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?
(a) Then mean of the sample and the value of Z with an area of 10% in right tail.
Find the area under the standard normal curve between z = 1.6 and z = 2.6.
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
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.
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.
Rounded to the closest hundreth, the standard deviation for the set is approximately 19.52. Juxtaposed
With the 95% Confidence Interval for Mean, Median, and St Dev are as described above.
C. Using a table to compare the difference between problem #1 and problem #2, respectively, we can see the obvious differences between the optimal stocking quantity and daily expected profit figures.
normally distributed, the 95% of all values will be within 2 standard deviations from the mean.
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
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.