Chapter 15: Problem 34
A golfer keeps track of his score for playing nine holes of golf (half a normal golf round). His mean score is 85 with a standard deviation of \(11 .\) Assuming that the second 9 has the same mean and standard deviation, what is the mean and standard deviation of his total score if he plays a full 18 holes?
Short Answer
Step by step solution
Define the Problem
Calculate the Mean of 18 Holes
Calculate the Variance for 18 Holes
Find the Total Variance for 18 Holes
Calculate the Standard Deviation for 18 Holes
Summarize the Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Deviation
To compute the standard deviation of the total 18 hole score from the original exercise, we start by finding the variance for each 9-hole game, and since the games are independent, we add these variances together. The total variance is 242, so the standard deviation—being the square root of variance—is approximately 15.56. This indicates a moderate spread in his total golf scores over 18 holes.
Variance
In our golfer example, the variance for each 9-hole segment is found by squaring the standard deviation 11, resulting in 121. Since each 9-hole game score is independent, we sum these variances to find the total variance over 18 holes, which turns out to be 242. By understanding variance, we can see that it gives us the "spread squared" and helps in reaching the standard deviation.
Independent Variables
In the exercise, the scores from the first and second 9-hole rounds are treated as independent. This means the outcome of one does not affect the other. Because of this, the variance of the total score is simply the sum of the individual variances from each half of the game. Without independence, we'd need additional methods to compute the total score's statistics.
Normal Distribution
In this exercise, the assumption of a normal distribution is useful because it suggests the golfer's scores are evenly spread around the mean (85 per 9 holes). Given a normal distribution, we can confidently apply the rules for calculating mean and standard deviation for combined data. The second 9-hole round having the same distribution properties allows for tidy computations and explanations of the golfer's full 18-hole performance.