3. Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15, and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob? Lower than Bob?
z=100
To find the proportion of the population that would score higher or lower than Bob, we need to convert his IQ score into a z-score and then refer to the standard normal distribution table.
To convert Bob's IQ score to a z-score, we use the formula:
z = (X - μ) / σ
Where:
X = Bob's IQ score (134 in this case)
μ = Mean IQ (100 in this case)
σ = Standard deviation (15 in this case)
Substituting the values into the formula:
z = (134 - 100) / 15
z = 34 / 15
z ≈ 2.27
Now, let's look up the z-score of 2.27 in the standard normal distribution table. The table provides the proportion of data below (or to the left of) a given z-score.
Looking up 2.27 in the table, we find that the proportion below this z-score is approximately 0.9884.
To find the proportion above (or to the right of) Bob's z-score, we subtract the proportion below the z-score from 1:
proportion above = 1 - 0.9884
proportion above ≈ 0.0116
Therefore, approximately 1.16% of the population would score higher than Bob (proportion above), and approximately 98.84% would score lower than Bob (proportion below).