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3-3 example5

# 3-3 example5 - Student Grady Simonton Course Math119...

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Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:31 ANI Fuel consumption is commonly measured in miles per gallon (mifgal). An agency designed new fuel consumption tests to be used starting with 2008 car models. Listed below are randomly selected amounts by which the measured MPG ratings decreased because of the new 2008 standards. Find the range, variance, and standard deviation for the sample data. Is the decrease of 4 mir’ gal unusual? Why or why not? 14333222413332322221 First, ﬁnd the range of the sample. The range of a set of data values is the difference between the maximum data value and the minimum data value. _ (maximum data value) — (minimum data range _ value) = (4 - l) = 3 Therefore, the range of the sample is 3 mifgal. The variance of a sample is found by using the formula , where E is the mean of the sample, and n is the Z (x — i) 2 n _ size of the sample. The mean, i, 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 mean, i, of the sample. _ _1+4+3+...+2+2+1 X_ 20 E 20 2.4 hours Calculate (X — E) for the ﬁrst value in the sample data. 0—2.4) —1.4 (ac—E) Calculate (x - E) for the next 3 values. Page 1 Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:31 ANI (4—2.4) = 1.6 (3—2.4) = 0.6 (3—2.4) = 0.6 The remaining values of (x - E) are shown below. Calculate (x — i) 2 for the ﬁrst value in the sample data. (pl—if = (1—2.4)2 = (—1.4)2 = 1.96 — 2 Calculate (x — x) for the next 3 values. (4—2.4)2 = (1.6)2 = 2.56 (3—2.4)2 = (0.6)2 = 0.36 (3—2.4)2 = (0.6)2 = 0.36 The remaining values of (x — i) 2 are shown below. (x—;)(x-;)2 (x—;)(x—E)2 (x—;)(x—;)2 - 1.4 1.96 - 0.4 0.16 0.6 0.36 1.6 2.56 — 0.4 0.16 0.6 0.36 0.6 0.36 — 0.4 0.16 0.6 0.36 0.6 0.36 1.6 2.56 - 0.4 0.16 0.6 0.36 - 1.4 1.96 0.6 0.36 Now ﬁnd the sum of the values of (x — E) 2. — 2 2(X - x) = 14.8 Finally, divide the sum of the values of (x — 1:) 2 by (n - 1) to ﬁnd the variance, where n is the sample size. Round to Page 2 Student: Grady Silnonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e Time: 11:31 ANI three decimal places. Sh=za—az n - l _ 14.8 _ 20—1 = 0.779 The standard deviation can be found by using the formula below. SZIZa—92 n-l Notice that the standard deviation is the square root of the variance. Since the variance has already been calculated, take its square root, rounding to three decimal places. 5 = V 0.779 = 0.883 Therefore, the standard deviation is approximately 0.883 mifgal. A data value is considered unusual if it is more than 2 standard deviations away ﬁ‘om the mean. If s = 0.883, calculate 25. Two standard deviations is equal to 1.766 mix’gal. If a value is more than 1.766 mifgal away from the mean, it is considered to be unusual. Use this information to answer the question. Page 3 ...
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3-3 example5 - Student Grady Simonton Course Math119...

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