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# Ch 3 2 measures of variation

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Module 2: Describing Data Using Numerical Measures
ch03. Describing Data Using Numerical Measures
03.2 Measures of variability
Range, Variance, Standard deviation, Coefficient of Variation

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### Ch 3 2 measures of variation

1. 1. The salaries for the staff of the XYZ Manufacturing Co. are shown here. Find the range. - Find the ranges for the paints. Solution W.P(1) : The range is: R = highest value - lowest value = \$100,000 - \$15,000 = \$85,000 The range is: R = highest value - lowest value = 10 – 5 = 5 The range is: R = upper limit of the last class – lower limit of the first class = 14 – 5 = 9 Or The range is: R = 14.5 - 4.5 = 10  Range W.P(1) : 3-2 Example (3-20/p-125) : (Employee Salaries) Crude Data Staff Salary Owner \$100,000 Manager 40,000 Sales representative 30,000 Workers 25,000 15,000 18,000 W.P(2) : 3-2 Example Grouped Data (Without Class Interval) Data (x) Frequency (f) f.x 5 1 5 6 3 18 7 4 28 8 8 64 9 3 27 10 5 50 Total = 24 192 W.P(3) : 3-2 Example Grouped Data (With Class Interval) Class x Frequency (f) f.x 5 - 6 5.5 2 11 7 - 8 7.5 5 37.5 9 - 10 9.5 8 76 11 - 12 11.5 4 46 13 - 14 13.5 1 13.5 Total = 20 184 Page of1 8 ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES YANBU INDUSTRIAL COLLEGE DEPARTMENT OF GENERAL STUDIES SEMESTER I, 1437-38 (H) / 2016-17 (G) Exercises # Ch3-2 Measures of Variation Student’s ID: _______________ Student’s Name: ______________________________________________ Course: STAT-201 Section: Date: 24 – 10 – 2016 Instructor’s Name: Mr. Osama Alwusaidi Instructor’s Signature:
2. 2. The formula for the population variance is: The formula for the variance & standard deviation for data obtained from samples are as follows: Shortcut or Computational Formulas for the S & S2: Find the sample variance and standard deviation for the amount of European auto sales for a sample of 6 years shown. The data are in millions of dollars. Solution W.P(4) : Step 1: ! Step 2: Step 3: ! - The variance is 1.28 rounded. - Hence, the sample standard deviation is 1.13. Variance and Standard Deviation for Ungrouped Data Formula and Shortcut or Computational Formulas Ungrouped Data σ 2 = X −µ( ) 2 ∑ N ⇒ σ = σ 2 = X −µ( ) 2 ∑ N s2 = X − X( ) 2 ∑ n −1 ⇒ s = s2 = X − X( ) 2 ∑ n −1 s2 = n X2 ∑( )− X∑( ) 2 n n −1( ) ⇒ s = s2 = n X2 ∑( )− X∑( ) 2 n n −1( ) W.P(4) : 3-2 Example (3-23/p-129) : (European Auto Sales) 11.2 11.9 12.0 12.8 13.4 14.3 * Source: USA TODAY. Xi i=1 n ∑ = 75.6 ⇒ X = Xi i=1 n ∑ n = 75.6 6 =12.6 (xi − x) (xi − x)2 xi − x( )2 i=1 n ∑ = Data (xi) 11.2 -1.4 1.96 11.9 -0.7 0.49 12.0 -0.6 0.36 12.8 0.2 0.04 13.4 0.8 0.64 14.3 1.7 2.89 75.6 6.4 S2 = (Xi − X)2 i=1 n ∑ n −1 = 6.4 6 −1 =1.28 ⇒ S = S2 = 1.28 =1.13 Page of2 8
3. 3. Solution W.P(4) : (use the shortcut formula) Step 1: Step 2: ! - The variance is: 1.28 rounded. - Hence, the sample standard deviation is: 1.13. Using the frequency distribution for Example 2–7, ﬁnd the mean. The data represent the number of miles run during one week for a sample of 20 runners. Solution W.P(5) : (use the shortcut formula) Step 2: ! - The variance is: 68.7 rounded. - Hence, the sample standard deviation is: 8.3.  Xi 2 Data (Xi) 11.2 125.4 11.9 141.6 12.0 144.0 12.8 163.8 13.4 179.6 14.3 204.5 75.6 958.9 s2 = n X2 ∑( )− X∑( ) 2 n n −1( ) = 6 958.9( )− 75.6( ) 2 6 6 −1( ) =1.28 ⇒ s = s2 = 1.28 =1.13 Variance and Standard Deviation for Grouped Data Formula and Shortcut or Computational Formulas Grouped Data (With Class Interval) W.P(5) : 3-2 Example (3-24/p-129) : (Miles Run per Week) Class limits Class boundaries Frequency (f) Midpoints (Xm) f . Xm f . X2 m 6 - 10 5.5 - 10.5 1 8 8 64 11 - 15 10.5 - 15.5 2 13 26 338 16 - 20 15.5 - 20.5 3 18 54 972 21 - 25 20.5 - 25.5 5 23 115 2645 26 - 30 25.5 - 30.5 4 28 112 3136 31 - 35 30.5 - 35.5 3 33 99 3267 36 - 40 35.5 - 40.5 2 38 76 2888 Total = 20 161 490 13310 s2 = n f ⋅ Xm 2 ∑( )− f ⋅ Xm∑( ) 2 n n −1( ) = 20 13,310( )− 490( )2 20 20 −1( ) ≈ 68.7 ⇒ s = s2 = 68.7 ≈ 8.3 Page of3 8
4. 4. The coefﬁcient of variation, denoted by CVar, The formula for CVar is: The mean of the number of sales of cars over a 3-month period is 87, and the standard deviation is 5. The mean of the commissions is \$5225, and the standard deviation is \$773. Compare the variations of the two. Solution W.P(6) : The coefficients of variation are: ! Since the coefficient of variation is larger for commissions, the commissions are more variable than the sales. Coefficient of Variation (C.V) For samples, CVar = s X ⋅100 For populations, CVar = σ µ ⋅100 W.P(6) : 3-2 Example (3-25/p-132) : (Sales of Automobiles) s1 =5 s2 = 773 X1 = 87 X2 = 5225 sales commissions CVar = 5 87 ⋅ 100 = 5.7% sales CVar = 773 5225 ⋅ 100 =14.8% commissions Page of4 8
5. 5. The number of yards gained in NFL playoff games by rookie quarterbacks is shown. a - ﬁnd the range, variance, and standard deviation and use the shortcut formula for the unbiased estimator to compute the variance and standard deviation. Solution W.P(7) : (use the shortcut formula) ! - The variance is: 1984.49 rounded. - Hence, the sample standard deviation is: 44.55 rounded. Exercises 3-2 W.P(7) : 3-2 Exercises (15/p-129) : (Football Playoff Statistics) 193 66 136 140 157 163 181 226 135 199 xi 2 Data (xi) 193 37249 66 4356 136 18496 140 19600 157 24649 163 26569 181 32761 226 51076 135 18225 199 39601 1596 272582 s2 = n X2 ∑( )− X∑( ) 2 n n −1( ) = 10 272582( )− 1596( ) 2 10 9( ) ≈ 1984.49 ⇒ s = s2 = 1984.49 ≈ 44.55 Page of5 8
6. 6. In a study of reaction times to a speciﬁc stimulus, a psychologist recorded these data (in seconds). Solution W.P(8) : (use the shortcut formula) Step 2: ! - The variance is: 0.847 rounded. - Hence, the sample standard deviation is: 0.92. W.P(8) : 3-2 Exercises (22/p-139) : (Reaction Times) Class limits Class boundaries Frequency (f) Midpoints (Xm) f . Xm f . X2 m 2.1 - 2.7 2.05 - 2.75 12 2.4 28.8 69.12 2.8 - 3.4 2.75 - 3.45 13 3.1 40.3 124.93 3.5 - 4.1 3.45 - 4.15 7 3.8 26.6 101.08 4.2 - 4.8 4.15 - 4.85 5 4.5 22.5 101.25 4.9 - 5.5 4.85 - 5.55 2 5.2 10.4 54.08 5.6 - 6.2 5.55 - 6.25 1 5.9 5.9 34.81 Total = 40 134.5 485.27 s2 = n f ⋅ Xm 2 ∑( )− f ⋅ Xm∑( ) 2 n n −1( ) = 40 485.27( )− 134.5( ) 2 40 40 −1( ) = 0.847 ⇒ s = s2 = 0.847 = 0.92 Page of6 8
7. 7. The following sample data reﬂect electricity bills for ten households in San Diego in March. Compute the range, variance, and standard deviation for these sample data. Discuss which of these three measures you would prefer to use as a measure of variation. Solution W.P(9) : Step 1: ! Step 2: Step 3: ! Solution W.P(9) : (use the shortcut formula) Step 1: Step 2: !   W.P(9) : 3-2 Exercises (22/p-139) : (Reaction Times) \$118.20 \$67.88 \$133.40 \$88.42 \$110.34 \$76.90 \$144.56 \$127.89 \$89.34 \$129.10 Xi i=1 n ∑ = 1086.03 ⇒ X = X∑ n = 1086.03 10 = 108.603 (xi − x) (xi − x)2 xi − x( )2 i=1 n ∑ = Data (xi) \$118.20 \$9.60 \$92.10 \$67.88 -\$40.72 \$1658.36 \$133.40 \$24.80 \$614.89 \$88.42 -\$20.18 \$407.35 \$110.34 \$1.74 \$3.02 \$76.90 -\$31.70 \$1005.08 \$144.56 \$35.96 \$1292.91 \$127.89 \$19.29 \$371.99 \$89.34 -\$19.26 \$371.06 \$129.10 \$20.50 \$420.13 \$1086.03 \$6236.89 S2 = (xi − x)2 i=1 n ∑ n −1 = 6236.89 10 −1 = 692.988 ⇒ S = S2 = 3.623 = 1.903 xi 2 Data (xi) \$118.20 \$13971.24 \$67.88 \$4607.69 \$133.40 \$17795.56 \$88.42 \$7818.10 \$110.34 \$12174.92 \$76.90 \$5913.61 \$144.56 \$20897.59 \$127.89 \$16355.85 \$89.34 \$7981.64 \$129.10 \$16666.81 \$1086.03 \$124183.01 s2 = n X2 ∑( )− X∑( ) 2 n n −1( ) = 10 124183.01( )− 1086.03( )2 10 9( ) ≈ 692.988 ⇒ s = s2 = 692.988 ≈1.903 Page of7 8
8. 8. Ages of Accountants The average age of the accountants at Three Rivers Corp. is 26 years, with a standard deviation of 6 years; the average salary of the accountants is \$31,000, with a standard deviation of \$4000. Compare the variations of age and income. In a study of reaction times to a speciﬁc stimulus, a psychologist recorded these data (in seconds). Solution W.P(10) : The coefficients of variation are: ! Since the coefficient of variation is larger for Age, the Age are more variable than the salary of the accountants. W.P(10) : 3-2 Exercises (31/p-139) : (Reaction Times) s1 = 6 s2 = \$4000 X1 = 26 X2 = \$31,000 Age Salary CVar = 6 26 ⋅ 100 = 23.08 % years CVar = 4000 31,000 ⋅ 100 = 12.9 % \$ Page of8 8 Mr. Osama A. Alwusaidi | Yanbu University College | GS Dept. | Saudi Arabia | instructor (B.Ed.), Mathematics. (Ed.M.), Testing, Measurement and Statistics (Psychology) Mobile: +966-544115001 | Phone: +966-43932961 | Cisco:1574 | Fax: +966-43925394 Email: alwusaidio@rcyci.edu.sa | Site: www.rcyci.edu.sa | Address: P.O.Box 31387 Yanbu Industrial City 41912 Saudi Arabia