3-3 example6 - Student: Grady Silnonton Colu'se:...

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Unformatted text preview: Student: Grady Silnonton Colu'se: 1\-Iat11119: Elementary Statistics - Spring 3010 - C‘RN: 49339 Instructor: Shawn Pan-'ini - 16 weeks Date: 3.-"'18.-"'10 Book: T1‘iola: Elementary Statistics. 11e Time: 11:33 AM Listed below are the playing times (in seconds) of 16 popular songs. Find the range, variance, and standard deviation for the set of data. Does the standard deviation change much ifthe longest playing time is deleted? 462 20?V 280 230 239 297 239 243 229 310 29?V 219 199 212 320 291 First, find the range of the sample. The range ofa set of data values is the difference between the maximum data value and the minimum data value. _ (maximum data value) - (minimum data range — value) = (462-—l99) = 263 Therefore, the range of the sample is 263 seconds. ‘) x - x ' The variance, 32, of a sample is found by using the formula 52 = fl—l), where x is the variable used to represent 11 — the individual data values, ; is the mean of the sample, and n is the size of the sample. The mean, ;, of a sample is found with the formula below, where x is the variable used to represent the individual data values, and n is the number of data values in the sample. 2 x (— sum of all data values x = n <— number of data values Find the sample mean. 462 + 20? + 280+ ...+ 212 + 320 + 29] 16 g- 4214 16 267.125 Calculate (x - ;) for the first value in the sample data. (462-—267J25) 194.875 (x—E) Calculate (x - ;) for the remaining data values. Recall that ;= 262.125. Page 1 Student: Grady Simonton Colu'se: 1\-Iat11119: Elelnentaiy Statistics - Spring 3010 - C‘RN: 49339 Instructor: SlIau-‘11Pa1vini - 16 weeks Date: 3.-"'18.-"'10 Book: T1‘iola: Elelnentaiy Statistics. Me Time: 11:33 AM 462 194.875 297 29.875 291 23.875 207 —60.l25 219 —48.l25 280 12.875 199 — 68.125 230 —37.l25 212 —55.l25 239 — 28.125 320 52.875 Calculate (x — 2 for the first value in the sample data, rounding to four decimal places. (x —;)2 = (462 — 261125)2 = (194.875)2 = 379762656 — 2 . . . . . . Calculate (x — x) tor the rema1n1ng data values, rounding to tour dcc1mal places. (x—i) (x-i)2 (x—i) (x-i)2 (x—i) (x-if 194.875 37,976.2656 — 28.125 791.0156 —68.125 4,641.0156 —60.12 3,615.0156 —24.125 582.0156 —55.125 3,038.7656 12.875 165.7656 — 38.125 1,453.5156 52.875 2,795.7656 — 37.12 1,378.2656 42.875 1,838.2656 23.875 570.0156 —28.12 791.0156 29.875 892.5156 29.875 892.5156 —48.1252,316.0156 Now find the sum ofthe values of (x — 2. 2(x — if = 63,737.75 Finally, divide the sum of the values of (x — 2 by (n - l) to find the variance, where n is the sample size. Round to three decimal places. 26—92 n— l 63,737.75 16— 1 4249.183 Id The standard deviation can be found using the formula below. 52126—92 n—l Page 3 Student: Grady Si111o11ton Instructor: Shawn Pan-'illi Date: 3.-"'18..-"'10 Colu'se: 1\-Iat1111_9: Elementary Statistics - Spring 3010 - C‘RN: 49339 - 16 weeks Book: T1‘iola: Elementary Statistics. Me Time: 11:33 AM Notice that the standard deviation is the square root of the variance. Since the variance has already been calculated, take its square root. 5 = \l4,249.183 65.186 Therefore, the standard deviation is approximately 65.186 seconds. Now remove 462 from the sample and recalculate the standard deviation. Find the mean, E, of the sample after 462 is removed. _ 207+280+230+...+212+320+291 15 >4 || 254.1333 Calculate (x - ;) for the data values of the new sample, rounding to four decimal places. x (x-x) x (X—;) x (x-;) 207 —47.1333 239 — 15.1333 219 — 35.1333 280 25.8667 243 — 11.1333 199 — 55.1333 230 — 24.1333 229 — 25.1333 212 —42.1333 239 - 15.1333 310 55.8667 320 65.8667 297 42.8667 297 42.8667 291 36.8667 — 2 , , . . . Calculate (x - x) for the data values of the new sample, rounding to tour decrmal places. - 47.1333 2,221.5480 -15.1333 229.0168 -35.1333 1,234.3488 25.8667 669.0862 — 11.1333 123.9504 — 55.1333 3,039.6808 —24.l333 582.4162 —25.l333 631.6828 —42.1333 1,775.2150 — 15.1333 229.0168 55.8667 3,121.0882 65.8667 4,338.4222 42.8667 1,837.5540 42.86671.837.5540 36.8667 1,359.1536 Now find the sum ofthe values of (x — 2. . — 2 2(x - x) = 23,229.7338 . . . — 2 .. . . . D1v1de the sum or the values of (x - x) by (15 — l) to 11nd the variance, where n is the sample srze. Round to three decimal places. Page 3 Student: Grady Silnonton Colu'se: 1\-Iat11119: ElenIentaiy Statistics - Spring 3010 - C‘RN: 49339 Instructor: Shawn Pan-ini - 16 weeks Date: 3.-"'18.-"'10 Book: Triola: ElenIentaiy Statistics. Me Time: 11:33 AM Id 26; —;)Z n - l 23,229.7338 15 — 1 1,659.26? 0'1 || The standard deviation can be found using a formula or by taking the square root of the variance. Since the variance has already been calculated, take its square root, rounding to three decimal places. The square root of the variance is 40.?34. Therefore, the standard deviation of the sample without the longest playing time is approximately 40.34 seconds. Compare this to the standard deviation of the sample that includes the longest playing time, approximately 65.186 seconds. Note that there is a significant decrease in standard deviation when the longest playing time is deleted. Page 4 ...
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This note was uploaded on 05/21/2010 for the course MATH 49239 taught by Professor Parvini during the Spring '10 term at Mesa CC.

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3-3 example6 - Student: Grady Silnonton Colu'se:...

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