a standardised unit of measurement called a $z$z-score.

What is a $z$z-score?

A $z$z-score is a value that shows how many standard deviations a score is above or below the mean. Mathematically, it is the ratio of the distance a score is above or below the mean to the standard deviation. In other words, it's indicative of how an individual's score deviates from the population mean, as shown in the picture below.

• A positive $z$z-score indicates the score was above the mean.
• A $z$z-score of zero indicates the score was equal to the mean.
• A negative $z$z-score indicates the score was below the mean.

Remember a "population" just means every member of a group is counted. It doesn't have to be people. For example, it may be the Australian population, all the students in Year 10 in a school or all the chickens on a farm.

What are $z$z-scores used for?

$z$z-scores are used to compare various normally distributed data sets. For example, let's say Sam got $75$75 on his Biology exam and $80$80 on his Chemistry exam. At first glance, it would seem that he did better on his Chemistry exam. However, then he received this info from his teacher:

What does this mean for Sam?

To really understand how Sam performed in his exams, we need to calculate his $z$z-score for both of them. Let's do that now!

Calculating $z$z-scores

There is a formula was can use for calculating the $z$z-scores of a population.

So let's start by calculating Sam's $z$z-score for Biology:

This means he is $1.67$1.67 standard deviations above the mean in Biology.

Now let's calculate his $z$z-score for Chemistry:

This means he is $0.83$0.83 standard deviations above the mean in Chemistry.

His $z$z-score for Biology was larger than the $z$z-score for Chemistry, indicating that he performed better in Biology, relative to the class. Furthermore we can say that he was in the top $20.3%$20.3% of the class for Chemistry, but in the top $4.7%$4.7% for Biology. We'll take a look at how to calculate these probabilities in the sections that follow.

Worked example

example 1

A machine produces components whose weights are normally distributed. The intention is for this machine to be calibrated to produce components whose weights have a mean of $600$600 g, with only $0.95%$0.95% of components having a weight less than $590$590 g. Determine the standard deviation of the calibrated machine.

Think: When we do not know the mean or the standard deviation or both of these things for a general normal distribution, we can use the link between the general normal distribution and the standard normal distribution to help us calculate these.

The link between the two is the standardising formula: $z=\frac{x-\mu}{\sigma}$z=x−μσ​

In this situation we have $x=590$x=590 and $\mu=600$μ=600. Visually we can represent the information like this:

If we duplicate this diagram, but this time for the standard normal distribution, we can see that we can then use the inverse normal function of the calculator to calculate the associated $z$z score.

Do: Using the inverse normal function of the calculator we find $z_1=-2.3455$z1​=−2.3455. We can now put all our information into the standardising formula. Where the score of interest $x=590$x=590, the population mean is $\mu=600$μ=600 and the corresponding $z$z-score is $-2.3455$−2.3455.

The empirical rule in terms of $z$z-scores

We were introduced to the empirical rule in the previous lesson and we can now express the rule in terms of $z$z-scores.

Practice questions

question 1

question 2