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9.055 Empirical rule with z-scores

Lesson

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.
Careful!

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.

Formula for calculating $z$z-scores from a population

$z=\frac{x-\mu}{\sigma}$z=xμσ

This means:

$\text{standardised z score}=\frac{\text{raw score}-\text{population mean score}}{\text{standard deviation}}$standardised z score=raw scorepopulation 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 $=$= $\frac{x-\mu}{\sigma}$xμσ
  $=$= $\frac{75-70}{3}$75703
  $=$= $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 $=$= $\frac{x-\mu}{\sigma}$xμσ
  $=$= $\frac{80-75}{6}$80756
  $=$= $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.

Worked example

example 1

A machine produces components whose weights are normally distributed. The intention is for this machine to be calibrated to produce components whose weights have a mean of $600$600 g, with only $0.95%$0.95% of components having a weight less than $590$590 g. Determine the standard deviation of the calibrated machine.

Think: When we do not know the mean or the standard deviation or both of these things for a general normal distribution, we can use the link between the general normal distribution and the standard normal distribution to help us calculate these.

The link between the two is the standardising formula: $z=\frac{x-\mu}{\sigma}$z=xμσ

In this situation we have $x=590$x=590 and $\mu=600$μ=600. Visually we can represent the information like this:

If we duplicate this diagram, but this time for the standard normal distribution, we can see that we can then use the inverse normal function of the calculator to calculate the associated $z$z score.

Do: Using the inverse normal function of the calculator we find $z_1=-2.3455$z1=2.3455. We can now put all our information into the standardising formula. Where the score of interest $x=590$x=590, the population mean is $\mu=600$μ=600 and the corresponding $z$z-score is $-2.3455$2.3455.

$z$z $=$= $\frac{x-\mu}{\sigma}$xμσ

The formula for a $z$z-score

$-2.3455$2.3455 $=$= $\frac{590-600}{\sigma}$590600σ

Substitute in known values

$\sigma$σ $=$= $\frac{-10}{-2.3455}$102.3455

Rearrange for $\sigma$σ

  $=$= $4.26$4.26 g ($2$2 d.p)

 

 

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.

The empirical rule
  • $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.

  1. If Paul received a score of $102$102 in the test, what was his $z$z-score correct to two decimal places?

  2. 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.

  1. Find Marge’s $z$z-score in Mathematics.

  2. Find Marge’s $z$z score in Philosophy.

  3. Which exam did Marge do better in, compared to the rest of her class?

    Philosophy

    A

    Mathematics

    B

 

Outcomes

4.4.1.4

understand standardised normal variables (z-values, z-scores) and use these to compare samples

4.4.2.2

recognise features of the graph of the probability density function of the normal distribution with mean 𝜇 and standard deviation 𝜎 and the use of the standard normal distribution

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