Final answer:
The probability that a randomly selected adult has an IQ greater than 120.4, with a mean of 95.5 and a standard deviation of 16.2, can be found by calculating the z-score and looking up this value in a z-table.
Step-by-step explanation:
The student is asking about finding the probability that a randomly selected adult from a normal distribution with a mean IQ of 95.5 and a standard deviation of 16.2 has an IQ greater than 120.4. To find this probability, we first need to convert the IQ score into a z-score.
The formula for converting an individual score to a z-score is:
z = (X - μ) / σ
Where X is the IQ score, μ is the mean IQ, and σ is the standard deviation.
So for an IQ of 120.4:
z = (120.4 - 95.5) / 16.2 ≈ 1.537
Once we have the z-score, we can use a z-table, a calculator, or statistical software to find the probability associated with this z-score.
The probability statement would read: P(X > 120.4) = P(Z > 1.537), and we would look up 1.537 on the z-table to find the probability.