Final answer:
Using the Central Limit Theorem and the standard normal distribution, we calculated the probability that the sample mean gain for a sample of 31 years is between 400 and 700 to be about 38.67%. Therefore, option D is correct.
Step-by-step explanation:
To find the probability that the mean gain for a sample of 31 years was between 400 and 700, we can use the Central Limit Theorem since the sample size is greater than 30. Given the population mean (μ) is 653 and the population standard deviation (σ) is 1541, we can find the standard error of the mean using the formula σ/√n, where n is the sample size.
The standard error (SE) of the mean for our sample size of 31 years is:
SE = 1541 / √31 ≈ 276.66
Next, we need to convert the range of interest (400 to 700) into z-scores. The z-score for a value X is given by:
Z = (X - μ) / SE
Calculating z-scores for X = 400 and X = 700:
Z400 = (400 - 653) / 276.66 ≈ -0.913
Z700 = (700 - 653) / 276.66 ≈ 0.17
Using the standard normal distribution table, we can find the probabilities associated with these z-scores:
P(Z < -0.913) and P(Z < 0.17).
Subtracting the smaller probability from the larger gives the probability of the sample mean being between 400 and 700. Let's assume the probabilities from the z-table are P(Z < -0.913) = 0.1808 and P(Z < 0.17) = 0.5675. Thus, the probability the sample mean is between 400 and 700 is 0.5675 - 0.1808 = 0.3867 or about 38.67%.
We can interpret this result as follows:
About 38.67% of samples of 31 years will have an annual mean gain between 400 and 700. Hence, option D is correct.