5.1.2 The z-Value

Mariana Gatzeva

Chapter 5 The Normal Distribution and Some Basics of Probability

5.1.2 The z-Value

In the previous section you discovered that we can “orient” ourselves about where a specific value lies along the normal distribution in relation to the average by means of the standard deviation. In Example 5.1 we saw that 68 percent of students’ test scores were between 55 and 75 (i.e., between -1 and +1 standard deviations from the mean), 95 percent of scores were between approximately 45 and 85 (i.e., between about -2 and +2 standard deviations from the mean), and that 99 percent of scores were between approximately 35 and 95 (i.e., between -3 and +3 standard deviations from the mean). Thus, if your score was, say, 60, you would know that it was below the mean, but within 1 standard deviation away, which wouldn’t be as bad as, say, had you scored 40, which is more than two standard deviations away from the mean.

Hmm, do we really need standard deviations to tell us that a test score of 40 is bad news, you ask. Everyone knows that.

In absolute terms, sure, a score of 40 (out of 100) would be considered a failing one. In relative terms, however — which is also known as grading on a curve — a score of 40 doesn’t tell you anything, unless you know the mean and the standard deviation.

To better illustrate this, imagine another set of test scores, and that on that test you get a score of 80. In absolute terms, a score of 80 (out of 100) would be quite good. What about in relative terms? Can you think of a situation where a score of 80 would be considered worse than a score of 40?

What if I told you that the mean in the first case (when we imagine you scored 40) was 35 with a standard deviation of 5, while the mean in the second case (when we imagined you scored 80) was 90 with a standard deviation of 2? (You might find it easier to see the point if you grab a pen and paper and simply draw a line with the mean in the middle, then add and subtract that many standard deviations away from it in each direction, above and below.)

A score of 40 (i.e., $35+5=40$ ) is 1 standard deviation above the mean of that test. A score of 80 (i.e., $90-5(2)=80$ ) is 5 standard deviations below the mean of that other test. In fact, 80 is well below the even 3 standard deviations away from the mean where 99 percent of scores are; it’s at the very far end of the left “tail” of the distribution, likely an outlier.

It turns out that the second test we imagined was so easy, scoring 80 on it was too low given how easy it was. On the other hand, scoring 40 on the first test we imagined was quite good given how hard it was.

This mental exercise shows you that expressing values in terms of standard deviations has its merits, as it puts the values into perspective — which allows us to make comparisons. A score/value in and of itself doesn’t tell you anything — not unless you know where it falls in relation to the mean and how far away it is. Now only if there was a way to express any value in terms of standard deviations without having to always calculate 1 standard deviation away, 2 standard deviations away, 3 standard deviations away from the mean (or to have to resort to pen and paper)…

Guess what? There is! Expressing a value in terms of standard deviations is a process aptly called standardization (as it produces scores that have a uniform, standard meaning allowing comparison) and the standardized values are called z-values (or z-scores). We standardize values by expressing the distance of the value from the mean in standard deviations, i.e.:

$\frac{\textrm{original score} - \textrm{mean}}{\textrm{standard deviation}}=\textrm{z-value}$

Or, in proper notation, where we denote the mean by μ^[1], the small-case Greek letter for m (from mean):

$\frac{x_i - \mu}{\sigma}=z$

Following this formula, a score of 40 when the mean is 35 and the standard deviation is 5 (i.e., when μ=35 and σ=5) has a z-score of

$\frac{x_i - \mu}{\sigma}=\frac{40-35}{5}=\frac{5}{5}=1=z$

and a score of 80 when the mean is 90 and the standard deviation is 2 (i.e., when μ=90 and σ=2) has a z-score of

$\frac{x_i - \mu}{\sigma}=\frac{80-90}{2}=\frac{-10}{2}=-5=z$

Thus, we formally found what we already knew from before: that in the former case, the score of 40 was 1 standard deviation above the mean (i.e., its $z=1$ ) and the score of 80 was 5 standard deviations below the mean (i.e., its $z=-5$ ). If this seems repetitive — after all, we reached the same conclusion without any fancy formulas — that’s only because I chose easily calculatable numbers to illustrate my point more easily. Perhaps an example with less “easy” numbers will convince you of the formula’s worth.

Example 5.2 Average Monthly Rent for a Two-Bedroom Apartment in Vancouver

The Vancouver Sun recently reported that the average monthly rent of a two-bedroom apartment in Vancouver, BC was $2,915, at the time of writing the highest in all Canada. (REFERENCE https://vancouversun.com/news/local-news/vancouver-two-bedroom-apartments-now-cost-close-to-3000-report) While the standard deviation was not reported, for the purposes of this exercise we can imagine it as $150.

What is the z-score of a family which pays $2,630 per month for their two-bedroom condo? How about the z-score of someone who pays $3,450 for theirs?

Of course, we could grab a pen and paper and draw the normal distribution demarcating where 1, 2, and 3 standard deviations away from the mean fall in order to see where the two listed rents are relative to the demarcations. However, using the z-score formula makes for a faster (and a more precise) answer.

In the first case, we have:

$\frac{x_i - \mu}{\sigma}=\frac{2630-2915}{150}=\frac{-285}{150}=-1.9=z$

In the second case, we have:

$\frac{x_i - \mu}{\sigma}=\frac{3450-2915}{150}=\frac{535}{150}=-5=3.6=z$

That is, the first family’s monthly rent of $2,630 is below the average but not that unusual: with a z-score of -1.9, it falls within 2 standard deviations away from the mean, which is within what 95 percent of renters in Vancouver pay for their two-bedroom apartments.

On the other hand, the second person’s rent of $3,450 is quite high: with its z-score of 3.6, it falls beyond 3 standard deviations away from the mean, i.e., it’s higher than what 99 of people pay monthly for a two-bedroom apartment.

Again, we see the use of standardization and z-scores, as it allows us to put values into perspective.

Now is your turn to try.

Do It! 5.1 Comparing Average Monthly Rent for a One-Bedroom Apartment in Vancouver, Toronto, and Montreal

According to the National Rent Rankings monthly report for July 2019 by Rentals.ca (REFERENCE https://rentals.ca/national-rent-report), the average monthly rent for a one-bedroom apartment was $2,028 in Vancouver, BC, $2,259 in Toronto, ON, and $1,231 in Montreal, QC. Assume the standard deviations are $140 in Vancouver, $180 in Toronto, and $125 in Montreal.

Using z-values, compare and analyze where in the distribution a rent of $1,950 will put a Vancouverite, a Torontonian, and a Montrealer who all pay the same rent but in different cities.

_{(Answer: Vancouverite’s z=-0.6, Torontonian’s z=-1.7, Montrealer’s z=5.8.)}

The Greek letter μ is pronounced as "MYU". The difference between using $\overline{x}$ and μ and the reason we use the latter here will be explained in Chapter 6. ↵

5.1.2 The z-Value

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