Final answer:
To determine the probability of an adult having an IQ score over 135.8, calculate the z-score by subtracting the mean from the IQ score and dividing by the standard deviation, then use a z-table or calculator to find the corresponding probability.
Step-by-step explanation:
To find the probability that a randomly selected adult has an IQ greater than 135.8 when adults have IQ scores that are normally distributed with a mean of 103.1 and a standard deviation of 17.9, we first need to convert the IQ score into a z-score. A z-score is given by the formula:
Z = (X - μ) / σ
where X is the value for which we are finding the probability, μ is the mean, and σ is the standard deviation.
Let's calculate the z-score:
Z = (135.8 - 103.1) / 17.9
Now, use a standard normal distribution table or a calculator to find the probability that corresponds to this z-score. The result will give us the probability that an IQ score is greater than 135.8. Remember, since standard normal distributions are symmetrical, you might need to subtract your result from 1 to get the probability of a score being above a certain z-score.