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2/8/2019 16. A private equity firm is evaluating two alternative investments. Although the returns are random, each investmen
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Q.16) Given that, the first investment has a mean return of $2500000 with a standard deviation of $150000. The second investment has a mean return of $2725000 with a standard deviation of $300000.

Let X1 ~ N($2500000, $150000)

X2 ~ N($2725000, $300000)

a) We want to find, P(X1 <= $2300000)

XI_μ 2300000-2500000 ) = P(Z <-1.33) P(Z S -1.33) PlX1 2300000)-P( 150000 0.0918

The probability is 0.0918

b) We want to find, P(X2 <= $2300000)

P(XI 230000)-P2-230000 P(X1 2300000)P( ,0778 ) = P(Z <-1.42) 300000

The probability is 0.0778

c) For 1st investment, P( X1 < $2200000)

X1_μ 2200000-2500000 P(X1< 2200000P 0.0228 150000

=> P(X1 < $2200000) = 0.0228

For 2nd investment, P(X2 < $2200000)

X2-μ 2200000-2725000 300000 ) = P(Z <-1.75) PlX2 < 2200000)-P(-_-< 0.0401

=> P(X2 < $2200000) = 0.0401

Therefore, the probability of a return being less than $2200000 is 0.0228 with the first investment and 0.0401 with the second investment, so the firm should make the 1) first investment.

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