ECON 221 Lecture Notes - Lecture 4: Standard Deviation
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The following table shows the probability distribution of the number of days (x) to complete a certain type of construction project by a contractor. | |||||||||
x | f(x) | ||||||||
1 | 0.05 | ||||||||
2 | 0.20 | ||||||||
3 | 0.35 | ||||||||
4 | 0.30 | ||||||||
5 | 0.10 | ||||||||
3 | The probability that it will take less than 3 days to complete a randomly selected project is ______. | ||||||||
a | 0.60 | ||||||||
b | 0.45 | ||||||||
c | 0.35 | ||||||||
d | 0.25 | ||||||||
4 | The standard deviation of the days to complete a project is _______. (Rounded to 2 decimal points.) | ||||||||
a | 1.03 | ||||||||
b | 1.24 | ||||||||
c | 1.35 | ||||||||
d | 1.41 | ||||||||
5 | The contractor's project cost is made up of a fixed cost of $20,000, plus $2,000 for each day taken to complete the project. Expected value of total project cost is. | ||||||||
a | $25,300 | ||||||||
b | $26,400 | ||||||||
c | $27,500 | ||||||||
d | $28,600 | ||||||||
6 | In the previous question the standard deviation of total project cost is: | ||||||||
a | $22,060 | ||||||||
b | $22,828 | ||||||||
c | $2,060 | ||||||||
d | $2,828 | ||||||||
7 | If you toss a balanced coin 20 times, the probability of getting 10 tails is: f(x = 10) = _______. | ||||||||
a | 0.5000 | ||||||||
b | 0.2461 | ||||||||
c | 0.1762 | ||||||||
d | 0.1026 | ||||||||
1.) Two projects A and B have the potential uniform annual benefits and their associated probabilities of occurrence as shown below.
EUAB | Probability | EUAB | Probability |
$1000 | 0.1 | $1500 | 0.2 |
$2000 | 0.3 | $2500 | 0.4 |
$3000 | 0.4 | $3500 | 0.3 |
$4000 | 0.2 | $4500 | 0.1 |
The expected value of these project are:
a. $1,650 and $1,987
b. $2,700 and $2,800
c. $3,650 and $3,245
d. $2,210 and $2,151
2.) An investment opportunity has the potential of generating yearly revenues with the associated probabilities for the next five years as shown below. The salvage value at the end of five years is 0. The potential revenue in any given year is independent of any other year. What are the mean and standard deviation of the present worth, using an interest rate of 12%?
Potential Revenue, $ Probability
20,000 0.30
30,000 0.40
50,000 0.20
60,000 0.10
Select one:
a. $222,480 and $81,503
b. $235,652 and $65,238
c. $111,070 and $38,802
d. $122,570 and $48,901
3.) What is the expected net present worth of the following project assuming a life of 10 years? The project has absolutely no salvage value at the end of the useful life. The company uses an interest rate of 12% to evaluate this project.
Initial Cost,$ | Probability | Annual Revenue, $ | Probability |
30,000 | 0.15 | 7,000 | 0.10 |
40,000 | 0.20 | 9,000 | 0.25 |
60,000 | 0.30 | 10,000 | 0.30 |
80,000 | 0.25 | 12,000 | 0.30 |
100,000 | 0.10 | 15,000 | 0.05 |
Select one:
a. $3,705
b. -$2,305
c. -$22,305
d. $4,506
4.) An investment of $18,000 is expected to generate annual revenue of $8.000 throughout the life of the investment. The risk is based on the life of the investment. The estimate of the probability for the duration of the investment is given in the table below.
Life, Years | 3 | 4 | 5 | 6 |
Probability | 0.1 | 0.4 | 0.3 | 0.2 |
No matter what the life of the investment might be, there will be no salvage value. Using a value of 15% MARR, what is the risk (standard deviation) associated with this investment?
a. $3.598
b. $4,536
c. $2,528
5.) A city engineer has compiled the following data on a flood damage project. to control the flood, the construction of a small dam will cost $100,000. Based on the estimated damage data collected, what is the expected damage?
Damage Estimate, $ | Probability |
200,000 | 0.30 |
100,000 | 0.50 |
50,000 | 0.10 |
0 | 0.10 |
Select one:
a. $225,550
b. $365,800
c. $215,005
d. $115,000
1. You are given only three quarterly seasonal indices and quarterly seasonally adjusted data for the entire year. What is the raw data value for Q4? Raw data is not adjusted for seasonality.
Quarter Seasonal Index Seasonally Adjusted Data
Q1 .80 295
Q2 .85 299
Q3 1.15 270
Q4 --- 271
2. One model of exponential smoothing will provide almost the same forecast as a liner trend method. What are linear trend intercept and slope counterparts for exponential smoothing?
A. Alpha and Delta
B. Delta and Gamma
C. Alpha and Gamma
D. Standard Deviation and Mean
3. When performing correlation analysis what is the null hypothesis? What measure in Minitab is used to test it and to be 95% confident in the significance of correlation coefficient.
A. Ho: r = .05 p < .5
B. Ho: r = 0 p >.05
C. Ho: r ? 0 p?.05
D. Ho: r = 0 p?.05
In decomposition what does the cycle factor (CF) of .80 represent for a monthly forecast estimate of a Y variable? |
A. The estimated value is 80% of the average monthly seasonal estimate.
B. The estimate is .80 of the forecasted Y trend value.
C. The estimated value is .80 of the historical average CMA values.
D. The estimated value has 20% more variation than the average historical Y data values.
5. A Wendy's franchise owner notes that the sales per store has fallen below the stated national Wendy's outlet average of $1,368,000. He asserts a change has occurred that reduced the fast food eating habits of Americans. What is his hypothesis (H1) and what type of test for significance must be applied? |
A. H1: u ? $1,368,000 A one-tailed t-test to the left.
B. H1: u = $1,368,000 A two-tailed t-test.
C. H1: u < $1,368,000 A one-tailed t-test to the left.
D. H1: p < $1,368,000 A one-tailed test to the right
A. The rejection region and the t-table value generally gets smaller for sample size below 31. |
A. Yes. The data are significantly correlated through the 12th lag. C. No. Only the 12 lag period is not correlated. D. You cannot tell since the number of sample observations is not provided. E. The p-value is above .05 so the data is correlated. |
A. Type 2 error |
A. Yes. They move in the same direction as statistical significance. |
A. The weight cannot be calculated since the data observation is not given. |
A. Yes. The correlation coefficient is .873 that is greater than .05. |
A. Yes, since the residuals randomly vary in magnitude. |
A. -101.0 |
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