# Measures of Variation and Dispersion report

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Measures of Variation and Dispersion Reporters: Juan Miguel Bañez Miguel Fernando Bilan Lorenz Angelo Moya Statistics

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The simplest and crudest measure of dispersion is the range. This is defined as the difference between the largest and the smallest values in the distribution. If are the values of observations in a sample, then range (R) of the variable X is given by: Formula: R = H.S. – L.S. R = U.B. – L.B. Range

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Example * The range of the set of scores : 11, 13, 12, 14, 15 R = H.S. – L.S. R = 15 – 11 R = 4 Class Interval F <CF 98 – 100 2 40 95 – 97 1 38 92 – 94 1 37 89 – 91 6 36 86 – 88 6 30 83 – 85 5 24 80 – 82 9 19 77 – 79 2 10 74 – 76 3 8 74 – 73 5 5 N = 40 * The range of the set of scores : 11, 13, 12, 14, 15 R = U.B. – L.B. R = 15 – 11 R = 4

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Quartile Deviation Formula: I.R. = Q 3 –Q 1 Q.D. = Q3 – Q1 or I.R. 2 2 The inter-quartile range is frequently reduced to the measure of semi-interquartile range, known as the quartile deviation (QD), by dividing it by 2. A measure similar to the special range (Q) is the inter-quartile range . It is the difference between the third quartile (Q 3 ) and the first quartile (Q 1 ).

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Mean Deviation Formula: M.D. = ∑ X – X N The mean deviation is an average of absolute deviations of individual observations from the central value of a series. Average deviation about mean

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Example a.) Find the mean deviation of the following ungroup frequency distribution: X 10 12 12 14 b.) X 2 3 4 5 6 7 8 F 17 12 19 28 19 9 2 a.) Calculate the mean: x = ∑X = 10+12+12+14 = 48 = 12 N 4 4 Add the column X – X X X – X 10 2 12 0 12 0 14 2 ∑ X – X = 4 M.D. = ∑ X – X 4 M.D. = 4 4 = 1 b.) Calculate the mean by using the formula x = ∑ fx . This means we are going to add the entries in column fx. X F F ∙ X 2 17 34 3 12 36 4 19 76 5 28 140 6 19 114 7 9 63 8 2 16 N = 106 ∑ fX = 479 X = ∑X N = 479 106 X ≈ 4.52 Add the entries in column X – X and f X – X X m – X F X m – X 2.52 42.84 1.52 18.24 0.52 9.88 0.48 13.44 1.48 28.12 2.48 22.32 3.48 6.96 12.48 ∑f X m – X = 141.8 MD = ∑f X – X N = 141.8 106 MD ≈ 1.32

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Standard Deviation Standard deviation is the positive square root of the mean-square deviations of the observations from their arithmetic mean. Formula: S 2 = ∑ (X – X) 2 N- 1

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Example The price of a 250 gram-soap powder of a leading brand was recorded from 10 supermarkets. The prices (in pesos) were: 90, 73, 78, 79, 83, 95, 77, 79, 74, 82 Find : a.) The mean price b.) the range of the prices c.) the standard deviation of the prices a.) The mean price is : x = ∑fX = 90+73+78+79+83+95+77+79+74+82 = 810 = P 81 N 10 10 b.) The range of the prices is 95 – 73 = P22 c.) Add the column ( x – x ) 2 and sum up the scores. X ( x – x ) 2 90 81 73 64 78 9 79 4 83 4 95 196 77 16 79 4 74 49 81 0 ∑(x – x) 2 = 427 S 2 = ∑( x – x) 2 = 427 ≈ 47.44 N – 1 10 – 1 S = √47.44 ≈ 6.89

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Standard Deviation for Group Data Formula: The formula used in the computation of the standard deviation which is the mean deviation method, will be very difficult to deal with when the mean is a decimal number as shown example in Mean Deviation. There is an easier way by which the variance and standard deviation can be derived from the previously used formula. This method of computing the variance and standard deviation is called the “raw score method” . For Ungrouped Frequency Distribution S 2 = N ∑fX 2 – ( ∑fX) 2 N (N – 1) For Grouped Frequency Distribution S 2 = N ∑fX m 2 – ( ∑fX m ) 2 N (N – 1) For Ungrouped Data S2 = N∑X2 – (∑X)2 N (N – 1)

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Example Calculate the standard deviation using the raw score method. a.) X 10 12 12 14 b.) X 2 3 4 5 6 7 8 F 17 12 19 28 19 9 2 Calculate the means of the set of data. X 10 12 12 14 ∑X = 48 X = ∑X = 48 = 12 N 4 Add the entries in column x 2 and square all the scores and get the sum. X 2 100 144 144 196 ∑X 2 = 584 Substitute ∑X = 48 and ∑X 2 = 584 in the formula. S 2 = N ∑X 2 – (∑X) 2 N ( N – 1) S 2 = 4(584) – (48) 2 = 2336 – 2304 4 ( 4 – 1) 12 ≈ 2.67 = √ 2.67 ≈ 1.63 X = ∑fX = 479 = 4.52 N 106 Calculate the means of the set of data. Add the entries in column x 2 and fx 2 . X F Fx 2 17 34 3 12 36 4 19 76 5 28 140 6 19 114 7 9 63 8 2 16 N = 106 ∑ fX = 479 X 2 Fx 2 4 68 9 108 16 304 25 700 36 684 49 441 64 128 ∑fx 2 = 2433 Substitute ∑fX = 479 and ∑fX 2 = 2433 in the formula. S 2 = N ∑fX 2 – (∑fX) 2 N ( N – 1) S 2 = 106(2433) – (479) 2 = 257 898 – 229 441 11130 11130 = 2.56 = √ 2.56 ≈ 1.59

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