Final answer:
The distribution of IQ in a population of adults is normal, and so is the distribution of the sample mean IQ. To find the probabilities requested, we can use the z-score formula and the standard normal distribution table.
Step-by-step explanation:
(a) The distribution of IQ: The distribution of IQ in a large population of adults is normal. This means that the majority of adults have an IQ score close to the mean, with fewer adults having IQ scores further away from the mean.
(b) The distribution of the sample mean IQ is also normal. The mean of the sample mean IQ is equal to the population mean IQ, which is 114. The standard deviation of the sample mean IQ is equal to the population standard deviation divided by the square root of the sample size, which is 21/sqrt(60).
(c) To find the probability that the sample mean IQ is less than 113, we need to calculate the z-score of 113 using the formula z = (x - μ) / (σ / sqrt(n)). Plugging in the values, we get z = (113 - 114) / (21 / sqrt(60)). Once we have the z-score, we can look up the corresponding probability in the standard normal distribution table.
(d) To find the probability that the sample mean IQ is greater than 113, we can subtract the probability found in part (c) from 1.
(e) To find the probability that the sample mean IQ is between 113 and 116, we need to calculate the z-scores for both 113 and 116, and look up the corresponding probabilities in the standard normal distribution table. Then, we can subtract the probability of the lower z-score from the probability of the higher z-score.