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1. The risk of a portfolio of financial assets is sometimes called investment risk. In general,
investment risk is typically measured by computing the variance or standard deviation of the
probability distribution that describes the decision maker’s potential outcomes. The greater the
variation in potential outcomes, the greater the uncertainty faced by the decision maker; the
smaller the variation in potential outcomes, the more predictable the decision maker’s gain and
loss. The two discrete probability distributions given in the table were developed from historical
data. The describe the potential total physical damage losses next year to the fleets of delivery
trucks of two different firms.
a) Verify that both firms have the same expected total physical damage loss.
Firm A
Loss next year
$
$500.00
$1,000.00
$1,500.00
$2,000.00
$2,500.00
$3,000.00
$3,500.00
$4,000.00
$4,500.00
$5,000.00
Total =
Firm B
Expected
Probability Value
0.01
0
0.01
$5.00
0.01
$10.00
0.02
$30.00
0.35
$700.00
0.3
$750.00
0.25
$750.00
0.02
$70.00
0.01
$40.00
0.01
$45.00
0.01
$50.00
1
$2,450.00
Expected
Loss next year
Probability Value
$
0
0
$200.00
0.01
$2.00
$700.00
0.02
$14.00
$1,200.00
0.02
$24.00
$1,700.00
0.15
$255.00
$2,200.00
0.3
$660.00
$2,700.00
0.3
$810.00
$3,200.00
0.15
$480.00
$3,700.00
0.02
$74.00
$4,200.00
0.02
$84.00
$4,700.00
0.01
$47.00
Total =
1
$2,450.00
b) Compute the standard deviation of each probability distribution, and determine which firm
faces the greater risk of physical damage to its fleet next year and why?
Firm A
Loss next year
$
$500.00
$1,000.00
$1,500.00
$2,000.00
$2,500.00
$3,000.00
$3,500.00
$4,000.00
$4,500.00
$5,000.00
Total =
Firm B
Probability
0.01
0.01
0.01
0.02
0.35
0.3
0.25
0.02
0.01
0.01
0.01
1
Expected
Value
0
5
10
30
700
750
750
70
40
45
50
2,450
p*x^2
Loss next year
0
2,500
10,000
45,000
1,400,000
1,875,000
2,250,000
245,000
160,000
202,500
250,000
6,440,000
$
$200.00
$700.00
$1,200.00
$1,700.00
$2,200.00
$2,700.00
$3,200.00
$3,700.00
$4,200.00
$4,700.00
Total =
Probability
0
0.01
0.02
0.02
0.15
0.3
0.3
0.15
0.02
0.02
0.01
1
Expected
Value
0
2
14
24
255
660
810
480
74
84
47
2,450
p*x^2
0
400
9,800
28,800
433,500
1,452,000
2,187,000
1,536,000
273,800
352,800
220,900
6,495,000
E(X) =
STDEV(X) =
$2,450.00
$661.44
E(X) =
STDEV(X) =
$2,450.00
$701.78
Clearly, firm B has a greater associated risk because its standard deviation is higher.
2. The random variable x has a normal distribution with µ=1000 and σ = 10.
a. Find the probability that x assumes a value more than 2 standard deviations from its
mean.
P(X > 1000+2*10) = P(X > 1020) = 0.02275
b. More than 3 standard deviations from µ.
P(X > 1000+3*10) = P(X > 1030) = 0.001350
c. Find the probability that x assumes a value within 1 standard deviation of its mean.
P(1000-10 <= X <= 1000+10) = P(990 <= X <= 1010) = 0.68269
d. Within 2 standard deviations of µ.
P(1000-20 <= X <= 1000+20) = P(980 <= X <= 1020) = 0.9545
e. Find the value of x that represents the 80th percentile of this distribution.
X = 1008.42
f.
The 10th percentile.
X = 987.1845
3. Health insurers and the federal government are both putting pressure on hospitals to shorten
the average length of stay of their patients. In 1999, the average length of stay for men in the
US was 5.4 days and the average for women was 4.7 days. A random sample of 20 hospitals in
one state had a mean length of stay for women in 2003 of 3.8 days and a standard deviation of
1.2 days.
a. Use a 90% confidence interval to estimate the population mean length of stay for
women for the state’s hospitals in 2003
Standard Error
UCL =
LCL =
0.441361
4.241361
3.358639
b. Interpret the intervals in terms of this application
We are 90% confident that the true population mean lies somewhere in between 3.36
days and 4.24 days.
c. What is meant by the phrase 90% confidence interval
90% CI gives a level of confidence of 90% with which we can state that the true
population parameter lies between any two values.
4. Suppose you are interested in conducting the statistical test of H0: µ = 255 against Ha : µ > 255,
and you have decided to use the following decision rule: Reject H0 if the sample mean of a
random sample of 81 items is more than 270. Assume that the standard deviation of the
population is 63.
a. Express the decision rule in terms of z
Reject the null hypothesis if p-value < alpha.
b. Find α, the probability of making a Type I error, by using this decision rule.
Z = (270 – 255)/(63/√81) = 2.143
Therefore,  = 0.0161 = 1.61%