Question

Asempa, the account manager for Northern Securities, has a portfolio that includes 20
shares of Albert Information Systems (AIS) and 30 shares of Beta Cyber Analytics (BCA).
Both firms provide Web access devices that compete in the consumer market. The price of
AIS stock is normally distributed with mean ?? = 25 and variance ??
2 = 81. The price of
BCA stock is also normally distributed with mean ?? = 40 and variance ??
2 = 121. The
stock prices have a negative correlation ??? = −0.40. Asempa has asked you to determine
the probability that the portfolio value;
a. Will exceed 2,000. (2 marks)
b. Will be less than 1,800. (2 marks)
c. Will be between 1,600 and 2,200. (2 marks)
(HINT USE THE TRANSFORMATION PROPERTIES OF MEAN& VARIANCE)
B)The number of phone calls, Y, received per day by Sarah has the following probability
distribution:
Y 0 1 2 3 4 ≥ 5
P(Y = y) 0.24 0.35 2w w 0.05 0
a. Find the value of w. ( 1 mark)
b. Find the mode of Y. ( 1 mark)
c. Find the probability that the number of phone calls received by Sarah on any particular
day is more than the mean number of phone calls received per day. ( 2 marks)

Answer 1

$A) \\ \textrm{AIS(X) return follows Normal(} \mu_x=25 , \sigma^2_x = 81) \\ \textrm{BCA(Y) return follows Normal(} \mu_y=40 , \sigma^2_Y = 121)$

PXY -0.4

Portfolio V = 20X+30Y follows normal distribution with

$\mu_V = 20*\mu_X+30*\mu_Y \\= 20*25+30*40 =1700$

$\sigma_V^2 = 20^2 *\sigma_X^2+ 30^2*\sigma_Y^2+ 20*30*p_{XY}*\sigma_X*\sigma_Y\\ =400*81+900*121-0.4*9*11*600 \\ =117540 = 342.84^2$

a)

$P(V>2000) \\ = P(z>\frac{2000-1700}{342.84}) \\= P(z>0.875) \\= 1-P(z<= 0.875) \\ = 1-0.809 \\=0.19$

b)

$P(V<1800) \\ = P(z<\frac{1800-1700}{342.84}) \\= P(z<0.292) \\ =0.615$

c)

$P(1600<V<2200) \\ = P(\frac{1600-1700}{342.84}<z<\frac{2200-1700}{342.84}) \\= P(-0.29<z<1.46) \\ P(z<1.46)-(1-P(z<0.292))\\ =0.928 - (1-0.615) \\=0.543$