ISOM 5510 Quiz II
1. A financial analyst uses a random variable X to represent the daily return of a stock. Which statement is false?
The outcome of random variable X can always be interpreted as a random selection from a population of the same shape as the probability distribution of X
The graph of the probabiliy distribution of X is one way to anticipate the histogram for future returns
The expected value of X equals the return of the stock for a future trading day.
The probability distribution of X can be viewed as the population of future returns by repeating the experiment of X a larger number of times.
2. Two investments J and K have annual mean gross return equal 1.08 and 1.14 respectively. The variance of J and Kare 0.04 and 0.12, respectively. In the long run, which one is more likely to perform better?
Investment K because the mean return is higher.
Investment J because the variance is smaller and hence the risk is lower.
Investment K beause the long-term multi-period return is higher.
Investment J because the volatility drag is lower.
3. Which statement is false?
The chi-square statistics calculated from two completely different contingency tables have the same value, then the strength of assocation in these two tables must be the same.
We can fill in the cells of the contingency table from the marginal counts alone if the two variables are not assocaited.
The percentage of cases in the first column within each row of a contingency table are the same if the variables are not assocaited.
The value of chi-square statistics will not change, if we turn the columns into rows and rows into columns.
4. Which statement is correct?
The probablity that the total sales for Models I and II is at least 700,000 units equal 0.03.
The probability that the sales for Model I is 200,000 units equals 0.26
The expected sales for Model II equals 350,000 units
The probability that Model I sells 200,000 units given that Model II sells 300,000 units equals 0.20
5. Which if the following statements is true?
Cov (X,Y) > 0
Cov (X, Y)/ σx σy < -1 (σx σy are the standard deviations of X and Y, respectively)
Cov (X,Y) = 0
Cov (X, Y) < 0
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