• Qnt Week 6
    Ch. 3, p. 95 Exercise 88 88. Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season. A. Select the variable team salary and find the mean, median, and the standard deviation. B. Select the variable that refers to the age the...
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  • Practice Question
    normal distribution. Since the standard deviation is so high (and going even one standard deviation below average takes you outside of the range of the data: 5 hours below the mean of 4 hours equals –1 hours and is obviously outside the range of the data), it is pretty clear that the histogram for the...
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  • Qnt561 One and Two Samples Sets Week 5
    Weight Reducers is less than 10 pounds. Exercise Question 32: Dole Pineapple, Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found...
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  • hw3 stats
    0 1 Standard Normal Q (b) 2 −3 −2 −1 0 1 Standard Normal Q 2 (c) 7. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Let X be the variable...
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  • Thesis
    to his school? 59 Student John Ali GPA 2.85 77 School Mean GPA 3.0 80 School Standard Deviation 0.7 10 (Solution to Exercise 2.6 on p. 72.) Your concentration should be on what the standard deviation does, not on the arithmetic. The standard deviation is a number which...
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  • Exam3
    minimum specification are (50,000) × (0.159) = 7950. This exercise may also be solved using Excel or MINITAB: (1) Excel Function NORMDIST(X, µ, σ, TRUE) (2) MTB > Calc > Probability Distributions > Normal Cumulative Distribution Function Normal with mean = 40 and standard deviation = 5 x P( X 48...
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  • Qnt 561 Week One
    up to 140 12 140 up to 160 16 160 up to 180 7 180 up to 200 4 Total 50 a. Estimate the mean cost. The estimated mean cost would be $141.2 b. Estimate the standard deviation. The standard deviation is 26.23 c. Use the Empirical Rule to estimate the proportion of costs within two...
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  • Chapter 9 Qnt/561 Week Five
    a standard deviation of 13 pounds. A random sample of 40 firefighters taken six months after the program was implemented revealed that the mean of the sample was 192 pounds. Construct a 96% confidence interval for the mean weight of all firefighters six months after the exercise rooms were...
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  • Fitness
    Exercise 6 Find the mean, standard deviation and variance of (a) The annual rainfall data for the years 1971 – 1990 Year 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 Rain (mm) 1 340 990 1 120 1 736 260 1 100 1 379 1 125 1 430 1...
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  • Mastering a Calculator
    for the weights (kg) of a random sample of 30 first year university female students. Find the standard deviation, the variance and the mean. Graduate’s weight (kg) Frequency Cumulative frequency 60 2 2 61 14 16 62 8 24 63 1 25 64 5 30 The...
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  • One- and Two-Sample Tests of Hypothesis, Variance, and Chi-Squared Analysis Problem Sets
    Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of...
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  • LEAD 6341 Research Methods and Statistics Midterm Exam
    are 3 and 8. 2) You have been asked to give a student two different tests; an intelligence test and a creativity test. The student scored 123 on the intelligence test and 123 on the creativity test. The mean for the intelligence test is 100, and the standard deviation is 16...
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  • Addmaths
    intervals are called grouped data; whereas ungrouped data are data that are distributed randomly without class intervals.    In statistics, mean, mode, median, range, interquartile range, variance and standard deviation are used to represent the data collected. The mean of ungrouped data is the...
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  • Gfhjt
    Exercise 6 Find the mean, standard deviation and variance of (a) The annual rainfall data for the years 1971 – 1990 Year Rain (mm) Year Rain (mm) 1971 1 340 1972 990 1973 1 120 1974 1 736 1975 260 1976 1 100 1977 1 379 1978 1 125 1979 1 430 1980 1 446 1981 1 459...
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  • Ya Ya Ya
    observations will be below 16 and 25% will be higher than 16. * * 4.47 interquartile range * 2, 5, 6, 8, 9, 10, 11, 14, 18, 21 * * Coefficient of variation – is the standard deviation of a set of observations divided by their mean Measure of Linear Relationship Two...
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  • Mba510 Week 5 Problem Set
    GRADE 9/9 Lind Chapter 9; Exercise 12 The American Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 60 pounds with a standard deviation of 20 pounds. a. What is the value of the population mean? What...
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  • Whatever
    = 6 Data Value (# cars) | Frequency | Cumulative Frequency | 3 | 14 | 14 | 4 | 19 | 33 | 5 | 12 | 45 | 6 | 9 | 54 | 7 | 11 | 65 | 25 | 65 |   | P7 |   | 45.5 | Exercise 16 Find the value that is 3 standard deviations: a. Above the mean = 8.89 b. Below the mean = 0.60...
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  • Normal Distribution
    above and below the mean? 2 standard deviations? 3 standard deviations? For Exercises 6 through 25, find the area under the standard normal distribution curve. 6. Between z ϭ 0 and z ϭ 1.89 7. Between z ϭ 0 and z ϭ 0.75 8. Between z ϭ 0 and z ϭ Ϫ0.46 9. Between z ϭ 0 and z ϭ Ϫ2.07 10. To the...
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  • Qnt561 Week 6 Comprehensive Problem Sets
    -60000)/2000 = 5 * It is unlikely that one truck was above 5 stdevs above the mean. The prob is almost 0. * * Exercise 38 (Ch. 8) 38. The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50 with a standard deviation of $5.00. Assume the distribution of amounts...
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  • investment
    . Investment Expected Return Standard Deviation 1 0.12 0.30 2 0.15 0.50 3 0.21 0.16 4 0.24 0.21 (5 points) 16. Which investment would you select if you are risk averse (A=4)? U1 = 0.12 – 0.5*4*0.32 = -0.06 (1) U2 = 0.15 – 0.5*4*0.52 = -0.35 (1...
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