Final answer:
The probability that the average IQ of 100 randomly selected adults is less than 97, with an IQ distribution mean of 100 and standard deviation of 15, can be determined by calculating the z-score of -2 and then using the standard normal distribution to identify the cumulative probability, which is 0.0228.
Step-by-step explanation:
The question asks what the probability is that the average IQ of 100 randomly selected adults is less than 97, given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. To solve this, we use the Central Limit Theorem, which states that the distribution of the sample means will be normally distributed with the same mean as the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size (known as the standard error).
Therefore, for a sample size of 100, the standard error is 15 / sqrt(100) = 1.5. Next, we convert the IQ score to a z-score by subtracting the population mean from the sample mean and dividing by the standard error: (97 - 100) / 1.5 = -2. Then we use standard normal distribution tables or software to find the probability of a z-score less than -2. This probability corresponds to the area under the normal curve to the left of the z-score.
The correct answer can be found by looking up the cumulative probability for a z-score of -2, which is approximately 0.0228. Therefore, the probability that the average IQ of 100 randomly selected adults is less than 97 is 0.0228.