How to Calculate a \(z\)-score
A \(z\)-score measures how many standard deviations a data point is from the mean, allowing for direct comparisons between different datasets. This helps identify unusually high or low values in contexts such as test scores, physical measurements, and financial data. This section explains \(z\)-scores, their formulas, and their applications using real-world examples.
\(z\)-scores
What is a Z-Score?
A \(z\)-score, also called a standard score, measures how many standard deviations a data point is above or below the mean.
Example 1
We have a normal distribution of heights with mean 175 cm and standard deviation 7 cm.
| Height (cm) | Empirical Rule Calculation | \(z\)-score |
|---|---|---|
| 154 | \[ \begin{align*} \mu + (-3 \times \sigma) &= 175 + (-3 \times 7) \\ &= 175 - 21 \\ &= 154 \end{align*} \] | -3 |
| 161 | \[ \begin{align*} \mu + (-2 \times \sigma) &= 175 + (-2 \times 7) \\ &= 175 - 14 \\ &= 161 \end{align*} \] | -2 |
| 168 | \[ \begin{align*} \mu + (-1 \times \sigma) &= 175 + (-1 \times 7) \\ &= 175 - 7 \\ &= 168 \end{align*} \] | -1 |
| 175 | \[ \begin{align*} \mu + (0 \times \sigma) &= 175 + (0 \times 7) \\ &= 175 + 0 \\ &= 175 \end{align*} \] | 0 |
| 182 | \[ \begin{align*} \mu + (1 \times \sigma) &= 175 + (1 \times 7) \\ &= 175 + 7 \\ &= 182 \end{align*} \] | 1 |
| 189 | \[ \begin{align*} \mu + (2 \times \sigma) &= 175 + (2 \times 7) \\ &= 175 + 14 \\ &= 189 \end{align*} \] | 2 |
| 196 | \[ \begin{align*} \mu + (3 \times \sigma) &= 175 + (3 \times 7) \\ &= 175 + 21 \\ &= 196 \end{align*} \] | 3 |
Notice that the number we multiply \(\sigma\) by always ends up being the value of
the \(z\)-score. Therefore, a \(z\)-score counts how many standard deviations
a data point is above or below the mean.
The image below shows both the
\(z\)-score and the values from the empirical rule for this distribution.
How to Calculate a \(z\)-score
Formula for a Z-Score
The formula for a z-score depends on the context, but all follow the general structure:
\[ z = \dfrac{\text{{data}} - \text{{mean}}}{\text{standard deviation}} \]
Z-Score for Different Contexts
- Individuals in a Population:
If an individual is randomly chosen from a population, the z-score is calculated using the formula:
\[ z = \dfrac{x - \mu}{\sigma} \]
- Individuals in a Sample:
If an individual is randomly chosen from a sample, the z-score is calculated using the formula:
\[ z = \dfrac{x - \bar{{x}}}{{s}} \]
Important Features of \(z\)-Scores
- \(z\)-scores measure the distance from the mean but have no units.
- A positive z-score means the value is above the mean.
- A negative z-score means the value is below the mean.
- A data point is significantly low if \( z < -2 \).
- A data point is significantly high if \( z > 2 \).
- \(z\)-scores allow comparisons of relative standing between different datasets.
Example 2
The height of men and the height of women both follow a bell-shaped distribution. The mean and standard deviation for each group are given below. Who is relatively taller: A man or a woman who is 5’8”?
| Group | Mean Height (in) | Standard Deviation (in) |
|---|---|---|
| Men | 69.9 | 3.0 |
| Women | 64.3 | 2.6 |
Solution
- Step 1: Compute the Z-Scores
Convert 5'8" (or 68 inches) into z-scores for both men and women.
\[ z_{\text{{men}}} = \dfrac{68 - 69.9}{3.0} = \dfrac{-1.9}{3.0} \approx -0.63 \]
\[ z_{\text{{women}}} = \dfrac{68 - 64.3}{2.6} = \dfrac{3.7}{2.6} \approx 1.42 \]
- Step 2: Interpret the Results
The z-score for men is negative, meaning 5’8” is below average for men. The z-score for women is positive, meaning 5’8” is above average for women. Since the z-score for women is higher in magnitude than the z-score for men, a 5’8” woman is relatively taller than a 5’8” man.
$$\tag*{\(\blacksquare\)}$$
Example 3
The ACT has an average score of 20.8 with a standard deviation of 5.8. The SAT has an average score of 1500 with a standard deviation of 300. Who performed worse relative to their peers: A student who scored 1075 on the SAT or a student who scored 15 on the ACT?
Solution
- Step 1: Compute the Z-Scores
Convert both scores into z-scores.
\[ z_{\text{{SAT}}} = \dfrac{1075 - 1500}{{300}} = \dfrac{-425}{{300}} \approx -1.42 \]
\[ z_{\text{{ACT}}} = \dfrac{15 - 20.8}{5.8} = \dfrac{-5.8}{5.8} = -1.00 \]
- Step 2: Interpret the Results
The SAT student has a z-score of -1.42, meaning they scored 1.42 standard deviations below the mean. The ACT student has a z-score of -1.00, meaning they scored 1.00 standard deviation below the mean. Since -1.42 is lower than -1.00, the SAT student performed worse relative to their peers.
$$\tag*{\(\blacksquare\)}$$
Extreme Values
What Does it Mean for One Value to be More Extreme than Another?
Let \(x\) and \(y\) be two values from one or more normal distributions. The value \(x\) is more extreme than \(y\) if its distance from the mean, measured in standard deviations, is greater.
How Do We Compute This?
To determine which value is more extreme, recall that absolute value measures distance, and standard deviation quantifies how far a data point is from the mean. Therefore, taking the absolute value of the \(z\)-scores will tell us which value is more extreme. In other words, the larger the absolute value of the \(z\)-score, the more extreme the data point is.
The formal procedure is:
- Step 1: Compute the \( z \)-score for both data points.
- Step 2: Take the absolute value of both \( z \)-scores.
- Step 3: The value with the largest absolute \( z \)-score is the more extreme value.
Example 4
The weight of men and women in their 30s follows a normal distribution. The average weight for men is 208 lbs with a standard deviation of 25 lbs, while the average weight for women is 175 lbs with a standard deviation of 25 lbs. Determine which weight is more extreme: a man who weighs 145 lbs or a woman who weighs 224 lbs.
Solution
To determine which weight is more extreme, we calculate the absolute value of the z-score for each case:
- Man who weighs 145 lbs
\[ z = \left| \dfrac{155 - 208}{{25}} \right| = \left| \dfrac{-53}{{25}} \right| = \left| -2.12 \right| = 2.12 \]
- Woman who weighs 224 lbs
\[ z = \left| \dfrac{224 - 175}{{20}} \right| = \left| \dfrac{{49}}{{20}} \right| = \left| 2.45 \right| = 2.45 \]
Since \(2.45\) is the larger value, the woman who weights 224lbs has the more extreme weight.
Man \(z=-2.12\)
Woman \(z=2.45\)
$$\tag*{\(\blacksquare\)}$$
Conclusion
\(z\)-scores provide a standardized way to compare data points within and across distributions by measuring how many standard deviations a value is from the mean. This concept is essential for identifying extreme values, assessing relative standing, and making meaningful comparisons between datasets. In future topics, we will expand on z-scores by exploring the standard normal distribution, probabilities associated with z-values, and confidence intervals, and hypothesis testing. Understanding z-scores now will help build a strong foundation for these future topics!