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Description

In the week following the first part of the assignment complete the following

peer review

(Links to an external site.)

Links to an external site.

of the other members of your assigned Canvas group. You

must include feedback

in the reviews that will enable the writers to improve their final submittal. As a minimum, address the following after considering the approach and justification of your peers’ work:

- Did your classmate use the appropriate header for their memorandum? Is the information in it correct? If not, specify what is incorrect. Is the subject line clear and concise? If not, offer an idea of what needs to be stated differently.
- Are there parts of their approaches and justifications that are unclear to you? Specify what part(s) of their memorandum, and why it is unclear. What changes would help to make it clearer?
- What aspects of their responses were particularly helpful in making their memorandum clear?
- Did your peers utilize a professional writing style as outlined in the assignment? Provide examples of how they did or did not meet this requirement.
- Did their use of visuals in their document help make their ideas more clear?

At the end of your review, offer an honest assessment of how your peers completed their assignments. Utilizing the exact language below, indicate whether they:

- Met minimum writing requirements
- Exceeded minimum writing requirements
- Did not meet minimum writing requirements

To complete reviews, reply to the posted work of each of your group members. Feedback MUST be actionable, that is something specific that they can continue to do, do more of, change or add to their work. For example “your analysis was good” or “your analysis was weak” does not provide feedback they can work with; Use “How to write a business memorandum” as a resource to determine specifically what makes the document strong, what may be missing that needs to be added or when a part of the document is not formatted correctly or achieving what it should. Specific feedback may look like:

- Including the provided data in text form is long and difficult to read, consider using a table instead.
- The second sentence of your analysis does not read clearly to me, consider rewording it.
- You omitted the recommendation in the summary, it should be included there.

GRADING RUBRIC – 10 points total

(assignments 2.3 and 2.4 will be graded together)

Memorandum header is included & formatted per reading – 1 point

Executive Summary is included – 1 point

Memorandum is divided into sections with titles – 1 point

A minimum of one graphic is included – 1 point

A conclusion is included – 1 point

Reviews are completed – 1 point

Reviews include actionable feedback – 2 points

The specific required evaluative language is included at the end of each review – 2 points

BUSINESS MEMORANDUM

Al Sawafi1

Alazhar Al Sawafi

To: Chief Financial Officer, Aileen Kitts

From: Chief Operating Officer, Lin Gerard

Date: 26/3/2019

Re: Vendor Recommendations

Business memorandum

Executive summary

Two vendors are trying to sell their equipment. The preferred diameter for the shaft is

2.5±0.05 inches. Therefore, the one that will have a diameter of 2.5.the vendor with shafts of

2.51 diameters is the most preferred.

Recommendations

For one to get the best option, it will be right to have metrics such as median, range or

quartiles. Besides, there will be need to have distribution fit statistics including fit goodness for

normal distribution. This will be fitted toward the data that will get added to Q-Q plots for

normal distribution that will get compared to the data.

Vendor A has a variance of 2.48 and 0.001 variance and a standard deviation of 0.032.

On the other hand, B has a mean of 2.51, variance of 0.002 and a standard deviation of 0.045. As

a result, they have similarities when it comes to magnitude difference. The desired level, in this

BUSINESS MEMORANDUM

Al Sawafi 2

case, is 2.50 making standard deviation to play a crucial role. In fact, this is because 0.045 is

somehow higher than the 0.032. As such, B will be the preference for this case.

The tolerance intervals, in this case, will be given by normal distribution x-stigma. As a

result, there will be provided by 1, 2, 3, 4, 5, 6. The increase i the interval is widening even

though the variability or the width will be more for the case of A than in the case of B.

The mean overall in terms of tolerance intervals will be somehow close toward the

desired level of 2.5. However, in terms of unbiasedness or accuracy, there is a similarity between

A and B and therefore no significant difference. All the same, there is a difference for

consistency, precision and variability between A and B. Since A has better accuracy, it becomes

a better choice than B. Therefore, the selection of vendor will not need additional information

case of B like it will be in case of A.

BUSINESS MEMORANDUM

Al Sawafi 3

Conclusion

Vendor A has a variance of 2.48 and 0.001 while B had a mean of 2.52 and variance of

0.002. Therefore, considering these facts, vendor B is the most preferred.

Memorandum

To:

Lin Gerard, COO

Aileen Kitts, CFO

From: Thomas Knudsen, Regional Manager

Date: March 26, 2019

Subject:

Machine Equipment Vendor Decision

Summary

The vendors Ace Machines and Best Machinery have each submitted information on their

equipment for assessment to determine which would be best for this company to do business

with. I have personally analyzed the data provided, and based on my analysis and this company’s

equipment needs, I recommend purchasing equipment from Ace Machines.

Analysis

To analyze the data, I used the provided means and variances of each vendor’s gearbox shaft

diameters to create normal distribution information. From this I was able to determine which

vendor produced the most product that falls within our acceptable parameters of 2.5±0.05in and

created curves to represent the data.

Distribution 1: Ace Machines

14

10.5

7

3.5

0

2.2

2.325

2.45

2.575

2.7

This first normal distribution curve represents the data from Ace Machines, which produces

shafts with a mean diameter of 2.48in and a standard deviation of 0.0316.

pg. !1

Vendor Recommendation

Distribution 2: Best Machinery

9

6.75

4.5

2.25

0

2.2

2.325

2.45

2.575

2.7

This second normal distribution curve represents the data from Best Machinery, which produces

shafts with a mean diameter of 2.51in and a standard deviation of 0.0447.

The data from Ace Machines is more concentrated around its mean value than that of Best

Machinery’s. This is represented graphically by a narrower peak in Ace Machinery’s distribution.

As a result of this concentration, more of the data falls within our working parameters.

Additional Considerations

I believe a key part of this decision should be the price each manufacturer is selling our company

their product at. Before a final decision is made, perhaps this should be looked into.

Conclusion

Based on the data provided by each vendor and my analysis of it, I recommend purchasing the

product of Ace Machines, as it conforms more closely with our company’s working

requirements.

pg. !2

Vendor Recommendation

Addendum 1: Mathematical Analysis

Normal Distribution data, Ace Machines:

Standard Shaft

Number of

Deviation Diameter Shafts

-4

2.3536

0.00423513

-3.75

2.3615

0.01115809

-3.5

2.3694

0.02761654

-3.25

2.3773

0.06421038

-3

2.3852

0.14024837

-2.75

2.3931

0.28777097

-2.5

2.401

0.55469305

-2.25

2.4089

1.00441936

-2

2.4168

1.70857489

-1.75

2.4247

2.7302949

-1.5

2.4326

4.09865809

-1.25

2.4405

5.78003435

-1

2.4484

7.65730141

-0.75

2.4563

9.52966557

-0.5

2.4642

11.1413078

-0.25

2.4721

12.2363328

0

2.48

12.6247557

0.25

2.4879

12.2363328

0.5

2.4958

11.1413078

0.75

2.5037

9.52966557

1

2.5116

7.65730141

1.25

2.5195

5.78003435

1.5

2.5274

4.09865809

pg. !3

Vendor Recommendation

1.75

2.5353

2.7302949

2

2.5432

1.70857489

2.25

2.5511

1.00441936

2.5

2.559

0.55469305

2.75

2.5669

0.28777097

3

2.5748

0.14024837

3.25

2.5827

0.06421038

3.5

2.5906

0.02761654

3.75

2.5985

0.01115809

4

2.6064

0.00423513

Normal Distribution data, Best Machinery:

Standard Shaft

Number of

Deviation Diameter Shafts

-4

2.3312

0.00299396

-3.75

2.342375

0.00788805

-3.5

2.35355

0.0195231

-3.25

2.364725

0.04539257

-3

2.3759

0.0991465

-2.75

2.387075

0.2034354

-2.5

2.39825

0.392132

-2.25

2.409425

0.71005933

-2

2.4206

1.2078516

-1.75

2.431775

1.93014136

-1.5

2.44295

2.89748536

-1.25

2.454125

4.08610929

pg. !4

Vendor Recommendation

-1

2.4653

5.41321531

-0.75

2.476475

6.73685531

-0.5

2.48765

7.87618181

-0.25

2.498825

8.65029344

0

2.51

8.92488323

0.25

2.521175

8.65029344

0.5

2.53235

7.87618181

0.75

2.543525

6.73685531

1

2.5547

5.41321531

1.25

2.565875

4.08610929

1.5

2.57705

2.89748536

1.75

2.588225

1.93014136

2

2.5994

1.2078516

2.25

2.610575

0.71005933

2.5

2.62175

0.392132

2.75

2.632925

0.2034354

3

2.6441

0.0991465

3.25

2.655275

0.04539257

3.5

2.66645

0.0195231

3.75

2.677625

0.00788805

4

2.6888

0.00299396

pg. !5

Vendor Recommendation

____________________________________________________________________________________

Memorandum

To: Lin Gerard, COO, Automotive Parts Corporation

CC: Aileen Kitts, CFO, Automotive Parts Corporation

From: Ibrahim Imran, Project Lead, Automotive Parts Corporation

Date: March 20, 2019

Subject: Recommendation for the purchase of gearbox shafts

Executive Summary

To decide upon a vendor for the purchase of shafts for gearboxes, 100 samples from each vendor

were analyzed. Using the mean and normal distribution of diameters for each, the probability of

the shaft produced being within the acceptable limits of diameter was calculated. Findings

indicate that steel shafts from vendor A are closer to the desired length of 2.5±0.05, with fewer

anomalies.

Analysis

To compare the performances of both vendors, the mean and normal distribution of the diameters

was used to plot a distribution curve. The curves (Figure 1.0) show the distribution of values

around the mean. For vendor A, the steeper curve represents lesser deviation, which is more

ideal.

Figure 1.0

Using the same values, the probability of the shafts being within the desirable limits was

calculated for each vendor. Values calculated were 0.815 and 0.725 for vendors A and B

respectively, indicating that vendor A produced a larger number of acceptable shafts. Thus,

purchasing equipment from vendor A would be more appropriate.

Additional Considerations

Further input, for example, production and maintenance costs must be taken into account when

making a transaction with the vendor. Useful life of the equipment, and reliability must also be

monitored over time. As an alternative analysis option, output capacity of the vendor is another

way to help our company with supply and demand.

Conclusion

As indicated by the findings, vendor A is better suited to our current needs, with lesser deviation

and more ideal equipment produced. This is proved by the plotted graphs as well as the

probability calculations.

Recommendation for the purchase of gearbox shafts

1

Addendum 1: Statistical Calculations

Data:

Probability Equations

Ace Machines

!P (2.45 ≤ x ≤ 2.55)

P (2.45 ≤ x ≤ 2.55)

!φ

(

0.001 )

2.55 − 2.48

−φ

0.815

(

Best Machinery

0.001 )

2.45 − 2.48

!φ

(

0.002 )

2.55 − 251

−φ

(

0.725

0.002 )

2.45 − 2.51

Memorandum

To:

Lin Gerard, COO Automotive Parts Corporation

CC:

Aileen Kitts, CFO Automotive Parts Company

From: Ashton Rodriquez, Senior Engineer, Automotive Parts Company

Date: March 22, 2019

Subject: Recommended Vendor for Automotive Shaft

Summary

I have been assigned with making the decision between Ace Machines and Best Machinery to

provide us with equipment for our shaft-machining task. As the Senior Engineer it is my

responsibility to ensure that this company receives the highest quality products that allow for us

to produce the best parts. To ensure that we receive high quality products I have asked both

companies, Ace Machines and Best Machinery, to provide me with statistical data on their

respective machines to help me decide which company’s machine would be more reliable in

making our 2.5inch diameter shaft. After completing the statistical analysis, I found that Best

Machinery is the better choice due to its closer expected value to 2.5inches. This would provide

us with a closer desired diameter on a more consistent and constant basis.

Analysis

Based upon the research completed and the calculations from the data provided, the best

company for our company to collaborate with on our steel shafts is Best Machinery. Over the

100 samples obtained, Best Machinery was able to create shafts with a statistical mean of 2.51

and variance of 0.002. The competing company Ace Machines produced a statistical mean of

2.48 and a variance of 0.001.

As stated above the statistical mean of Best Machinery was 2.51 and the standard variance was

0.002, while the statistical mean for Ace Machines was 2.48 and the standard variance was

0.001. The acceptable dimensions for the shaft are between 2.45 and 2.55 inches in diameter with

the most desirable product being exactly 2.5 inches. From this we can decipher that the standard

variance for Best Machinery appears to be better than the standard variance for Ace Machines.

This, however is not enough for me to make a decision and therefore I went further with the

provided data and completed further analysis.

I went on to calculate the probability of each equipment producing a shaft with acceptable

dimensions. I calculated the probability for each machine separately given the allowed tolerances

Vendor Recommendation

of .05inches and found that Best Machinery has a probability of 0.0725 while Ace Machines has

a probability of 0.815. With probabilities similar to one another I decided to perform further test

in order to inform my decision. I decided to look at the normal distribution graphs. The normal

distribution graph for Best Machinery proved to be better and suggested that the data values of

Machinery will not fall far from mean. After completing all of these calculations, I concluded

that Best Machinery would provide a higher quality product.

Normal Distrubution

400

300

200

100

0

2.44

2.468

2.495

2.523

2.55

Diameter (IN)

Ace Machinery

Best Machinery

Additional Considerations

Based on my current calculations and the data provided thus far I believe Best machinery is the

best choice of the two companies. However, additional information would be useful to help

differentiate the two companies from one another and allow me to make a decision that would be

best for our company. Information that could be provided includes the median, outliers and the

range. The median would help to justify if the majority of the shafts fell above or below the mean

diameter. The range would be used to display how many shafts are inconsistent with the ideal

diameter, and the outliers would help to further justify this. Another piece of information that

would be valuable to obtain is the cost of the machinery as it is important that our company stays

within the means of our budget. In order to make a more informed and justified decision I will

require this information from our vendors.

Conclusion

After completing my calculations, I have concluded that Best Machinery will provide a higher

quality of equipment for our company. The calculations prove that Best Machinery’s equipment

will provide us with a more consistent 2.5inch shaft based on its standard deviance, normal

Vendor Recommendation

distribution curve and probability equations. Therefore, I believe that it would be in the best

interest of the company to do business with Best Machinery.

Vendor Recommendation

Addendum 1: Statistical Calculations

Data:

Dimensions

!2.5 ± 0.05 = [2.45, 2.55]

Ace Machines

Tolerance

2.45

Mean

2.48

Variance

0.001

Normal Distribution

1

Standard Deviation

0.03162278

Best Machinery

Tolerance

2.45

Mean

2.51

Variance

0.002

Normal Distribution

1

Standard Deviation

0.04472136

Probability Equations

Ace Machines

!P (2.45 ≤ x ≤ 2.55)

!φ

(

P (2.45 ≤ x ≤ 2.55)

Vendor Recommendation

0.001 )

2.55 − 2.48

−φ

0.815

(

Best Machinery

0.001 )

2.45 − 2.48

!φ

(

0.002 )

2.55 − 251

−φ

(

0.725

0.002 )

2.45 − 2.51

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