6.5 – Finding Proportions or Probabilities from an Individual Score (X)
In this section, we are going to start with skills we learned in the previous section:
- draw a normal distribution
- locate a z-score
- shade the distribution
- identify the type of shaded area (e.g., body, tail, slice)
- lookup the z-score in the Unit Normal Table
- identify the appropriate proportion (e.g., body, tail, slice)
And then, we are going to add:
- use z-score formulas from the previous chapter.
This combination will allow us to answer more typical questions from the real world of research and practice. To do this, we just need to add a couple more steps to what we used in the previous section:
How to find the proportion/probability for a given range of individual scores
- Draw a picture of the normal distribution curve.
- Draw a vertical line in the middle of the distribution.
- Label this middle line with the Mean score ([latex]\mu[/latex] or M)
- Locate the individual score (X) on the curve (approximately) and label it with a vertical line.
- Shade the appropriate part of the distribution (e.g., above the score, below the score, between the mean and the score).
- Look at the shaded chunk and decide if it is a body, a tail, or a slice.
- Calculate the z-score for the individual score using the formula: [latex]z=\frac{X-\mu}{\sigma}[/latex]
- Create a second x-axis below the individual-score axis, label the middle vertical line as z = 0, and label the vertical line for the real-world score as the z-score you calculated from the previous step.
- Look up your z-score in the first column of the Unit Normal Table.
- Then go across that row to the appropriate column (Body, Tail, or Slice) from step 6 to get your answer.
Let’s say we want to answer the question:
Being “normally distributed,” tells us that if you were to create a graph of a population’s scores on an IQ test, it would form a normal distribution which means that most people tend to score near the mean. As you get farther from the mean, in either direction, you get fewer and fewer people. It also means that we will be able to use the Unit Normal Table to look up proportions. We also know that the average score on an IQ test is 100, and people in the population deviate from that mean by an average of 15 points.
Note: In this class, questions in homework and exams will always provide you with enough information to answer the question. Thus, you do not need to memorize the population means and standard deviations for various standardized tests. You can also assume that scores in our questions are normally distributed.
Outside of this class, researchers and practitioners cannot always assume the distribution shapes and have to examine the shapes of their distributions.
So now we can draw a normal distribution curve:
Next, we draw a vertical line in the middle of the distribution and label it with the mean for IQ tests: μ = 100:
Now, we can examine this distribution and see that many people tend to score near 100, and as you get farther from the mean of 100, in either direction, you get fewer and fewer people.
Our next step is to locate the score of X = 120 on the curve. Since the score of 120 is above the mean of 100, we can put a vertical line to the right of the mean. It doesn’t matter exactly how far we put the line above the mean because only using it to determine what proportion column (body, tail, or slice) to use.
To find the probability of people who score higher than 120, we want to know the proportion (remember, proportion = probability) of the normal distribution that lies above (or is greater than) the score of X = 120. Thus, we shade the part of the distribution that includes IQ scores higher than 120 (or to the right of the vertical line at X = 120). In other words, we want the following proportion: P (X > 120).
Looking at that picture, we see that the shaded area is a “tail,” which tells us that we will find the proportion/probability in the Tail column of the Unit Normal Table.
However, the cutoff point for that tail is not a z-score but is instead an individual score (X). Thus, we need to calculate the z-score for the individual score of X = 120. To do that, we simply use the z-score formula that solves for z:
[latex]z = \frac{X - \mu}{\sigma}[/latex]
When we plug in everything we were given:
- X = 120
- μ = 100
- σ = 15
We find that the IQ score of 120 corresponds to a z-score of +1.[latex]\overline{3333}[/latex]:
[latex]z = \frac{X - \mu}{\sigma} = \frac {120 - 100}{15} = \frac{20}{15} = +1.\overline{3333}[/latex]
Next, we go to the Unit Normal Table and look for our z-score of z = +1.[latex]\overline{3333}[/latex] in the z-score column. An important thing to note is that z-scores in the Unit Normal Table only use 2 decimal places, so we need to round our z-score of z = +1.[latex]\overline{3333}[/latex] to z = 1.33. Then we look across that row until we come to the Tail column, and we find a proportion of P (z > +1.33) = 0.0918, or 9.18% of the normal distribution lies above a z-score of z = +1.33.
Because that z-score of +1.33 corresponds to an IQ score of 120, we know the proportion of people who score above 120 on an IQ test is 0.0918, or 9.18% of people score above 120 on an IQ test.
And then, because proportion equals probability, we now know that the probability of an individual having an IQ score greater than 120 is 0.0918.
Another way of putting it is that if you were to randomly grab someone and have them take a standardized IQ test, there is a 9.18% chance that their score is above 120.
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