Find the indicated probability and interpret the result. From 1975 through 2020, the mean annual gain of the Dow Jones Industrial Average was 653 . A random sample of 35 years is selected from this population. What is the probability that the mean gain for the sample was between 400 and 700 ? Assume \( \sigma=1539 \). The probability is \( \square \) 0.4063 . (Round to four decimal places as needed.) Interpret the result. Select the correct choice and fill in the answer box to complete your choice. (Round to two decimal places as needed.) A. About \( \square \) \( \% \) of samples of 46 years will have an annual mean gain between 400 and 700 . B. About \( \square \) \( \% \) of samples of 35 years will have an annual mean gain between 653 and 700 . C. About \( \square \) \( \% \) of samples of 35 years will have an annual mean gain between 400 and 653 . D. About \( \square \) \( \% \) of samples of 35 years will have an annual mean gain between 400 and 700 .

To find the probability that the mean gain for the sample was between 400 and 700, we can use the z-score formula and the standard normal distribution. Given: - Mean annual gain of the Dow Jones Industrial Average: \( \mu = 653 \) - Standard deviation: \( \sigma = 1539 \) - Sample size: \( n = 35 \) - Lower limit: \( 400 \) - Upper limit: \( 700 \) First, we need to calculate the z-scores for the lower and upper limits: \[ z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}} \] For the lower limit (400): \[ z_{lower} = \frac{400 - 653}{\frac{1539}{\sqrt{35}}} \] For the upper limit (700): \[ z_{upper} = \frac{700 - 653}{\frac{1539}{\sqrt{35}}} \] Next, we will calculate the z-scores and then use the standard normal distribution to find the probability that the mean gain for the sample was between 400 and 700. Calculate or simplify the expression \( (700 - 653) / (1539 / sqrt(35)) \). The z-scores for the lower and upper limits are approximately -0.9726 and 0.1807, respectively. Now, we can use the standard normal distribution to find the probability that the mean gain for the sample was between 400 and 700. The probability is given by the area under the standard normal curve between the z-scores -0.9726 and 0.1807. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.4063. Therefore, the probability that the mean gain for the sample was between 400 and 700 is approximately 0.4063. Now, we need to interpret the result. The correct choice is: D. About 40.63% of samples of 35 years will have an annual mean gain between 400 and 700.