The Knowles/Armitage (KA) group at Merrill Lynch advises clients on how to create a diversified investment portfolio. One of the investment alternatives they make available to clients is an All World fund composed of global stocks with good dividend yields. One of their clients is interested in a portfolio consisting of investment in the All World fund and a Treasury bond fund. The expected percent return of an investment in the All World fund is 7.80% with a standard deviation of 18.90%. The expected percent return of an investment in a Treasury bond fund is 5.50% and the standard deviation is 4.60%. The covariance of an investment in the All World fund with an investment in a Treasury bond fund is -12.4.

a. Which of the funds would be considered the more risky?

Why?

b. If KA recommends that the client invest 75% in the All World fund and 25% in the Treasury bond fund, what is the expected percent return and standard deviation for such a portfolio?

Expected return (Round your answer to three decimal places)
Standard deviation (Round your answer to two decimal places)
What would be the expected return and standard deviation, in dollars, for a client investing $10,000 in such a portfolio?

Expected return $
Standard deviation $
c. If KA recommends that the client invest 25% in the All World fund and 75% in the Treasury bond fund, what is the expected return and standard deviation for such a portfolio? Round your answer to three decimal places.

Expected return (Round your answer to three decimal places)
Standard deviation (Round your answer to four decimal places)
What would be the expected return and standard deviation, in dollars, for a client investing $10,000 in such a portfolio?

Expected return $
Standard deviation $
d. Which of the portfolios in parts (b) and (c) would you recommend for an aggressive investor?

Which would you recommend for a conservative investor?

Why?

Step-by-step explanation:

hello,

we will denote the expected return of investment in the all world fund as E(A).

we will also denote the standard deviation of investment in the world fund as S(A).

we will denote the expected return of investment in the treasury bond fund as E(T)

we will also denote the standard deviation of investment in the treasury bond fund as S(A).

A. since S(A) = 18.90% and S(T) = 4.60%. We see that the standard deviation for all world is greater than that of treasury fund, thus the fluctuation in all world fund is higher than that of treasury fund, hence the all world fund is more risky.

B. expected percent return of portfolio = (percentage investment in the all world × E(A) ) + (percentage investment in the treasury fund × E(T) )

       =( 75% × 7.80%) + ( 25% × 5.5%)

       = 7.225%

standard deviation percent of portfolio =  (percentage investment in the all world × S(A) ) + (percentage investment in the treasury fund × S(T) )

      =  (75% × 18.90%) + ( 25% × 4.60%)

      = 15.325%

expected return in dollars for a client investing $10,000

           = 7.225% × $10,000

          =  $722.500

standard deviation return in dollars for a client investing $10,000

       = 15.325% × $10,000

       = $1,532.500

C. similarly, we simply follow the same procedure in (b) above.

  expected percent return of portfolio =  ( 25% × 7.80%) + ( 75% × 5.5%)

                                                                  = 6.075%

standard deviation percent of portfolio = (25% × 18.90%) + ( 75% × 4.60%)

                                                                   = 8.175%

expected return in dollars for a client investing $10,000

           = 6.075% × $10,000

            = $607.5

standard deviation return in dollars for a client investing $10,000

       = 8.175% × $10,000

        = $817.5

D.   i will recommend portfolio (b) for an aggressive investor because it more risky since its standard deviation is $1,532.500 and it generates more expected return of $722.500 but i will advice a conservative investor to go for portfolio (c) because it is less risky since its standard deviation is  $817.5

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )

a

b

d

step-by-step explanation: