Understanding Z-Scores and the Empirical Rule in Statistics

36 36. Consider a sample with a mean of 500 and a standard deviation of 100. What are the z-scores for the follow mean squared std dev Mean 500 520 280 500 -220 48400 100 Std Dev 100 650 450 500 -50 2500 100 500 500 500 0 0 100 450 520 500 20 400 100 280 650 500 150 22500 100 2400 73800 z scores z=(data-mean)/std dev std dev 520 0.2 650 1.5 500 0 450 -0.5 280 -2.2 38 a. 20 to 40 so with the mean of 30 and std dev of 5, 95% will fall between this ranges (2 std dev +/-_ b. 15 to 45 68% as 3 std dev +/- c. 25 to 35 almost all as only 1 dev +/1 std dev 5 mean 30 68% is within 1 dev of mean sample data sorted data dev from mean sample data Suppose the data have a bell-shaped distribution with a mean of 30 and a standard deviation of 5. Use the empirical rule to determine the percentage of data within each of the following ranges:

wing data values: 520, 650, 500, 450, and 280? z-score -2.2 -0.5 0 0.2 1.5 The Empirical Rule states that 99.7% of data observed following a normal distribution lies within 3 standard deviations of the mean. Under this rule, 68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean.

40 a. What percentage of regular grade gasoline sold between $3.33 and $3.53 per gallon? 68% mean std dev -2 3.33 3.43 3.53 2 0.1 b. What percentage of regular grade gasoline sold between $3.33 and $3.63 per gallon? mean std dev -3 3.33 3.43 3.53 3 0.1 c. What percentage of regular grade gasoline sold for more than $3.63 per gallon? 1.5% if 95% of the data is between 3 std dev of the mean, then 5% is remaining above and below, so 2.5% of gas sell 42 a. What is the z-score for a backyard structure costing $2300? z-score = (2300-3100)/1200 -0.666667 b. What is the z-score for a backyard structure costing $4900? z-score = (4900-3100)/1200 1.5 c. Interpret the z-scores in parts (a) and (b). Comment on whether either should be considered an outlier. mean is 3100, and the std dev is 1200. both structures are within 3 std(.67 below and 1.5 above) dev of the me d. If the cost for a backyard shed-office combination built in Albany, California, is $13,000, should this structure yes, this would be far outside 3 std dev of the mean. z-score = (13,000 - 3100)/1200 8.25 Price per Gallon of Gasoline. Suppose that the mean retail price per gallon of regular grade gasoline in the United States is $3.43 with a standard deviation of $.10 and that the retail price per gallon has a bell-shaped distribution. part a tells us that 68% is with in 3 dev from the mean. Half of 68% is above, half below the mean or 34%. The rule states a that 95% is with in 1 std dev from mean so 34%+(95/2)= 82.5%, so 81.5% of gas is sold between 3.33 and 3.63 Cost of Backyard Structure. Many families in California are using backyard structures for home offices, art studios, and hobby areas as well as for additional storage. Suppose that the mean price for a customized wooden, shingled backyard structure is $3100 . Assume that the standard deviation is $1200

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ls for higher than 3.63% ean so neither would be considered an outlier e be considered an outlier? Explain.

Winning Team Points Losing Team Points Winning Margin Arizona 90 Oregon 66 24 Duke 85 Georgetown 66 19 Florida State 75 Wake Forrest 70 5 Kansas 78 Colorado 57 21 Kentucky 71 Notre Dame 63 8 Louisville 65 Tennessee 62 3 Oklahoma State 72 Texas 66 6 Purdue 76 Michigan State 70 6 Stanford 77 Southern Cal 67 10 Wisconsin 76 Illinois 56 20 N 10 mean 76.5 median 76 mode 76 variance 49.17 std dev 7.01 std dev a. Compute the mean and standard deviation for the points scored by the winning teams. 7.0 c. Compute the mean and standard deviation for the winning margin. Do the data contain outliers? Explai Winning Margin mean squared 24 12.2 11.8 139.24 19 12.2 6.8 46.24 5 12.2 -7.2 51.84 21 12.2 8.8 77.44 8 12.2 -4.2 17.64 3 12.2 -9.2 84.64 6 12.2 -6.2 38.44 6 12.2 -6.2 38.44 10 12.2 -2.2 4.84 b. Assume that the points scored by the winning teams for all NCAA games follow a bell- shaped distribution. Using the mean and standard deviation found in part (a), estimate the percentage of all NCAA games in which the winning team scores 84 or more points. Estimate the percentage of NCAA games in which the winning team scores more than 90 points. z=(84-76.5 z=7.5/7=1. 68% of dat 32/2 =16, o z=(90-76.5 mean. 95 p 2.5% dev from mean

20 12.2 7.8 60.84 122 12.2 559.6 n 10 mean 12.2 variance 54.96 std dev 7.9 all winnig margins are within 3 std dev from the mean, so there are no outliers largest margin is 24 (24-12.2)/7.89 1.50 no outliers

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sorted data) Winning Team Points mean squared z-score 1 Louisville 65 76.5 -11.5 132.25 -1.64 2 Kentucky 71 76.5 -5.5 30.25 -0.78 3 Oklahoma State 72 76.5 -4.5 20.25 -0.64 4 Florida State 75 76.5 -1.5 2.25 -0.21 5 Purdue 76 76.5 -0.5 0.25 -0.07 6 Wisconsin 76 76.5 -0.5 0.25 -0.07 7 Stanford 77 76.5 0.5 0.25 0.07 8 Kansas 78 76.5 1.5 2.25 0.21 9 Duke 85 76.5 8.5 72.25 1.21 10 Arizona 90 76.5 13.5 182.25 1.93 765 442.5 mean 76.5 in. dev from mean 5)/7 .07 ta is within 1 std dev of the mean. so 100-68 =32, or 16% would score above 84 points 5)/7 = 1.93 which is approx 2 std dev from the percent is within 2 std dev of mean so 100-95=5,

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