Bell computer forecast company week 3
Bell computer forecast company week 3
The assignment is to look over two separate cases using the excel spreadsheet. I am to then take and interpret the data to identify and discuss the decisions based on the statistics within. One case will be determined by the mean and standard deviation and the second normal distribution. Once the information is determined my paper will discuss the factors.
Case one: Bell computer company
The mean of the expected value is the sum of the numbers or probabilities that have been factored with the expected outcomes. Also known as the average of long run occurrences (Black). The numbers within are not acquired through using a finance formula like a future value formula that may give you the amount of a annuity payment. This is also true for the return on investment formula, which is ROI= total gain (or loss)/ total cost of investment (CHERYL CLEAVES). No this is a statistical formula used to average the gain or expected gain of probabilities in an occurrence, the formula being; µ= ∑ (x) = ∑ [ x* P (x) ] (Black). The goal in this scenario is to assess risk and describe why a low risk or hgh venture is preferred. Having shown the expected value formula the following paragraph will explain its use and why the proposed outcome is preferred.
Case One: The mean of the expected value and explanation
The mean of the expected value formula µ= ∑ (x) = ∑ [ x* P (x) ] (Black) is different from other ratio formulas as explained in the previous paragraph. If I were to search for the average cost of a product I again would negate the purpose of this report. Explaining that AVC (average variable cost) = Total variable cost/ Q (output) (McConnell & stanley Brue) will not lead to a discovery of why the Bell company should take a low-risk venture. It can be explained with the reasoning and summary of my gathered notes on the expected value formula.
µ= ∑ (x) = ∑ [ x* P (x)] (Black)says that the average of long run occurrences can be shown in that way. Using the excel spreadsheet when you look it over it gives values and percentages. A brief summary will look this way:
The idea here is to give an idea of what you may find looking over the excel spreadsheet. However, from this I can show how I have deduced using the mean of expected value formula that taking the Medium expansion will lead to the best outcome. Simply put no company likes volatility or the chance that they may lose money. We can see in most business classes that it may cost money to make money and here we can see that making x = an outcome and P probability of that outcome. May look tis way ∑[ x*P(x)] or Medium (150) * .50 = 75. This in my opinion is the safest and most profitable of each outcome. Showing the computations on the excel spreadsheet.
Case 2 Kyle’s bits and bytes Using the normal distributionformula as P(x>.06| µ=200 and σ =30 how would I solve the question? The problem may be deduced for Kyle’s bits and bytes in this way; you want x to represent the number of printer’s available form week to week. With variable demand or the average being known as 30 or σ =30 (standard deviation). I would then simplify the matter and change x to any possible outcome now x -µ =? If I used the z score formula, then it may look like this how can I now deduce the inventory point I chose the 6% of the inventory. I see that 6% of 200 is 12, I let that represent x in the formula. Then I subtracted µ or 200 as 12-200= 188. Lastly, I divided the standard deviation σ = 30 by the sum of 188. With 5.33 (rather 5), So I looked over the Z score .4999997. So I think the reorder point should be every 5 or 6 sales or when there is at least halve of the starting inventory (no matter the number) because of the z score .4999997 or 50%.
I would want my numbers to be accurate in my business. I think that having the security of understanding would give greater confidence in my abilities. In this report I feel that my accuracy for what this is may suffice in what I used it for but there needs to be a more finite definition to my work. I feel that having used the table and the other formulas for both cases leaves me with the satisfaction of possibly doing the work right but as a manager I would want my work checked over and applied. The familiarity in doing the statistics may one day grow. At this point seeing that the z score leaves me with the sense of having done the work and seeing that no matter what I plugged in for x, I should still be able to see a point or figure. Along with the z score or table itself that gave me confidence in having used the functions properly. I used the z score as a percentage which seemed like it’s best use.
Typos that I may have corrected may have appeared in this one, Sorry about that.