Introduction
Standard deviation is used as a measure of how widely dispersed a set of measurements are about their mean value. In educational contexts, standard deviation provides a way of comparing test scores from different subject areas. Different subjects will have different mean test scores and the ranges of results obtained by the student will differ from subject to subject. To overcome this difficulty, test scores are often presented in terms of standard deviation units measured from the mean of each subject.
We can formalise such comparisons using the statistic known as a z-score. This is often used in relation to the normal probability distribution , though it can be used with other distributions. The process of obtaining a z-score is called standardisation and for this reasonz-scores are sometimes referred to as standardised scores.
To obtain a z-score, it is necessary to know the mean \mu and the standard deviation \sigma of a complete population. (Notice that we use different symbols for mean and standard deviation of a population than we do for a sample.)
Ideas
What is a z-score?
A z-score is a value that shows how many standard deviations a score is above or below the mean. It is indicative of how an individual's score deviates from the population mean, without actually referring to the particular value of the mean.
A positive z-score indicates that the score was above the mean. For example, a z-score of 1.43 indicates the score is 1.43 standard deviations above the mean.
A z-score of 0 indicates the score was equal to the mean.
A negativez-score indicates that the score was below the mean. For example, a z-score of -0.56 indicates the score is 0.56 standard deviations below the mean.)
Remember that z-scores can only be defined if the population mean \mu and standard deviation \sigma are known.
We cannot find z-scores using only a sample standard deviation and mean
z-scores are used to compare two or more different normally distributed data sets. For example, suppose that Sam scored 75 on his biology exam and 80 on his chemistry exam. On first glance, it would seem that he did better on his chemistry exam. He also received this information from his teacher, however:
| Mean | Standard Deviation | |
|---|---|---|
| \text{Chemistry} | 75 | 6 |
| \text{Biology} | 70 | 3 |
What does this mean for Sam?
To really understand how Sam performed in his exams, we can calculate his z-score for both exams. Let's do that now.
There is a formula was can use for calculating the z-scores of a population.
For a set of data with a population mean \mu and standard deviation\sigma, the z-score corresponding to a value x is: z=(x-\mu)\sigma.
Let's now calculate Sam's z-scores, starting with biology:
| \displaystyle z | \displaystyle = | \displaystyle \dfrac{x-\mu}{\sigma} | Write the formula |
| \displaystyle = | \displaystyle \dfrac{75-70}{3} | Substitute the values | |
| \displaystyle = | \displaystyle 1.67 | Evaluate and round to 2 decimal places |
So Sam is 1.67 standard deviations above the mean in biology.
Now let's calculate his z-score for chemistry:
| \displaystyle z | \displaystyle = | \displaystyle \dfrac{x-\mu}{\sigma} | Write the formula |
| \displaystyle = | \displaystyle \dfrac{80-75}{6} | Substitute the values | |
| \displaystyle = | \displaystyle 0.83 | Evaluate and round to 2 decimal places |
This means he is 0.83 standard deviations above the mean in chemistry.
So Sam's z-score for biology was nearly twice what it was for chemistry. The larger z-score for Biology than Chemistry, indicates that he performed better in Biology, relative to the class.
Examples
Example 1
Eileen scored 95.8 in her English exam, in which the mean score and standard deviation were 87 and 4 respectively. She also scored 53 in her Biology exam, in which the mean score was 42 and the standard deviation was 2.
Find Eileen's z-score in English. Give your answer to one decimal place if needed.
Find Eileen’s z-score in Biology. Give your answer to one decimal place if needed.
Which exam did Eileen do better in?
Example 2
In an entrance exam, applicants completed two papers.
On average, students performed better in Paper 1, but their marks were less spread out in Paper 2. The standard deviation of Paper 2 could be:
| Mean | Standard Deviation | |
|---|---|---|
| \text{Paper }1 | 77 | 13 |
| \text{Paper }2 | 57 | ?? |
Calculating z-scores: