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S-Problems/Soutions
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Problems

1. The Orange Octagon Discount Department Store sells snowblowers during the winter months. Snowblowers cost $175.00 each and sell for $395.00. The manager knows that the volume of sales is not large, as indicated by the following data on the number of snowblowers that were sold in each week during the past two selling seasons, which contained 20 weeks each: 0, 1, 3, 1, 2, 0, 3, 4, 2, 1, 4, 2, 0, 2, 0, 0, 2, 0, 1. 3, 1, 1, 1, 3, 3, 2, 2, 0, 4, 1, l, 0, 2, 2, 0, 1, 3, 1, 3 snowblowers. The manager has decided to start the season with four snowblowers on hand. Only snowblowers on hand at the beginning of the week can be sold. Additional snowblowers can be ordered from the supplier, with a two-week lead time, and the manager plans to order two of them whenever the quantity on hand falls to two or less at the end of the week. Use the random numbers that follow to simulate this policy for 10 weeks: 491, 631, 139, 502, 006, 856, 904, 91, 821, 757.

  1. What is the empirical probability distribution for the weekly sales of snowblowers?
  2. What random numbers go with each value of the random variable?
  3. Construct a table in the style of Example 19S-2b(2) in your textbook, showing the history of the inventory of snowblowers for the 10-week period.
  4. How many snowblowers were sold? How much gross margin was made?
  5. How many sales were lost? What was the dollar amount of the loss?
  6. What is your evaluation of the policy?

2. The number of ambulance calls can be represented by a Poisson distribution with a mean of four per day. Obtain the Poisson distribution from Table C in the Appendix and use the following random numbers to simulate six days of ambulance calls: 005, 820, 471, 187, 177, 341.

  1. a. Which random numbers go with each value of the random variable?
  2. Construct a table in the style of Solved Problem I in your textbook to show the history of the six days.

3. The daily sales of unleaded gasoline at the self-serve pump can be represented by a normal curve with a mean of 300 gallons and a standard deviation of 30 gallons. Use row 8, columns 1, 2, 3, 4, in Table 19S-2 in your textbook to simulate four days of gasoline sales.

4. (One step beyond) Use an EXCEL spreadsheet to simulate ten days of gasoline sales for the self-serve pump in problem #3. This will require the use of EXCEL's Analysis ToolPak VBA. To see if it is installed in your system, click on TOOLS and look for Data Analysis at the bottom of the list. If it is there, you are ready to proceed. If not, click on Add-Ins and select Analysis ToolPak VBA. Data Analysis should now appear in your TOOLS menu.

  1. How should the problem be entered on the spreadsheet?
  2. What are the daily sales?

5. The length of time which a customer spends pumping gasoline at the self-serve pump can be represented by a negative exponential distribution with a mean of 2 minutes. Use these four random numbers to simulate four customers pumping gasoline: 84275, 07523, 33362, 64270.

  1. What is the value of <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image398::/sites/dl/free/0072443901/24520/Image398.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image398 (0.0K)</a>Image398 ?
  2. What are the four times?
6. (One step beyond) The plant manager believes that in the production of fluorescent light tubes, 6 percent will turn out to be defective; the remainder will be satisfactory. Use rows 10-12 in Table 19S-1 in your textbook to simulate the production of 36 fluorescent light tubes.
  1. Which random numbers go with each type of fluorescent tube?
  2. What is the sequence of satisfactory (S) and defective (D) tubes.
  3. What percent of the 36 tubes are defective?
  4. Is the plant manager mistaken in the 6% figure?

 

1. a, b.

x

f

p

Cumulative Probability

Random

Numbers

 

0

1

2

3

4

9

12

9

7

3

.225

.300

.225

.175

.075

.225

.525

.750

.925

1.000

001-225

226-525

526 750

751-925

926-000

 

Totals

40

1.000

  

 

 

c.

Week

Random

Number

Demand

Sales

Quantity Ordered

Quantity Received

 

1

2

3

4

5

6

7

8

9

10

491

631

139

502

006

856

904

091

821

757

1*

2

0

1

0

3

3

0

3

3

1

2

0

1

0

3

3

0

0***

2

0

2

2**

2

0

0

2

2

0

2

0

0

0

2

2

2

0

0

2

2

 

Totals

 

16

12

  

*491 comes between 226 and 525 and designates X = 1.

**Since the inventory contains 1 snowblower at the end of the third week.

***Because the beginning inventory is 0.

 

Week

Beginning

Inventory

Sales

Receipts

Ending

Inventory

 

1

2

3

4

5

6

7

8

9

10

4

3

1

1

2

4

3

0

0

2

1

2

0

1

0

3

3

0

0

2

0

0

0

2

2

2

0

0

2

2

3

1

1

2

4

3

0

0

2

2

d. The total of the sales column is 12 snowblowers. The gross margin = 12(395 - 175) = $2,640.e. The difference between the total of the demand column and the total of the sales column is 4 snowblowers. The gross margin lost = 4(395 - 175) = $880.
f. While 10 weeks is a very small sample, a tentative hypothesis is that more snowblowers could be sold if the order quantity were increased.

 

2. a.

 

Number

Cumulative

Random

 

of Calls

Probability

Number Range

 

0

.018

001 018

 

1

.092

019 - 092

 

2

.238

093 - 238

 

3

.433

239 - 433

 

4

.629

434 - 629

 

5

.785

630 - 785

 

6

.889

786 - 889

 

7

.949

890 - 949

 

8

.979

950 - 979

 

9

.992

980 - 992

 

10

.997

993 - 997

 

11

.999

998 - 999

 

12

1.000

1,000

b.

 

Random

Number

 

Day

Number

of Calls

 

1

005

0*

 

2

820

6

 

3

471

4

 

4

187

2

 

5

177

2

 

6

341

3

*005 comes between 001 and 018 and designates X = 0.

3.

Day

Random

Number

Sales

 

1

2

3

4

2.40

0.38

- 0.15

- 1.04

372.00*

311.40

295.50

268.80

*Sales = Mean + (R.N.)(standard deviation) = 300 + (2.40)(30) = 372.00 gal.

  1. a. Click on TOOLS; DATA ANALYSIS; RANDOM NUMBER GENERATOR. Number of Variables = 1; Number of Random Numbers = 10; Distribution = Normal; Mean = 300; Standard Deviation = 30; Random Seed = 1776*; Output Range = C3:C12.

b.

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*NOTE: The choice of a value for the Seed Number is arbitrary. Experiment with some other seeds, such as your birthday in the form of 62180, perhaps.

5. a. The service time: 1/l = 2 minutes, and l = .50.

b. p(T) = exp(- 0.5t), and t = - (1/l)(ln(.RN)).

c.

Week

Random

Number

ln(.RN)

Pumping Time (min.)*

 

1

2

3

4

.84275

.07523

.33362

.64270

- 0.1711

- 2.5872

- 1.0978

- 0.4421

0.3422

5.1744

2.1958

0.8842

*Pumping Time = - 2×ln(.RN)

6. a.

Type of

Tube

Random

Number Range

 

Satisfactory

Defective

01-94

95-100

b. Reading Table 19S-1 horizontally: S S S S S S S S S S S S S S S S S S S S S S S D S S S S S S S S S S S S.
c. p = 1/36 = 2.78%.
d. Not necessarily; the 6% defective tubes will occur over the long run but not for any very short period. (For the statistician: the standard error of the sample percent, if the true percent of defective tubes is, in fact, 6% is 3.96% and it is easy to draw a sample with p = 2.78% from a population with 6% defective items.)








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