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Final answer:To determine the probability that Fred's and Leona's ACT scores differ by 5 points, we calculate the standard deviation of the difference and the corresponding z-scores for a 5-point difference. The cumulative probabilities for these z-scores give us the total probability of approximately 0.077 that their scores will differ by 5 or more points in either direction.Explanation:The goal is to find the probability that Fred's and Leona's scores on the ACT college entrance exam differ by 5 or more points. Since both have equal ability, we assume their scores are normally distributed with a mean (μ) of 24 and a standard deviation (σ) of 2. The difference in two independent normal random variables also follows a normal distribution. The mean of the difference is 0 (since Fred and Leona have the same expected score), and the standard deviation of the difference is the square root of the sum of the squares of the individual standard deviations, which is √(2^2 + 2^2) = √8 ≈ 2.83.To find the probability of a difference of 5 or more points, we can calculate the z-scores for +5 and -5. The z-score for 5 is z=(5-0)/√8 ≈ 1.77, and for -5, it is z=(-5-0)/√8 ≈ -1.77. We then use the standard normal distribution to find the probabilities for z-scores less than -1.77 and greater than +1.77.The cumulative probability for a z-score of -1.77 is approximately 0.0385, and for a z-score greater than +1.77 it is 1 minus the cumulative probability for +1.77, which is also approximately 0.0385. Therefore, the total probability that the scores differ by 5 or more points in either direction is about 0.0385 + 0.0385 = 0.077....

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