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.
As mentioned above, a $z$z-score is a value that shows how many standard deviations a score is above or below the mean. In other words, it's indicative of how an individual's score deviates from the population mean, as shown in the picture below.
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.
If only the sample mean and sample standard deviation are known, we need to calculate a $t$t-statistic which we will learn about later.
$z$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. On 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 | 6 |
Biology | 70 | 3 |
What does this mean for Sam?
To really understand how Sam performed in his exams, we need to calculate his z-score for both of them. Let's do that now!
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
So let's start 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 nearly twice what it was for chemistry! We'll learn more about comparing scores later but if you think about the empirical rule, this is a really significant jump!
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?
Kathleen scored $83.4$83.4 in her Biology exam, in which the mean score and standard deviation were $81$81 and $2$2 respectively. She also scored $60$60 in her Geography exam, in which the mean score was $46$46 and the standard deviation was $4$4.
Find Kathleen’s $z$z-score in Biology. Give your answer to one decimal place if needed.
Find Kathleen’s $z$z-score in Geography. Give your answer to one decimal place if needed.
Which exam did Kathleen do better in?
Biology
Geography
The number of runs scored by Maximilian in each of his innings is listed below.
$34,33,31,33,32,32,33,31,33,33$34,33,31,33,32,32,33,31,33,33
What was his batting average correct to two decimal places?
What was his (sample) standard deviation correct to two decimal places?
What was $z$z-score of his final innings score, correct to two decimal places?
What was the $z$z-score of his highest score, correct to two decimal places?