8 Data Analysis 2
8.4 Z-Scores and the Normal Curve
Learning Objectives
By the end of this section it is expected that you will be able to:
- Convert a data item to a z-score
- Solve applications using z-score tables
The Normal Curve
When a set of data values is normally distributed, the 68-95-99.7 Rule can be used to determine the percentage of values that lie one, two or three standard deviations from the mean. We will shift gears and explore how to determine where a specific data value lies in relation to all other values. As an example, a student who has written a college entrance exam may want to know where they placed in comparison to all other students. This section will explore how to determine this.
Consider the normal curve which is an idealized representation of a normally distributed population. The normal curve, also called a bell-shaped curve, is represented in Figure 1. The area under the curve represents 100% (or 1.00) of the data (or population) and the mean score is 0.
We have seen that the standard deviation plays an important role in the normal distribution.
Refer to Figure 2 for the visual representation of the 68 – 95 – 99.7 Rule. For a normally distributed set of data:
- Approximately 68% (68.26%) of the data items fall within one standard deviation of the mean.
- Approximately 95% (95.44%) of the data items fall within two standard deviations of the mean.
- Approximately 99.7% of the data items fall within three standard deviations of the mean.
Z–Scores
When a data set is normally distributed we can use a standardized score, called the z-score, to determine the number of standard deviations that a data value is from the mean.
Reconsider an example from the previous section. Acertain segment of the economy has a normally distributed salary, with a mean salary of $45,000 and a standard deviation of $4000. Refer to Figure 3.
With this information we are able to determine that a salary of $49,000 lies exactly one standard deviation above the mean since $45,000 + $4000 = $49, 000. In turn, using the 68-95-99.7 Rule we can determine that a salary of $49,000 is higher than 84% of the other salaries for this segment of the economy. The calculation would be 50% + (68%/2) = 84%. This calculation was possible since $49,000 was exactly one standard deviation away from the mean.
Consider a salary which does not lie exactly one, two or three standard deviations from the mean, such as $38,500. The calculation does not appear so straightforward but as it turns out we can use a z-score for situations such as this. A z-score converts a data value and standardizes it so that we are able to determine how many standard deviations a specific data value will lie above or below the mean.
Z-scores can be used in situations with a normal distribution. Consider a chemistry class with a set of test scores that is normally distributed. The average score is 76% and one student receives a score of 55%. Converting the 55% to a z-score will provide the st