IQ scores are normally distributed and assume that the average IQ for all Economics majors is 108 with a standard deviation of 11. What percentage of Economics majors would have an IQ of more than 118?
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 Z score.
To find the percentage of Economics majors with an IQ score of more than 118, we can use the concept of the standard normal distribution.
Step 1: Standardize the score
We need to standardize the IQ score of 118 to a standard score (also known as the z-score) using the formula:
z = (x - μ) / σ
where:
- x is the IQ score (118 in this case)
- μ is the mean (average IQ for Economics majors, 108)
- σ is the standard deviation (11)
Plugging in the values, we get:
z = (118 - 108) / 11
z = 10 / 11
z ≈ 0.9091
Step 2: Find the percentage
Once we have the z-score, we can use a standard normal distribution table (also known as a z-table) to find the corresponding percentage.
Using the z-table, we find that the percentage of data beyond a z-score of 0.9091 is approximately 0.1814.
Step 3: Convert to percentage
Since we are interested in the percentage of Economics majors with an IQ score higher than 118, we subtract the percentage we just found from 1 (to account for the area under the whole curve):
Percentage = 1 - 0.1814
Percentage ≈ 0.8186
Therefore, approximately 81.86% of Economics majors would have an IQ score of more than 118.