Final answer:
To calculate the proportion of people with IQ scores above 100, the Z-score formula is used and then the standard normal distribution table to find the necessary probability. The details of the calculation are not provided; the values in the example are hypothetical.
Step-by-step explanation:
The question is focused on the normal distribution of IQ scores and involves finding the proportion of individuals with scores above a certain value. Since IQ scores are normally distributed with a mean (μ) and standard deviation (σ), we can use the Z-score formula to find the proportion of people aged 20 to 34 who have IQ scores above 100:
Z = (X - μ) / σ
Where:
- X is the score in question (100)
- μ is the mean IQ score (in this case, 110)
- σ is the standard deviation (in this case, 25)
Once we have the Z-score, we can use the standard normal distribution table to find the proportion of the population above this score. For example, if the Z-score is calculated to be -0.4, we would look up the probability associated with Z = -0.4, which would give us the proportion of individuals scoring below 100. Subtracting this from 1 gives us the proportion scoring above 100.
However, the specifics of the calculation are not provided in the question, and the values used in the example may not represent the actual Z-score for an IQ of 100 given the mean and standard deviation outlined in the questio