Questions: Assume that adults have IQ scores that are normally distributed with a mean of 97.4 and a standard deviation of 18.8, Find the probability that a randomly selected adult has an IQ greater than 116.3. (Hint: Draw a graph.) The probability that a randomly selected adult from this group has an IQ greater than 116.3 is (Round to four decimal places as needed.)

Solution Steps

We are given that the IQ scores of adults are normally distributed with a mean (\( \mu \)) of 97.4 and a standard deviation (\( \sigma \)) of 18.8. We need to find the probability that a randomly selected adult has an IQ greater than 116.3.

To find the probability, we first calculate the Z-score for the value 116.3 using the formula:

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

Substituting the values:

\[ Z = \frac{116.3 - 97.4}{18.8} \approx 1.0053 \]

The probability that a randomly selected adult has an IQ greater than 116.3 can be expressed as:

\[ P(X > 116.3) = 1 - P(X \leq 116.3) = 1 - \Phi(Z) \]

Where \( \Phi(Z) \) is the cumulative distribution function (CDF) of the standard normal distribution. Thus, we have:

\[ P(X > 116.3) = 1 - \Phi(1.0053) \]

Using the standard normal distribution table or a calculator, we find:

\[ \Phi(1.0053) \approx 0.8426 \]

Therefore, the probability is:

\[ P(X > 116.3) = 1 - 0.8426 = 0.1574 \]

Final Answer