Questions: IQ scores form a normal distribution with μ=100 and σ=15. Individuals with IQs above 140 are classified in the genius category. What proportion of the population consists of geniuses?

Solution Steps

To determine the Z-score for an IQ of 140, we use the formula:

$$z = \frac{X - \mu}{\sigma}$$

Substituting the values:

$$z = \frac{140 - 100}{15} = \frac{40}{15} \approx 2.6667$$

Thus, the Z-score for IQ 140 is $z \approx 2.6667$.

Next, we need to find the probability that an individual has an IQ greater than 140. This is represented mathematically as:

$$P(X > 140) = P(Z > 2.6667) = \Phi(\infty) - \Phi(2.6667)$$

Where $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution. Since $\Phi(\infty) = 1$, we have:

$$P(X > 140) = 1 - \Phi(2.6667)$$

From the calculations, we find:

$$P(X > 140) \approx 0.0038$$

Final Answer