The explanation for the answer to the given question is as follows: Hence, the current stock available is 180 units. To find out the probability of the company running out of the stock of 180 units within a time of next 3 weeks, let’s take ' z ' as a value that should be checked in the standard deviation table in order to find the probability... Let us first find the value of ' z ', for which we should know the mean demand of the period of 3 weeks: So to do this we can assume that the demand for week one is 65 units, for week two is 50 units and for week three is 45 units. Mean Demand of the period of 3 weeks is equal to the Sum of mean demands of per week that is,60+55+45 = 160 Now, let us determine the standard deviation for the period of 3 weeks, as it is mentioned in question. So, lets instead find variance of per week, and take …show more content…
Hence, First week difference = 60 - 55 = 5. Second week difference = 55 - 45 = 10. Third week difference = 45 - 60 = -15. Variance for week one is 25. Variance for week two is 100.Variance for week three is 225. (Since square of any negative number gives us a positive number). Hence, variance of 3 weeks is equal to the Sum variance of per week that is, 25+100+225=350 Standard Deviation of 3 weeks is equal to the Square root of the variance of 3 weeks that is, Square root of 350 = 18.71. Now that we have the found the mean and standard deviation of 3 weeks, let’s find ' z ' value by using the formula:-Z = ( total value - total mean ) / standard deviation that is,Z = 180-160/18.71 = 1.07. Now we have to check the value of 1.07 in the standard deviation table, On checking we get the value as 85.77% this is probability that the units will be sufficient for the period of next 3 weeks, Hence, the probability of the company running out of units = 100 - 85.77. Therefore, the probability of the company running out of units is
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.
We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:
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?
Let’s assume you have taken 1000 samples of size 64 each from a normally distributed population. Calculate the standard deviation of the sample means if the population’s variance is 49.
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?
What is the probability that a randomly selected order will require between three and six days?
5. Find the sample variance s2 for the following sample data. Round your answer to the nearest hundredth.
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
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.
Afterwards use the standard deviation equation to plug in your number's to get the standard deviation
30. The manager of the local National Video Store sells videocassette recorders at discount prices. If the store does not have a video recorder in stock when a customer wants to buy one, it will lose the sale because the customer will purchase a recorder from one of the many local competitors. The problem is that the cost of renting warehouse space to keep enough recorders in inventory to meet all demand is excessively high. The manager has determined that if 90% of customer demand for recorders can be met, then the combined cost of lost sales and inventory will be minimized. The manager has estimated that monthly demand for recorders is normally distributed, with a mean of 180 recorders and a standard deviation of 60. Determine the number of recorders the manager should order each month to meet 90% of customer demand.
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)