Math 221 Quiz Review for Weeks 5 and 6 1. Find the area under the standard normal curve between z = 1.6 and z = 2.6. First we look for the area of both by doing “2nd ‚Vars‚ NORMALCDF” and inputting “-1000‚ “Z‚” 0‚ 1 then find the difference between both. 2. A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61‚400 with a standard deviation of $2‚200. Find a 95% confidence interval for the true mean annual
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squared deviations‚ (d) variance‚ and (e) standard deviation: 2‚ 2‚ 0‚ 5‚ 1‚ 4‚ 1‚ 3‚ 0‚ 0‚ 1‚ 4‚ 4‚ 0‚ 1‚ 4‚ 3‚ 4‚ 2‚ 1‚ 0 Answer: A) Mean = 2 B) Median = 2 C) Sum of squared deviations =-52 D). Variance = -2.47 E). Standard deviation = 1.57 12. For the following scores‚ find the (a) mean‚ (b) median‚ (c) sum of squared deviations‚ (d) variance‚ and (e) standard deviation: 1‚112; 1‚245; 1‚361; 1‚372; 1‚472 Answer: A) Mean = 1.361 B) Median = 1.3124 C) Sum of squared deviations = 0.07608920
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Using the Standard Deviation You made a number of observations about the data sets for the school activities. You used mean and median to measure the center of the data‚ and you used the interquartile range (IQR) to measure the spread. When outliers are present‚ the median and IQR are used to measure center and spread because they are unaffected by extreme values. When the data appears to be symmetric and there are no known outliers‚ the mean and standard deviation (another measure of spread)
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caffeinated beverages consumed by Americans each day‚ based on data from the National Sleep Foundation. Determine whether or not the data represent a probability distribution. *If the data represent a probability distribution‚ find its mean and standard deviation. * If the data don’t represent a probability distribution‚ identify the requirement(s) for a probability distribution that is not satisfied. The data does not represent a probability distribution because all the probabilities are
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– Difference between Depreciation by Straight Line Method and Depreciation by Reducing Balance Method 6 2.0 The Difference 6 Question 3 - Standard Deviation and Quartile Deviation 7 Standard Deviation 7 Quartile Deviation 8 3.0 Purpose of Calculating Standard Deviation and Quartile Deviation 8 3.1 Calculation of Standard Deviation and Quartile Deviation 8 Reference 9 Question 1 – Difference between Simple Interest and Compound Interest To know the difference between simple and compound
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FINANCE 411 – SUMMER 2012 ASSIGNMENT 2 DUE DATE: June 4‚ 2012 [Submit a Hard Copy and via Drop Box; Keep a copy ] http://www.globefund.com/ http://www.globefund.com/review/ http://globefunddb.theglobeandmail.com/gishome/plsql/gis.co_search 1. Choose 1 fund from the Globe and Mail 15-Year Mutual fund Review issue [on internet] as follows: 1. fund names start with the same letter as your surname [or as close as possible] 2. 1 Canadian
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$3‚291.00 up to $15‚906.00 for the period of the said year. Its average price is $6‚165.26 with a standard deviation of $2‚949.50. It can be seen that prices are not close by to one another. With regards to mileage‚ the majority of the automobiles runs 41 miles for every gallon of gasoline‚ while the least runs only for 12 miles. The mean of mileage has resulted to 21.30 mpg‚ with a standard deviation of 5.79 mpg. As to the variable repair record it can be seen that only 69 were observed out of the
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c) Sum of square deviations is 56. d) Variance is 2.666 or 2.7 e) Standard deviation is 1.632 or 1.6 12. a) Mean is 1312.4 or 1312 b) Median is 1361 c) Sum of square deviations is 76092.2 d) Variance is 15218.44 e) Standard deviation is 123.363 13. a) Mean is 3.166 b) Median is 3.25 c) Sum of square deviations is 0.44738 d) Variance is 0.074 e) Standard deviation is 0.272 16. a) Governor-Mean is 43 and Standard deviation is 5.916 CEO- Mean
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skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately
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at different temperatures. After this‚ equipment called colorimeter will be used to take a reading for absorbency. The readings for the absorbency will be repeated three times in order to find the mean of each sample and to also find out the standard deviation of each set of data which will be recorded in a table format. The temperature rising will cause damage to the cell membrane due to the temperature rising above the membranes temperature. The mean absorbency will be calculated from five different
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