The scores on a personality test are normally distributed with a mean of 250 and a standard deviation of 30. What proportion of people taking the test can be expected to score above 229?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
-0.7
To find the proportion of people who can be expected to score above 229 on the personality test, we can use the concept of the standard normal distribution.
Step 1: Calculate the z-score:
The z-score measures the number of standard deviations a particular value is from the mean. It is calculated using the formula:
z = (X - μ) / σ
Where:
X = the value we want to find the proportion for (229 in this case)
μ = mean of the distribution (250 in this case)
σ = standard deviation of the distribution (30 in this case)
Substituting the values, we have:
z = (229 - 250) / 30 = -0.7
Step 2: Determine the proportion using a z-table:
A z-table gives the area under the standard normal distribution curve for different z-scores. We need to find the proportion of values above the z-score of -0.7.
Looking up the z-score of -0.7 in the z-table, we find that the proportion is 0.7580. This proportion represents the area to the left of the z-score; therefore, we subtract it from 1 to get the proportion above the z-score.
Proportion above z-score = 1 - 0.7580 = 0.2420
So, approximately 24.20% of people taking the test can be expected to score above 229.