\fFor each part of the Moving-Up Ceremony, only 10 sections will be called to receive their diploma on stage. For the first batch, wherein sections Afghanite 45 Ammetrine 52 are arranged in alphabetical order, the sections are listed below. Agate 43 Ammolite 52 Alexandrite 36 Andalusite 51 Amber- 38 Apatite- 53 Amethyst - 52 Aquamarine 52 Variance measures how far a data set is X X - P spread out 36 -11,40 11 40 129.96 38 -9.40 9.40 88.36 E(x-X) $2 _E(X-x2 43 -4.40 4 40 19.36 n -1 n-1 45 -2.40 240 5.76 $2 _372.40 51 3.60 9 3,60 12.96 52 4.60 4.60 21.16 Standard deviation $2 = 41.38 52 4.60 4 60 21.16 differentiates sets of scores 52 4.60 21.16 with equal averages. 52 4.60 4.60 21.16 53 5.60 5,60 31.36 S = VE(X X)2 E(x - X) n -1 S= 1- 1 E( X-X) = 372.40 S = V 41.38 S = 6.43For each part of the Moving-Up Ceremony, only 10 sections will be called to receive their diploma on stage. For the first batch, wherein sections Afghanite 45 Ammetrine 52 are arranged in alphabetical order, the sections are listed below. Agate 43 Ammolite 52 Alexandrite 36 Andalusite 51 The RANGE is the simplest measure of variability, It is the Amber- 38 Apatite- 53 difference between the largest value and the smallest value. Amethyst - 52 Aquamarine 52 RANGE = Highest Score - Lowest Score. 36, 38, 43, 45, 51, 52, 52, 52, 52, 53 R = HS - LS R = 17 Mean deviation is used to compute how far the R = 53 - 36 values in a data set are from the center point. The absolute value of MD = Elx - XI Find the value of x- I a number in is the distance of the n 36 -1140 11.40 number n from zero. x = 47.40 38 -9.40 9.40 E| X- X = 55.20 Variance measures Standard deviation 43 -4.40 4.40 how far a data set is 45 differentiates sets of scores -2.40 2 40 MD EX-x 51 3.60 3/60 spread out with equal averages. n 52 1 60 4.60 MD 55.20 E( x- x) 57 10 S = E(x - x)= 4 60 4 60 n-1 S= 52 11 - 1 MD= 5.52 57 1.60 4.60 5.60 5.60Example 2. The final grade of Moira, Jason, Kylie, Aljur, and Xian are 81, 85,92,79 and 94 respectively. Find the range, mean deviation, variance and standard deviation of their final grade. n -1 $2 174.80 5 79, 81, 85, 92, 94 $2 = 34.96 R = HS - LS R = 94 - 78 R = 16 Find the value of x S = VE(X-*)2 X X- X (x-x) n-1 x = 86.20 79 -7.20 7.20 51.84 81 -5.20 5.20 27.04 S = V 34.96 85 -1.20 1.20 1.44 92 5.80 5.80 93.64 S = 5.91 94 7 80 7.80 60.84 E| X- x | = E( x x) = 27.20 174.80 MD EX-x MD 27.20 MD= 5.44 7 5