Questions: Section 6.2: Applications of Normal Distributions 13th Edition Section 6.3: Applications of Normal Distributions 12th Edition Ex: Assume the adults have IQ scores that are normally distributed with a mean of 115 and a standard deviation of 20. Find the probability that a randomly selected adult has an IQ greater than 130.

Solution Steps

To find the Z-score for an IQ of 130, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ z = \frac{130 - 115}{20} = 0.75 \]

Thus, the Z-score for an IQ of 130 is \( z = 0.75 \).

We need to find the probability that a randomly selected adult has an IQ greater than 130. This can be expressed as:

\[ P(X > 130) = P(Z > 0.75) \]

Using the cumulative distribution function \( \Phi \), we can express this as:

\[ P(X > 130) = \Phi(\infty) - \Phi(0.75) \]

Since \( \Phi(\infty) = 1 \), we have:

\[ P(X > 130) = 1 - \Phi(0.75) \]

From standard normal distribution tables or calculations, we find:

\[ \Phi(0.75) \approx 0.7734 \]

Thus, the probability becomes:

\[ P(X > 130) = 1 - 0.7734 = 0.2266 \]

Final Answer