3-3 example4

# 3-3 example4 - Student Grady Sinlonton Colu'se 1-Iat11119...

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Unformatted text preview: Student: Grady Sinlonton 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: 11:30 AM Listed below are the durations (in hours) of a simple random sample of all ﬂights of a space shuttle program. Find the range, variance, and standard deviation for the sample data. Is the lowest duration time unusual? Why or why not? 94 104 218 125 172 220 199 368 255 252 373 358 218 263 0 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) = (373 —0) = 313 Therefore, the range of the sample is 373 hours. The variance, 52, ot‘a sample is found by using the formula below, where x is the variable used to represent the individual data values, x is the mean of the sample, and n is the size ofthe sample. ,= Est—if 11-1 5 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, rounding to four decimal places. _ 94+104+218+...+263+0 15 >4 | 3319 1 5 221.2667 hours Calculate (x — ;) for the first value in the sample data. (94 — 221.266?) — 121.2667 (x—E) Calculate (x - ;) for the remaining data values. Recall that ;= 221.266? Page 1 Student: Grady Silnonton Instructor: Shawn Pan-ini Date: 3.-"'18..-"'10 x (x-x) 270 48.7333 373 199 —22.2667 358 218 —3.2667 368 146.7333 218 175 —46.2667 255 33.7333 263 172 —49.2667 252 30.7333 0 Colu'se: 1\-Iat11119: ElenIentaiy Statistics - Spring 3010 - C‘RN: 49339 - 16 weeks Book: Triola: ElenIentaiy Statistics. 11e Time: 11:30 AM (x—i) 151.7333 136.7333 — 3.2667 41.7333 — 221.2667 Calculate (x — 2 for the ﬁrst value in the sample data. rounding to four decimal places. (x —;)2 = (94—221.2667)2 (—1212667)2 16,196.8129 — 2 . . . . . . Calculate (x — x) tor the remammg data values, rounding to tour deCImal places. — 127.266 16,196.8129 48.7333 2,374.9345 — 117.266 13,751.4789 — 22.2667 495.8059 —3.2667 10.6713 146.7333 21,530.6613 —46.2667 2,140.6075 33.7333 1,137.9355 — 49.2667 2,427.2077 30.7333 944.5357 - . - 2 Now tmd the sum of the values ot (x - x) . 2(x —§)2=153,440.9329 151.7333 23,022.9943 136.7333 18695.9953 — 3.2667 10.6713 41.7333 1,741.6683 — 221.266 48,958.9525 Finally, divide the sum of the values of (x — 2 by (n - 1) to ﬁnd the variance. where n is the sample size. Round to three decimal places. 2 20‘ ‘92 n — l 153,440.9329 15 - l 10,960.067 The standard deviation, 5, can be found by using the formula below. 5: ’Z(x-;)2 11-1 Page 3 Student: Grady Sinlonton Colu'se: 1\-Iat11119: Elementary Statistics - Spring 3010 - C‘RN: 49339 Instructor: Shawn Pan-'ini - 16 weeks Date: 3.-"'18.-"'10 Book: Triola: Elementary Statistics. 11c Time: 11:30 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 = t} 10,960.06? 104.690 Therefore, the standard deviation is approximately 104.690 hours. A data value is considered unusual if it is more than 2 standard deviations away from the mean. Since 5 = 104.690, calculate 25, rounding to three decimal places. Two standard deviations is equal to 209.380. Therefore, if a value is more than 209.380 hours away from the mean, it is considered to be unusual. Page 3 ...
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3-3 example4 - Student Grady Sinlonton Colu'se 1-Iat11119...

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