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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%