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. Just like in every branch of mathematics, 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.
What's really important to remember is that $z$z-scores can only be defined if the population parameters (ie. the mean and standard deviation of the population) are known.
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:
| Mean | S. D. | |
|---|---|---|
| Chemistry | $75$75 | $6$6 |
| Biology | $70$70 | $3$3 |
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.
$z=\frac{x-\mu}{\sigma}$z=x−μσ
This means:
$\text{standardised z score}=\frac{\text{raw score}-\text{population mean score}}{\text{standard deviation}}$standardised z score=raw score−population mean scorestandard deviation
Note: for sample (not population) data, we use $\overline{x}$x for the mean and $s$s for the standard deviation to estimate the population parameters.
So let's start by calculating Sam's $z$z-score for Biology:
| $z$z | $=$= | $\frac{x-\mu}{\sigma}$x−μσ |
| $=$= | $\frac{75-70}{3}$75−703 | |
| $=$= | $1.6666$1.6666... | |
| $z$z | $=$= | $1.67$1.67 (to 2 d.p.) |
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:
| $z$z | $=$= | $\frac{x-\mu}{\sigma}$x−μσ |
| $=$= | $\frac{80-75}{6}$80−756 | |
| $=$= | $0.8333$0.8333... | |
| $=$= | $0.83$0.83 (to 2 d.p.) |
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.
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.
- $68%$68% of scores have a $z$z-score between $-1$−1 and $1$1.
- $95%$95% of scores have a $z$z-score between $-2$−2 and $2$2.
- $99.7%$99.7% of scores have a $z$z-score between $-3$−3 and $3$3.
Remember, since the normal distribution is symmetric, we can halve the interval at the mean to halve the percentage of scores.
Practice questions
question 1
A general ability test has a mean score of $100$100 and a standard deviation of $15$15.
If Paul received a score of $102$102 in the test, what was his $z$z-score correct to two decimal places?
If Georgia had a $z$z-score of $3.13$3.13, what was her score in the test, correct to the nearest integer?
question 2
Marge scored $43$43 in her Mathematics exam, in which the mean score was $49$49 and the standard deviation was $5$5. She also scored $92.2$92.2 in her Philosophy exam, in which the mean score was $98$98 and the standard deviation was $2$2.
Find Marge’s $z$z-score in Mathematics.
Find Marge’s $z$z score in Philosophy.
Which exam did Marge do better in, compared to the rest of her class?
Philosophy
AMathematics
B