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Suppose you have some money to invest - for simplicity, $$ 1$ -and you are planning to put a fraction $w$ into a stock market mutual fund and the rest, $1 - w$ , into a bond mutual fund. Suppose that $$ 1$ invested in a stock fund yields $R_{s}$ after 1 year and that $$ 1$ invested in a bond fund yields $R_{b}$ , suppose that $R_{s}$ is random with mean $0.1 (10 %)$ and standard deviation 0.09 , and suppose that $R_{b}$ is random with mean $0.06 (6 %)$ and standard deviation 0.05 . The correlation between $R_{s}$ and $R_{b}$ is 0.31 . If you place a fraction $w$ of your money in the stock fund and the rest, $1 - w$ , in the bond fund, then the return on your investment is $R = w R_{s} + (1 - w) R_{b}$ .
Suppose that $w = 0.61$ . Compute the mean and standard deviation of $R$ .
The mean is 0.084 . (Round your response to three decimal places.)
The standard deviation is 0.064 . (Round your response to three decimal places.)
10 of 10 questions
Suppose that $w = 0.92$ . Compute the mean and standard deviation of $R$ .
The mean is . (Round your response to three decimal places.)
The standard deviation is (Round your response to three decimal places.)
What value of $w$ makes the mean of $R$ as large as possible?
$w =$
maximizes $μ$ . (Round your response to two decimal places.)
What is the standard deviation of $R$ for this value of $w$ ?
$σ =$
for this value of $w$ . (Round your response to two decimal places.)
What is the value of $w$ that minimizes the standard deviation of $R$ ? $w = 1$ minimizes the standard deviation of $R$ . (Round vour resoonse to two decimal olaces.)

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Final Answer: The mean of $R$ when $w = 0.92$ is $0.097$ and the standard deviation of $R$ when $w = 0.92$ is $0.084$ .

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Step 1 ：Given that the fraction of money invested in the stock fund is $w = 0.92$ , the mean return of the stock fund is $0.1$ , the standard deviation of the stock fund is $0.09$ , the mean return of the bond fund is $0.06$ , the standard deviation of the bond fund is $0.05$ , and the correlation between the stock and bond returns is $0.31$ .

Step 2 ：We can calculate the mean return on the investment $R$ using the formula $R = w R_{s} + (1 - w) R_{b}$ . Substituting the given values, we get $R = 0.92 * 0.1 + (1 - 0.92) * 0.06 = 0.0968$ .

Step 3 ：We can calculate the standard deviation of the return on the investment $R$ using the formula $\sqrt{w^{2} σ_{s}^{2} + (1 - w)^{2} σ_{b}^{2} + 2 w (1 - w) ρ σ_{s} σ_{b}}$ . Substituting the given values, we get $\sqrt{{0.92}^{2} * {0.09}^{2} + (1 - 0.92)^{2} * {0.05}^{2} + 2 * 0.92 * (1 - 0.92) * 0.31 * 0.09 * 0.05} = 0.084126$ .

Step 4 ：Rounding to three decimal places, the mean return on the investment $R$ when $w = 0.92$ is $0.097$ and the standard deviation of the return on the investment $R$ when $w = 0.92$ is $0.084$ .

Step 5 ：Final Answer: The mean of $R$ when $w = 0.92$ is $0.097$ and the standard deviation of $R$ when $w = 0.92$ is $0.084$ .

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