Two friends, Andy and Bob, participate in a game of bowling every week. From past experiences, it is knownthat both friends’ scores are approximately normally distributed, where Andy has a mean score of 150 with astandard deviation of 30, and Bob has a mean score of 165 with a standard deviation of 15. Assuming that theirscores are independent, which of the following values is closest to the probability that Andy will have a greaterscore than Bob in a single game?(A) 0.16(B) 0.28(C) 0.31(D) 0.33(E) 0.37
Question
Two friends, Andy and Bob, participate in a game of bowling every week. From past experiences, it is knownthat both friends’ scores are approximately normally distributed, where Andy has a mean score of 150 with astandard deviation of 30, and Bob has a mean score of 165 with a standard deviation of 15. Assuming that theirscores are independent, which of the following values is closest to the probability that Andy will have a greaterscore than Bob in a single game?(A) 0.16(B) 0.28(C) 0.31(D) 0.33(E) 0.37
Solution
To solve this problem, we need to find the probability that Andy scores higher than Bob. Since the scores are normally distributed and independent, we can use the properties of the normal distribution to solve this.
Step 1: Find the mean difference and standard deviation of the difference.
The mean difference is the difference in their average scores, which is 150 - 165 = -15.
The standard deviation of the difference is the square root of the sum of their variances (since they are independent), which is sqrt((30^2) + (15^2)) = sqrt(900 + 225) = sqrt(1125) = 33.54 (approximately).
Step 2: Convert the problem into a standard normal problem.
We want to find P(Andy - Bob > 0), which is equivalent to P(Z > (0 - (-15))/33.54) = P(Z > 0.4472), where Z is a standard normal random variable.
Step 3: Look up the value in the standard normal table.
The value of P(Z < 0.4472) is approximately 0.6726 from the standard normal table. However, we want P(Z > 0.4472), so we subtract the value from 1 to get 1 - 0.6726 = 0.3274.
So, the probability that Andy will score higher than Bob in a single game is approximately 0.33. Therefore, the answer is (D) 0.33.
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