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 many
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help analyse the accuracy of the data. There are an infinite number of internal and external factors that contribute to the outcome of a single exam result. Beaty‚ & Barling (1982) explains how factors such as stress and anxiety can contribute to low test results and they give several self help solutions of how to boost ones success. This report will focuses mainly on quantitative data that can be easily analysed and allows for clear observations to be given about the correlation each evaluated factor
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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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vertical axis and 1 cm for 10 marks on the horizontal axis. (5) (c) Use your graph to answer parts (i)–(iii) below‚ (i) Find an estimate for the median score. (2) (ii) Candidates who scored less than 35 were required to retake the examination. How many candidates had to retake? (3) (iii) The highest-scoring 15% of candidates were awarded a distinction. Find the mark above which a distinction was awarded. (3) (Total 16 marks) 2. At a conference of 100 mathematicians there are 72 men and
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Standard Deviation in the Business World QRB/501 Abstract On Standard Deviations in Job Performance The purpose of this study was to compare the expected payoffs from personnel programs based on standard deviation of job performances in dollars‚ the Global Estimation model‚ and the CREPID procedure. The study was done for route salesmen of a large soft drink bottling company. The Global Estimation model and the CREPID procedure were behaviorally based‚ where the standard deviation of job performance
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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 tar content of the sample is greater than 4.15 milligrams? [0.00621] 2. The safety limit of a crane is known to be 32 tons. The mean weight and the standard deviation of a large number of iron rods
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problem 1‚ how much would cash flow be if there were only $10‚000 in depreciation? All other factors are the same. b. How much cash flow is lost due to the reduced depreciation between Problems 1 and 2a? 3. Cash flow (LO2) Assume a firm has earnings before depreciation and taxes of $500‚000 and no depreciation. It is in a 40 percent tax bracket. a. Compute its cash flow. b. Assume it has $500‚000 in depreciation. Recompute its cash flow. c. How large
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score decreases by -0.347 average number of hours per week. Next‚ we have to interpret the random error component ( ) of the least squares line. The estimate of error standard deviation for this particular data set is 12.28455 which suggests that the work-life balance score (y) values should be two times of the standard deviation. This can be best expressed as in the equation of 2s = 2(12.28455) which equals 24.57 average hours per week. Then‚ we can analyze the accuracy of hypothesis model to
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circuit contains three resistors wired in series. Each is rated at 6 ohms. Suppose‚ however‚ that the true resistance of each one is a normally distributed random variable with a mean of 6 ohms and a standard deviation of 0.3 ohm. What is the probability that the combined resistance will exceed 19 ohms? How "precise" would the manufacturing process have to be to make the probability less than 0.005 that the combined resistance of the circuit would exceed 19 ohms? n = 3 µ =
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to obtain the highest overall reliability? 17. A major television manufacturer has determined that its 19-inch color TV picture tubes have a mean service life that can be modeled by a normal distribution with a mean of six years and a standard deviation of one-half year. a. What probability can you assign to service lives of at least (1) Five years? (2) Six years? (3) Seven and one-half years? b. If the manufacturer offers service contracts of four years on these picture tubes‚ what percentage
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