The problem you're asking about involves finding the probability of a normally distributed IQ score.

Given Information:

• Mean $\mu = 105$
• Standard deviation $\sigma = 15$
• We need to find the probability that an adult has an IQ score less than 123.

Steps to solve:

1. First, we calculate the z-score for $X = 123$ using the formula: $z = \frac{X - \mu}{\sigma} = \frac{123 - 105}{15} = \frac{18}{15} = 1.2$

2. Now, we will use a standard normal distribution table (or a calculator) to find the cumulative probability associated with a z-score of 1.2.

   Using a z-table or standard normal distribution function: $P(Z < 1.2) \approx 0.8849$

Thus, the probability that a randomly selected adult has an IQ less than 123 is 0.8849.

Would you like more details on this process or have any questions?

Here are 5 related questions to explore this further:

1. What is the probability that an adult has an IQ score higher than 123?
2. What is the probability that an adult has an IQ score between 90 and 123?
3. How do you find the z-score for values less than the mean, say for an IQ of 90?
4. What percentage of adults have IQ scores more than 2 standard deviations from the mean?
5. How would the probability change if the standard deviation was larger or smaller?

Tip: When dealing with normal distributions, the z-score helps standardize any normal random variable, allowing you to use tables or technology to calculate probabilities.