To find the probability that a randomly chosen person's IQ score will be higher than 127, we can follow these steps:

Step 1: Standardize the IQ score

Since IQ scores are normally distributed with a mean ($\mu$) of 100 and a standard deviation ($\sigma$) of 15, we first convert the IQ score of 127 to a standard normal score (also called a z-score) using the formula:

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

Where:

• $X = 127$ (the IQ score we are interested in),
• $\mu = 100$ (the mean),
• $\sigma = 15$ (the standard deviation).

$z = \frac{127 - 100}{15} = \frac{27}{15} = 1.8$

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

Next, we find the probability that a standard normal variable $Z$ is greater than 1.8. This is the area to the right of $z = 1.8$ on the standard normal distribution.

From standard normal distribution tables or using a calculator, the cumulative probability up to $z = 1.8$ is approximately 0.9641. This means that 96.41% of the population has an IQ score less than 127.

Step 3: Calculate the probability for IQ > 127

To find the probability that a person's IQ score is greater than 127, we subtract this value from 1:

$P(X > 127) = 1 - P(Z \leq 1.8) = 1 - 0.9641 = 0.0359$

Thus, the probability that a randomly chosen person will have an IQ score higher than 127 is approximately 0.036, or 3.6%.

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Would you like more details, or do you have any questions? Here are some related questions you may find helpful:

1. What is the probability that a person has an IQ less than 90?
2. How do you find the probability for an IQ range between 85 and 115?
3. What is the z-score for an IQ of 140?
4. How can we calculate the percentage of people with IQ scores within one standard deviation from the mean?
5. How does changing the standard deviation affect the probability calculations?

Tip: Always standardize a score to a z-score when working with normal distributions, as it simplifies the use of probability tables or calculators.