DSS 560: Midterm. Free Response portion
- East West Distributing is in the process of trying to determine where they should schedule next year’s production of a popular line of kitchen utensils that they distribute. Manufacturers in four different countries have submitted bids to East West. However, a pending trade bill in Congress will greatly affect the cost to East West due to proposed tariffs, favorable trading status, etc.
After careful analysis, East West has determined the following cost breakdown for the four manufacturers (in $1,000’s) based on whether or not the trade bill passes:
| Bill Passes | Bill Fails | |
| Country A | 260 | 210 |
| Country B | 320 | 160 |
| Country C | 240 | 240 |
| Country D | 275 | 210 |
| a. | If East West estimates that there is a 40% chance of the bill passing, which country should they choose for manufacturing? (12 points, 2 per EV and 4 for winner) |
| b. | Over what range of values for the “bill passing” will the solution in part (a) remain optimal? ( 4 points) To determine the range for probability values over which country B will be optimal, compare choosing country B versus countries A, C, and D. Let P = probability of the bill passing Country B would be preferred to country A as long as EV(B) < EV(A) Or P(320,000) + (1-p)160,000 < p(260,000) + (1-p)210,000 160,0000p + 160,000 < 50,000p + 210,000 110,000p < 50,000 P < .455 Country B would be preferred to country C as long as EV(B) < EV(C) P(320,000) + (1-p)160,000 < p(240,000) + (1-p)240,000 160,000p < 80,000 P < .50 Country B would be preferred to country D as long as EV(B) < EV(D) P(320,000) + (1-p)160,000 < p(275,000) + (1-p)210,000 P < .526 Thus, as long as the probability of the bill passing is less than .455, the East West should choose country B. |
- A manufacturing company is considering expanding its production capacity to meet a growing demand for its product line of air fresheners. The alternatives are to build a new plant, expand the old plant, or do nothing. The marketing department estimates a 35 percent probability of a market upturn, a 40 percent probability of a stable market, and a 25 percent probability of a market downturn. Georgia Swain, the firm’s capital appropriations analyst, estimates the following annual returns for these alternatives:
| Market Upturn | Stable Market | Market Downturn | |
| Build new plant | $690,000 | $(130,000) | $(150,000) |
| Expand old plant | 490,000 | (45,000) | (65,000) |
| Do nothing | 50,000 | 0 | (20,000) |
- What should the company do? (12 points/ 2 pts per calculation/decision)
Build New Plant
(0.35*690,000) + (0.4*-130,000) + (0.25*150,000) = $152,000
Expand Old Plant
(0.35*490,000) + (0.4*-45,000) + (0.25*-65,000) = $137,250
Do Nothing
(0.35*50,000) + (0.4*0) + (0.25*-20,000) = $12,500
The expected value of building a new plant exceed the value of expanding old plant and doing nothing.
Answer – The company should Build New Plant
- What returns will accrue to the company if your recommendation is followed? (4points)
Market Upturn = 0.35*690,000 = $241,500
Stable Market = 0.4*130,000 = ($52,000)
Market Downturn = 0.25 * 150,000 = ($37,500)
If the company decides to build a new plant, the return will either be
$241,500, ($52,000), or ($37,500)
- A calculus instructor uses computer aided instruction and allows students to take the midterm exam as many times as needed until a passing grade is obtained. Following is a record of the number of students in a class of 20 who took the test each number of times.
| Students | Number of Tests |
| 10 | 1 |
| 7 | 2 |
| 2 | 3 |
| 1 | 4 |
| a. | Use the relative frequency approach to construct a probability distribution and show that it satisfies the required condition. ( 4 points) |
| b. | Find the expected value of the number of tests taken. ( 4 points) |
| c. | Compute the variance. ( 4 points) |
| d. | Compute the standard deviation. ( 4 points) |
Solution
a). Relative Frequency = Frequency (students) / Total Population * 100
Total population = 10 + 7 + 2 +1 = 20
10 students = 10 / 20 * 100 = .50
7 students = 7 / 20 * 100 = .35
2 students = 2 / 20 * 100 = .10
1 student = 1 / 20 * 100 = .05
b). Probability of each event frequency * number of tests taken
(.50*1) + (.35*2) + (.10*3) + (.05*4) = 1.7
c). Variance
| X | f | fx | X2 | Fx2 |
| 1 | 0.5 | 0.5 | 1 | 0.5 |
| 2 | 0.35 | 0.7 | 4 | 1.4 |
| 3 | 0.10 | 0.3 | 9 | 0.9 |
| 4 | 0.05 | 0.2 | 16 | 0.8 |
| Sum = 1 | Sum = 1.7 | Sum = 3.6 |
Variance = (Σfx2 / Σf ) – (Σfx / Σf)2
3.6 / 1 – (1.7 / 1)2 = 0.71
d). Standard deviation = √Variance
Variance = 0.71
√0.71 = 0.8426
Answers table for question 3
| a. | Relative frequency | Number of tests | |
| .50 | 1 | ||
| .35 | 2 | ||
| .10 | 3 | ||
| .05 | 4 | ||
| Each probability is between 0 and 1 and the sum is 1.00. | |||
| b. | m = 1.7 | ||
| c. | s 2 = .71 | ||
| d. | s = .8426 |
