Portfolio standard deviation and diversification: You are asked to give financial advice to a mutual fund manager. The fund has
had a history of investing in an equally-weighted portfolio of 50 stocks and is considering cutting down the number of stocks to 10 or 20. You assume that all stocks the fund is considering investing in have the same standard deviation of 40% and that the correlation between each pair of stocks is identical and equal to 0.25.
a. To start your analysis, you calculate the standard deviation of the equally weighted portfolios with 10, 20, and 50 assets.
b. Given your answers to part a. what is the advice that you give the mutual fund manager?
a. To start your analysis, you calculate the standard deviation of the equally weighted portfolios with 10, 20, and 50 assets.
b. Given your answers to part a. what is the advice that you give the mutual fund manager?
1 answer:
Answer:
a) check attachment below
b) Whenever we have a portfolio whatsoever, that minimizes the variance of the portfolio is always preferred. Therefore, the portfolio of 50 stock is better than portfolio of 20 stocks which is as well better than the 10 stocks portfolio. The fund manager especially the mutual fund manager should not consider cutting down the number of stocks. If reducing the number of stock is very urgent then he should probably consider reducing it down to 20 as it is variance is less than that of 10 stocks portfolio.
Explanation:
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Answer:
- The final temperature in the container with Gold is 27.49 ⁰C
- The final temperature in the container with Iron is 33.01 ⁰C
Therefore, the highest final temperature is obtained in the container with Iron.
Explanation:
Q = mcΔT
Where;
Q is the quantity of heat gained or lost
m is the mass of the metals or water
c is the specific heat capacity
ΔT is the change in temperature, T₂ - T₁
T₂ is the final temperature and T₁ is the initial temperature
Heat lost by metals at 100°C is equal to heat gained by water at 25°C
Specific heat capacity of water = 4.18 J/g°C
Specific heat capacity of gold = 0.129 J/g°C
Specific heat capacity of iron = 0.45 J/g°C
⇒For Gold
-20*0.129*(T₂ - 100) = 18*4.18 (T₂ - 25)
-2.58T₂ +258 = 75.24T₂ - 1881
77.82T₂ = 2139
T₂ = 2139/77.82
T₂ = 27.49 ⁰C
⇒For Iron
-20*0.450*(T₂ - 100) = 18*4.18 (T₂ - 25)
-9T₂ +900 = 75.24T₂ - 1881
84.24T₂ = 2781
T₂ = 2781/84.24
T₂ = 33.01 ⁰C
Therefore, the highest final temperature is obtained in the container with Iron.