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Sean and Evan are college roommates who have part time jobs are servers in restaurants. The distribution of Sean's weekly income is approximately normal with mean 225$ and SD of 25$. The distribution of Evan's weekly income is approximately normal with mean 240 $ and SD of 15 $. Assuming their weekly incomes are independent of each other, What is closest to the probability that Sean will have a greater income than Evan in a randomly selected week?
Sean and Evan are college roommates who have part time jobs are servers in restaurants. The distribution of Sean's weekly income is approximately normal with mean 225$ and SD of 25$. The distribution of Evan's weekly income is approximately normal with mean 240 $ and SD of 15 $. Assuming their weekly incomes are independent of each other, What is closest to the probability that Sean will have a greater income than Evan in a randomly selected week?
Final answer:Using the distributions of Sean and Evan's weekly incomes, we calculate the distribution of the difference and find there is approximately a 30.35% chance that Sean will earn more than Evan in a given week.Explanation:To determine the probability that Sean will have a greater weekly income than Evan, we must consider the distribution of the difference in their incomes.Since Sean and Evan's incomes are independent, normally distributed random variables, the difference will also be normally distributed with a mean equal to the difference of their means, μSean - μEvan, and a standard deviation equal to the square root of the sum of the squares of their standard deviations, √(σSean² + σEvan²).Let's calculate these values:μdifference = μSean - μEvan = $225 - $240 = -$15σdifference = √(σSean² + σEvan²) = √(25² + 15²) = √(625 + 225) = √850 ≈ $29.15To determine the probability that Sean earns more than Evan, we need the probability that the difference is greater than zero. We look for the z-score corresponding to a difference of zero using the distribution of the difference:Z = (0 - (-$15)) / $29.15 ≈ 0.514The probability corresponding to a z-score of 0.514 from z-tables is approximately 0.6965 (or 69.65%). However, since we are looking for the probability that Sean earnsmore, we need the complement of this, which is 1 - 0.6965 = 0.3035 (or 30.35%)....