To determine the probability that a person has an IQ greater than 115, we need to find the area under the normal distribution curve to the right of the IQ value of 115.

The problem tells us that IQ scores are normally distributed with:

• Mean $\mu = 100$
• Standard deviation $\sigma = 16$

We need to find $P(X > 115)$, where $X$ represents the IQ score.

Step 1: Standardize the IQ score (convert it to a z-score)

The z-score formula is: $z = \frac{X - \mu}{\sigma}$ For $X = 115$: $z = \frac{115 - 100}{16} = \frac{15}{16} = 0.9375$

Step 2: Use the z-score to find the probability

Now that we have the z-score $z = 0.9375$, we can use a z-table or a standard normal distribution calculator to find the probability corresponding to this z-score.

The z-table gives the probability that a value is less than a given z-score. So, we first find $P(Z < 0.9375)$.

From the z-table: $P(Z < 0.9375) \approx 0.8253$

Step 3: Subtract from 1 to get $P(X > 115)$

Since we want the probability that a person has an IQ greater than 115, we subtract the result from 1: $P(X > 115) = 1 - P(Z < 0.9375) = 1 - 0.8253 = 0.1747$

Final Answer:

The probability that a person has an IQ greater than 115 is approximately 0.1747, or 17.47%.

Would you like further details on the process, or have any questions?

Here are 5 related questions to explore further:

1. What is the probability of someone having an IQ less than 85?
2. What percentage of people have IQs between 85 and 115?
3. How would the probability change if the standard deviation were different?
4. What is the z-score for an IQ of 130?
5. What is the IQ score corresponding to the top 5% of the population?

Tip: The z-score helps standardize any normal distribution, making it easy to use universal z-tables or calculators!