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?

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?
Transcript text: IQ scores form a normal distribution with $\mu=100$ and $\sigma=15$. Individuals with IQs above 140 are classified in the genius category. What proportion of the population consists of geniuses?
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Solution

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Solution Steps

Step 1: Calculate the Z-Score

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

z=Xμσ z = \frac{X - \mu}{\sigma}

Substituting the values:

z=14010015=40152.6667 z = \frac{140 - 100}{15} = \frac{40}{15} \approx 2.6667

Thus, the Z-score for IQ 140 is z2.6667 z \approx 2.6667 .

Step 2: Calculate the Probability of Being a Genius

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

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

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

From the calculations, we find:

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

Final Answer

The proportion of the population classified as geniuses is

0.0038 \boxed{0.0038}

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