The money (in thousands of dollars) made from investing in stocks \"Y stock\" an

Question

The money (in thousands of dollars) made from investing in stocks "Y stock" and "Z stock" are modeled as the random variables Y and Z, respectively. Assume Y and Z are independent with respective probability density functions fy (t/) and fz (z) as shown below: You want to make as much money as possible, of course. Which stock would you buy? Suppose you have forgotten your ECE 314, so you decide to flip a coin to decide which stock to buy. What is the probability that you make more than $400? You decide to rely on some vague idea of the "law of averages", and thus by 20 independent stocks, each of which has the probability density function f_Y (y). What is the probability that you make more than $400 at least half of the time (greaterthanorequalto 10 times)?

Explanation / Answer

Solution

Back-up Theory

From the plot of fY(y) and fZ(z), we find that both are Uniform{a, b}, where

a = 0.25 and b = 0.5 for Y and a = 0.25 and b = 1 for Z. So,

fY(y) = 4dy …………………………………………………………………………..(1)

fZ(z) = (4/3)dx………………………………………………………………………..(2)

Now, to work out solution,

Note that the unit for both Y and Z are thousands of dollars.

=> all answers in fractional values must be multiplied by 1000 to get answer in terms of dollars.

Part (a)

To decide on the stock to invest, the criterion would be the expected value.

E(Y) = integral(0.25 to 0.5) of 4ydy = 4{(0.5)2 – (0.25)2}/2 = 0.375 = $375.

E(Z) = integral(0.25 to 1) of (4/3)zdz = (4/3){(1)2 – (0.25)2}/2 = 0.625 = $625.

Since 625 > 375, investment should be on Z-Stock ANSWER [Additional inputs: if detailed steps as above are not necessrily to be shown, the above answers could be obtained faster by the fact that mean of Uniform{a, b} is (a + b)/2.     

Part (b)

Since the decision on which stock to buy is based on the result of flipping a coin, each of Y and Z have equal chance of ½. Thus, probability of making more than $400 =

(½){P(Y > 400) + P(Z > 400)}

= (½){integral(0.4 to 0.5) of 4dy + integral(0.4 to 1) of (4/3)dz} [400 = 0.4 of 1000]

= (½)[(4x0.1) + {(4/3)(0.6}] = 0.6 ANSWER

Part (c)

If N = number of times Y-Stock would give more than $400 out of 20 times, then

N ~ B(20, p), where p = probability a Y-Stock would give more than $400, which has been already found to be 0.4 while working out Part (b) [in bold font]

So, we want P(N 10). This, using Excel Function, is found to be (1 – 0.7553)

= 0.2447 ANSWER

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