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Golf Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations. $$\text { Norb, } x_{1}: \mu_{1}=115 ; \sigma_{1}=12 \quad \text { Gary, } x_{2}: \mu_{2}=100 ; \sigma_{2}=8$$ In the tournament, Norb and Gary are not playing together, and we will assume their scores vary independently of each other. (a) The difference between their scores is \(W=x_{1}-x_{2} .\) Compute the mean, variance, and standard deviation for the random variable \(W.\) (b) The average of their scores is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation for the random variable \(W.\) (c) The tournament rules have a special handicap system for each player. For Norb, the handicap formula is \(L=0.8 x_{1}-2 .\) Compute the mean, variance, and standard deviation for the random variable \(L.\) (d) For Gary, the handicap formula is \(L=0.95 x_{2}-5 .\) Compute the mean, variance, and standard deviation for the random variable \(L.\)

Short Answer

Expert verified
(a) Mean: 15, Variance: 208, SD: 14.42. (b) Mean: 107.5, Variance: 52, SD: 7.21. (c) Mean: 90, Variance: 92.16, SD: 9.6. (d) Mean: 90, Variance: 57.76, SD: 7.6.

Step by step solution

01

Mean of the Difference of Scores

The mean of the difference of random variables is the difference of their means. Using \(\mu_1 = 115\) and \(\mu_2 = 100\):\[ \text{Mean of } W = \mu_1 - \mu_2 = 115 - 100 = 15. \]
02

Variance of the Difference of Scores

Since the scores are independent, the variance of the difference \(W = x_1 - x_2\) can be found by adding their variances: \[ \text{Variance of } W = \sigma_{1}^2 + \sigma_{2}^2 = 12^2 + 8^2 = 144 + 64 = 208. \]
03

Standard Deviation of the Difference of Scores

The standard deviation is the square root of the variance. For \(W\), \[ \text{Standard deviation of } W = \sqrt{208} \approx 14.42. \]
04

Mean of the Average of Scores

The mean of a linear combination such as the average \(W = 0.5x_1 + 0.5x_2\) is the linear combination of their means: \[ \text{Mean of } W = 0.5\mu_1 + 0.5\mu_2 = 0.5(115) + 0.5(100) = 57.5 + 50 = 107.5. \]
05

Variance of the Average of Scores

For independent variables, the variance of the linear combination \(W = 0.5x_1 + 0.5x_2\) is \(0.5^2\sigma_1^2 + 0.5^2\sigma_2^2\):\[ \text{Variance of } W = 0.25(144) + 0.25(64) = 36 + 16 = 52. \]
06

Standard Deviation of the Average of Scores

The standard deviation is the square root of the variance: \[ \text{Standard deviation of } W = \sqrt{52} \approx 7.21. \]
07

Mean of Norb's Handicap

The mean of a linear transformation \(L = 0.8x_1 - 2\) is transformed as \(0.8\mu_1 - 2\): \[ \text{Mean of } L = 0.8(115) - 2 = 92 - 2 = 90. \]
08

Variance of Norb's Handicap

The variance of a linear transformation is \((0.8^2)\sigma_1^2\): \[ \text{Variance of } L = 0.64(144) = 92.16. \]
09

Standard Deviation of Norb's Handicap

The standard deviation is the square root of the variance: \[ \text{Standard deviation of } L = \sqrt{92.16} \approx 9.6. \]
10

Mean of Gary's Handicap

The mean of a linear transformation \(L = 0.95x_2 - 5\) is transformed as \(0.95\mu_2 - 5\): \[ \text{Mean of } L = 0.95(100) - 5 = 95 - 5 = 90. \]
11

Variance of Gary's Handicap

The variance of a linear transformation is \((0.95^2)\sigma_2^2\): \[ \text{Variance of } L = 0.9025(64) = 57.76. \]
12

Standard Deviation of Gary's Handicap

The standard deviation is the square root of the variance: \[ \text{Standard deviation of } L = \sqrt{57.76} \approx 7.6. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
In statistics, a random variable is a variable whose outcomes depend on the result of a random event. For Norb and Gary, their golf scores are considered random variables because each round can vary based on many factors, even though they have average scores over time. These scores are represented as random variables with the notations \(x_1\) for Norb and \(x_2\) for Gary.

Random variables can be discrete or continuous, but in this context, they are treated as numerical data with specific mean and standard deviation values. The mean of a random variable is its expected average value based on a probability distribution. It reflects the center of the distribution.

The standard deviation helps understand the spread around the mean, illustrating how much variance or "spread out" the values can have from one game to another. This understanding becomes crucial when calculating the difference or sum of these random variables' scores for various statistical calculations.
Mean and Variance
The mean is a fundamental concept in statistics that provides the average or expected value of a random variable. In Norb's and Gary’s case, their respective means are 115 and 100. When dealing with combined quantities, such as the difference \(W = x_1 - x_2\) or the average \(W = 0.5x_1 + 0.5x_2\) of their scores, we use linear operations on the means.

To find the mean of the difference, we subtract their means: \( \text{Mean of } W = \mu_1 - \mu_2 = 15 \). For their average score, calculate using the weighted mean: \( \text{Mean of } W = 0.5\mu_1 + 0.5\mu_2 = 107.5 \).

Variance measures how spread out numbers are around the mean and is computed differently when random variables are dependent or independent. It acts as a stepping stone to standard deviation. For independent random variables like Norb's and Gary's scores, the variance of their differences and sums involves adding or scaling the individual variances: for the difference \(\sigma_1^2 + \sigma_2^2 = 208\) and for the average \(0.25\sigma_1^2 + 0.25\sigma_2^2 = 52\). These computations highlight the variability in the outcomes expected when Norb and Gary play independently.
Standard Deviation
The standard deviation is one of the most used metrics in statistics, providing insight into the reliability and consistency of an average score. It is derived from the variance and depicts the average distance of each data point from the mean. For Norb and Gary, knowing their standard deviations helps us understand how typical fluctuations in their scores look like in the context of golf tournaments.

This value requires taking the square root of the variance. For the random variable difference in scores \( W = x_1 - x_2 \), the standard deviation is \( \sqrt{208} \approx 14.42 \). Similarly, for their respective handicaps and score averages, we compute this way to represent the reliance on the spread of numbers: \( \sqrt{52} \approx 7.21 \) for their average scores and \( \sqrt{92.16} \approx 9.6 \) for Norb's handicap.

In essence, the lower the standard deviation, the more predictable and consistent the player’s performance is with their average. Conversely, a higher standard deviation can indicate a wider range of possible outcomes, which is crucial in planning strategies or setting expectations.

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