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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) (xi)2 (x—i) (xi)2 (x—i) (xif
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 (xx) 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 Panini  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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 Standard Deviation, data values, Grady Silnonton

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