time gap is called “lead time.” From past experience‚ the materials manager notes that the company’s demand for glue during the uncertain lead time is normally distributed with a mean of 187.6 gallons and a standard deviation of 12.4 gallons. The company follows a policy of placing an order when the glue stock falls to a predetermined value called the “reorder point.” Note that if the reorder point is x gallons and the demand during lead time exceeds x gallons‚ the glue would go “stock-out” and the
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of a frequency distribution c. Indicates the typical value of a frequency distribution d. Represents the midpoint of a frequency distribution e. Comparing quantities where x and y are completely independent of each other or x can be included in y f. Represents the simplest measure of spread (or variability) g. Indicates the most frequent observation in a frequency distribution h. Represents a theoretical family of distributions that may have any mean or any standard deviation i. Indicates how
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created on a graph when using a frequency distribution method for a set of data‚ splitting the mean symmetrically. There is a big difference between standard deviation and the bell curve! Standard deviation shows the difference in variation from the average; the bell curve‚ also normal distribution or Gaussian distribution‚ shows the standard deviation and is created by the normal or equal distribution of the mean among either half. The bell curve is an important distribution pattern occurring in
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be at least 10‚000 units. Ignore price differences among styles in your initial analysis. In order to decide how many units of each style Wally should make‚ we should think about their order’s range‚ we ignore price differences among styles so we need to think about the maximum order and minimum order for each style. First of all‚ we need to calculate the stock out probability based on benefit percentage and risk percentage. According to the case‚ Obermeyer earned 24 percent of wholesale price
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Problem Sheet - I 1. Researcher conducted by a tobacco company indicates that the relative frequency distribution of tar content of its newly developed low-tar cigarette has a mean equal to 3.9 milligrams of tar per cigarette and a standard deviation equal to 1.0 milligram. Suppose a sample of 100 low-tar cigarettes is randomly selected from a day’s production and the tar content is measured in each. Assuming that the tobacco company’s claim is true‚ what is the probability that the mean
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examination. The distribution of marks is shown in the following grouped frequency table. Marks|1–10|11–20|21–30|31–40|41–50|51–60|61–70|71–80|81–90|91–100| Number of candidates|15|50|100|170|260|220|90|45|30|20| (a) Copy and complete the following table‚ which presents the above data as a cumulative frequency distribution. (3) Mark|£10|£20|£30|£40|£50|£60|£70|£80|£90|£100| Number of candidates|15|65|||||905|||| (b) Draw a cumulative frequency graph of the distribution‚ using a scale of
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histogram of the 1990 returns. (ii) Produce a histogram of the 1998 returns. (iii) Find the mean‚ median‚ range and standard deviation for the 1990 returns. Annual Returns % (1990) Mean 12.91865979 Median 11.38 Standard Deviation 9.297513067 Range 75.01 (iv) Repeat part (iii) for the 1998 returns. Annual Returns % (1998) Mean 6.355463918 Median 5.4 Standard Deviation 5.170830853 Range 42.76 (v) Which was the better year for investors? • 1990 was the better year for investors in
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1.0 Introduction This report will cover the distribution of final exam results for BSB123 and what factors influence the results. Factors that will be considered are the gender of the student‚ whether the student is studying a double or single degree‚ the results from the weekly quiz’s and the grade achieved on the mid semester report. The presence of outliers will be determined to help analyse the accuracy of the data. There are an infinite number of internal and external factors that contribute
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Sampling distribution The sampling distribution is the distribution of the values of a sample statistic computed for each possible sample that could be drawn from the target population under a specified sampling plan. Because many different samples could be drawn from a population of elements‚ the sample statistics derived from any one sample will likely not equal the population parameters. As a result‚ the sampling distribution supplies an approximation of the true value’s population parameters
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(a) Suppose we take a random sample of size 100 from a discrete distribution in this manner: A green die and a red die are thrown simultaneously 100 times and let Xi denote the sum of the spots on the two dice on the ith throw‚ i = 1‚ 2‚...100. Find the probability that the sample mean number of spots on the two dice is less than 7.5. n = 100 µ = 7 µ[pic] = 7 σ = 2.41 σ[pic] = 2.41 /[pic] |X |2 |3
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