Case Study Specialty Toys Essay

Rounded to the closest hundreth, the standard deviation for the set is approximately 19.52. Juxtaposed

Standard Deviation for the mean column is 0.476Standard Deviation for the median column is 0.754Standard deviation for the mean column has least variability

In our second assumption, instead of using the cost of goods per cases in 1986, we try to use the percentage it counts in the total expenses which is 50.4% and to find the sales needed to break-even. The detail of the calculation is shown in the answer for questions d. The result is that 95,635, a little bit higher than the estimated sales of 90,000.

A. The simulated function given in the Excel spreadsheet “Hamptonshire Express: Problem_#1” allows the user to find the optimal quantity of newspapers to be stocked at the newly formed Hamptonshire Express Daily Newspaper. Anna Sheen estimated the daily demand of newspapers to be on a normal standard distribution; stating that daily demand will have a mean of 500 newspapers per day with a standard deviation of 100 newspapers per day. Using the function provided, the optimal stocking quantity, which maximizes expected profit, is determined to be approximately 584 newspapers. If 584 newspapers were to be ordered, Hamptonshire Express will net an

The mean, average of the population, the standard deviation, deviation in the sample, and a histogram, a graph to show the percentage of the population, was prepared. The means calculated as follows: blue 0.2072, orange 0.2271, green 0.1852, yellow 0.1165, red 0.1448, and brown 0.1192. The histogram was bell shaped with three outliers, numbers outside the given range, of 55 candies, or one bag.

9. According to a survey of the top 10 employers in a major city in the Midwest, a worker spends an average of 413 minutes a day on the job. Suppose the standard deviation is 26.8 minutes and the time spent is approximately a normal distribution. What are the times that approximately 95.45% of all workers will fall? A. [387.5 438.5]B. [386.2 439.8]C. [372.8 453.2]D. [359.4 466.6]E. [332.6 493.4]

Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000 units, and best case in which sales = 30,000 units.

Ref: CQI Study Guide Q13 21. In a normal curve, approximately what percentage of the area is included within 3 standard deviations of the mean? a. 50.0% b. 66.6% c. 95.0% Page 4 of 32

An intelligence test for which the scores are normally distributed has a mean of 100 and a standard deviation of 15. Use this information to describe how the scores are distributed.

Afterwards use the standard deviation equation to plug in your number's to get the standard deviation

Since the expected demand is 2000, thus, the mean µ is 2000. Through Excel, we get the z value given a 95% probability is 1.96. Thus, we have: z= (x-µ)/ σ=(30000-20000)/

Under either scenario, there is a 75% chance that the company will achieve the $4 million target profit. When demand is 150,000, then the profit is going to be below the $4 million mark in either case. There is a 25% that the demand will be 150,000. There is a 75% chance that the demand is going to be 180,000 or higher. At 180,000 or higher, the company would generate a profit in excess of $4 million under either scenario. Therefore, there is a 75% chance that the company is

The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.

values under the curve lie within one standard deviation of the mean and 95% lie within two standard deviations3. This is good for intervallic continuous variables.

Standard deviation based on the available information (January to June) is 41.83 or approximately 42.