Mat 540 Week 4 Case Study

The sample size is based on the 20 day sample. The probability of the event (.03765) was determined in Question #2, the cumulative probability of Four-D rejecting a shipment (based on a sample of 10).

Answer: The standard deviation can be calculated by subtracting the expected return from the actual return for each year and squaring the results. The squares are summed, and divided by the number of observances minus 1. The square root of that result is the standard deviation.

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

The mean birth weight of infants born at a certain hospital in the month of April was 128 oz. with a standard deviation of 10.2 oz. Which of the following is a correct interpretation of standard deviation?

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

What is the probability that a randomly selected order will require between three and six days?

Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of

We started the case analysis by the normality test for the current data to check if the data is normally distributed or not. Our objective is to calculate the current Total Cost and then come up with a better solution to lower the cost.

28. The Polo Development Firm is building a shopping center. It has informed renters that their rental spaces will be ready for occupancy in 19 months.  If the expected time until the shopping center is completed is estimated to be 14 months, with a standard deviation of 4 months, what is the probability that the renters will not be able to occupy in 19 months?

Compute the probability of a stock-out for the order quantities suggested by members of the management team.

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)/

2) Compute the standard deviation for each of the four samples. Does the assumption of .21 for the population standard deviation appear reasonable?

SUPPLY CHAIN MANAGEMENT CASE 2 Quantitative Practice Problems MSc12, 11 May 2004 Question 1. a) Which of the two products should the GAP carry at the stores and which at the central warehouse for the online channel? Khaki pants Information given : The mean μ 800 Standard deviation σ100 Cost$30,- Holding cost 25% Lead time 4 weeks Target cycle service level95% z-value 1,645 Safety stock Khaki pants Service level * σ * √( Lead time) 1,645 * 100 * √(4) Safety stock for Khaki pants = 329 Cost Khaki pants Holding cost * safety stock 0,25 * $30,- * 329 Total holding cost of Khaki pants is $2.467,50 Cashmere sweaters Information given : The mean μ 50 Standard deviation σ50 Cost$100,- Holding cost 25% Lead time 4 weeks Target cycle service

S(t) = Pr(some one will adopt at time t| he has not adopted the product till time t)* (Size of the segment that has not adopted the product till time t)

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by: