Value Range = 67 – 23 Range = 44 To find the standard deviation: σ = √∑ (X - µ)2 N σ = √∑(X – 40.84) 2 = √5291.36 = √211.65 = 14.55 25 25 Exercise 82* a. To find the mean cost: X = ∑fM = 7‚060 = $141.20 N 50 b. To find the standard deviation: s = √∑f (M - X)2 = √33‚728 = √688.33 = 26.24 n – 1 50 - 1 c. To find the proportion of costs within two standard deviations of the mean: X ± 2s 141.2 + 2(26.24) = 193.68 141.2 -
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mean to be 26.8 miles per gallon. From previous studies‚ the population standard deviation is known to be 5 miles per gallon. Could we reasonably expect that we could select such a sample if indeed the population mean is actually 28 miles per gallon? 2. Computing World has asserted that the amount of time owners of personal computers spend on their machines averages 23.9 hours per week and has a standard deviation of 12.6 hours per week. A random sampling of 81 of its subscribers revealed a sample
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FINAL EXAM PBHE525 Complete the final exam offline during the final exam week. Once you have complete the exam‚ input your exam into the final exam shell in the exam folder on the course webpage. Good luck 1. 2. 3. 4. 5. 6. 7. 8. US Census statistics show that college graduates make more than $254‚000 more in their lifetime than non-college graduates. If you were to question the validity of this observation‚ what would be your basis for doing so? A. Definition of a college graduate
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Report Introduction Markowitz (1952‚ 1956) pioneered the development of a quantitative method that takes the diversification benefits of portfolio allocation into account. Modern portfolio theory is the result of his work on portfolio optimization. Ideally‚ in a mean-variance optimization model‚ the complete investment opportunity set‚ i.e. all assets‚ should be considered simultaneously. However‚ in practice‚ most investors distinguish between different asset classes within their portfolio-allocation
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1 3 6 4 7 5 4 6 4 7 1 8 1 More 0 Q1 Descriptive Statistics Excel Worksheet b Q2 d Q3 e Q4 b 5. For the following sample of scores: 2‚ 6‚ 1‚ 4‚ 2‚ 2‚ 4‚ 3‚ 2 Mean variance standard deviation 2.889 2.099 1.537 Find the mean‚ variance and standard deviation 6. A sample of size 7 (n = 7) has a mean of M = 9. One of the sample scores is changed from x = 19 to x = 5. What is the value for the new sample mean? New Sample Mean 7 7. Using Excel create a histogram
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standard deviation of 0.018 inch. The customer will buy only those pins with lengths in the interval 1.00 ± 0.02 inch. In other words‚ the customer wants the length to be 1.00 inch but will accept up to 0.02 inch deviation on either side. This 0.02 inch is known as the tolerance. 1. What percentage of the pins will be acceptable to the consumer? In order to improve percentage accepted‚ the production manager and the engineers discuss adjusting the population mean and standard deviation of the
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delivery order and letting the customers inform their own pick-up times. The data is fine and appropriate for statistical use‚ and hence we can go through the statistical calculations. Statistical Efforts Using Excel‚ we found the standard deviation and mean of the sample data for the total delivery time‚ preparation time‚ waiting time and travel time. First‚ in order to discuss the viability of the company’s 29-minute delivery guarantee strategy‚ we computed the following statistical parameters
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highest annual average return of 17.7%. The Russell 2000 had the second highest annual return with a standard deviation of 14.1% & 17.2%. Treasuries and bond returns were a poor performer relative to the other assets which were in-line with our group’s anticipations. However‚ the MSCI Index reported a lower return than the treasuries and bonds along with the second greatest standard deviation. Events such as‚ Russia’s sovereign debt default in 1998‚ which ultimately brought down LTCM likely increased
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mean parts per million of dissolved solids in a 25-week sample to be 52 with a standard deviation of 32. Has there been a change in the average level of dissolved solids in this stream? 2 An exhaustive survey of all users of a wilderness park taken in 1960 revealed that the average number of persons per party was 2.6 In a random sample of 25 parties in 1985‚ the average was 3.2 persons with a standard deviation of 1.08. a. Test the hypothesis that the number of persons per party has changed in
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determine that g was 9.7993 ± 0.0010 m/s2 which was 0.0653 % from the known value. If we took into account the period difference‚ we found that g was 9.7982 ± 0.0083 m/s2 which is 0.0552 %. Both methods gave us results well within our goal of 0.2% deviation from the expected value. I. INTRODUCTION is equal to τ = I0 α (II.2) In the course of the author’s physics degree‚ no constant has been so ubiquitous as the local acceleration due to gravity. As such‚ it proves an important pedagogical
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