8.04 Making comparisons between distributions

When you get your marks back for an assessment, you might be curious about the class average, or the mean mark. With this information, you can see how well you performed compared to the 'average' student. But with some additional information about the class performance, that being the standard deviation, you have a better sense of how well you did compared to the rest of your class.

In 8.03 we looked at what a $z$z-score is and how to calculate it. Recall that to calculate the $z$z-score for a value of $x$x from a data set, we first need the mean $\overline{x}$x, and the standard deviation $s$s. Then we use the formula:

$z=\frac{x-\overline{x}}{s}$z=x−xs​

A $z$z-score represents how many standard deviations above or below the mean a value is. With this in mind, $z$z-scores can be seen as a common unit of measure to compare between data sets that are approximately normally distributed, even if they have different means and standard deviations.

Exploration

Let's say that you have completed two maths tests and received a mark of $70$70 in each. In each test, the marks were approximately normally distributed, with a mean mark of $60$60 in each test.

At first glance it seems like you performed just as well in both tests, but to know the full story we also need to consider the standard deviation of each test. The standard deviation for Test $1$1 was $10$10 and the standard deviation for Test $2$2 was $15$15. This information is summarised in the table below:

Test	Mark	Mean	Standard Deviation
Test $1$1	$70$70	$60$60	$10$10
Test $2$2	$70$70	$60$60	$15$15

Let's have a look at the breakdown:

• In Test $1$1: your mark was one whole standard deviation above the mean.
• In Test $2$2: your mark was just shy of one whole standard deviation above the mean.

Test $1$1 has a lower standard deviation, so most of the students performed closer to the mean score of $60$60. This tells us that you performed better than a larger percentage of students in Test $1$1 than in Test $2$2.

Another way to think about this would be to go back to 8.02 where we looked at the empirical rule. There we saw that about $16%$16% of scores lie more than $1$1 standard deviation above the mean. So in Test $1$1, only about $16%$16% of scores would have been higher than yours, whereas in Test $2$2, more than $16%$16% of scores would have been higher than yours.

 

Using z-scores for comparison

$z$z-scores provide a numerical answer to the situation above. The greater the $z$z-score, the greater the number of standard deviations you are from the mean.

$z$z-score for Test $1$1:

$z$z	$=$=	$\frac{x-\overline{x}}{s}$x−xs​	(the $z$z-score formula)
	$=$=	$\frac{70-60}{10}$70−6010​	(substitute in the values)
	$=$=	$1$1	(simplify the expression)

$z$z-score for Test $2$2:

$z$z	$=$=	$\frac{x-\overline{x}}{s}$x−xs​	(the $z$z-score formula)
	$=$=	$\frac{70-60}{15}$70−6015​	(substitute in the values)
	$=$=	$\frac{2}{3}$23​	(simplify the expression)

As you can see, the $z$z-score for Test $2$2 is less than the $z$z-score for Test $1$1, which means that you performed better in Test $1$1 compared to other students who took the same test.

Practice questions

Example 1

example 2

example 3