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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used to calculate: =NORMDIST(5.5‚1.88‚1.99‚True) X= 5.5 Mean= 1.88 Standard Deviation=1.19 b) 32 to 35 weeks = 43.83% The NORMDIST formula was used to calculate: =NORMDIST (5.5‚5.73‚1.48‚True) X= 5.5 Mean= 5.73 Standard Deviation=1.48 c) 37 to 39 weeks = 4.66% The NORMDIST formula was used to calculate: =NORMDIST(5.5‚7.33‚1.09‚ True) X= 5.5 Mean= 7.33 Standard Deviation=1.09 d) 42 weeks and over = 2.75% The NORMDIST formula was used
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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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statistics Mean: 2 Median: 2 sum of squared deviations: 56 Variance: 2.8 standard deviation: 1.67332 12. Calculate descriptive statistics Mean: 1‚112 the mean is 56.5; 1‚1245 the mean is 123; 1‚1361 the mean is 181; 1‚1372 the mean is 186.5; 1‚1472 the mean is 236.5 Median: 1‚112 the median is 56.5; 1‚1245 the median is 123; 1‚1361 the median is 181; 1‚372 the median is 186.5; 1‚1472 the median is 236.5 sum of squared deviations: 1‚112 is 6160.5; 1‚1245 is 29768; 1‚361 is 64800;
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client can quickly learn whether the process is operating satisfactorily or corrective actions needs to be taken. Summary of Statistics Sample 1 Sample 2 Sample 3 Sample 4 Mean 11.96 12.10 12.14 12.15 Standard Error 0.04 0.04 0.04 0.02 Standard Deviation 0.22 0.23 0.23 0.16 Sample Variance 0.05 0.05 0.05 0.03 Sum 358.69 362.90 364.08 364.46 From the summary of statistics we can see that mean has an upward trend. Mean value differ from sample to sample. Here
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predictive statistical research. Standard deviation is a measure of the data spread and it is signified by the Greek letter sigma (σ). The standard deviation requires calculation of the average‚ compare each respondent’s value to the average‚ and square that difference (Burns & Bush‚ 2012‚ pg. 252). Therefore‚ the formula for standard deviation is the square root of the variance. The variance is the average of the squared differences of the mean. The standard deviation explains the density of data scattered
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