Exercises (Measure of Variation and Shape) Given the following data: 11 15 24 33 10 35 Find : 1. Range 2. Mean Deviation 3. Variance 4. Standard Deviation 5. Skewness 6. Kurtosis Given : Weight distribution of 24 students indicated below; Weights(kg) no. of students 52-54 1 55-57 4 58-60 3 61-63 5 64-66 6 67-69 3 70-72 2 Find 7. Variance 8. Standard Deviation 9. Skewness 10. Kurtosis 23 25 40 ANSWERS/SOLUTIONS 1. 10, 11, 15, 23, 24, 25, 33, 35, 40 π ππππ = πππ₯πππ’π π£πππ’π − ππππππ’π π£πππ’π π = 40 − 10 π = 30 π₯ 10 11 15 23 24 25 33 35 40 Σx π= N 216 = 9 = 24 π₯2 100 121 225 529 576 625 1089 1225 1600 Σπ₯ 2 N = 6090 (π₯ − π) −14 −`13 −9 −1 0 1 9 11 16 Σ(|π₯ − π|) =0 (|π₯ − π|) 14 13 9 1 0 1 9 11 16 Σ(|π₯ − π|) = 74 (π₯ − π)2 196 169 81 1 0 1 81 121 256 Σ(π₯ − π)2 = 906 (π₯ − π)3 −2744 −2197 −729 −1 0 1 729 1331 4096 Σ(π₯ − π)3 = 486 ππππ π·ππ£πππ‘πππ Σ(|π₯ − π|) ππ· = π 74 ππ· = 9 ππ· = 8.22 ππππππ π₯Μ = 24 ππ‘ππππππ π·ππ£πππ‘πππ ππππ€πππ π πΎπ’ππ‘ππ ππ Σ(π₯ − π)3 π πππ€ = n β π3 ππ’ππ‘ = 486 24 β (10.03)3 π πππ€ = 0.4884 ππ’ππ‘ = Σ(π₯ − π)2 π=√ n 906 π= √ 9 π = 10.03 (π₯ − π)4 38416 28561 6561 1 0 1 6561 14641 65536 Σ(π₯ − π)4 = 160278 ππππππππ Σ(π₯ − π)2 π2 = N 906 9 π 2 = 100.67 π2 = π πππ€ = Σ(π₯ − π)4 π β π4 160278 24 β (10.03)4 ππ’ππ‘ = 0.66 2. ππππβπ‘π (π) ππ. ππ ππ‘π’ππππ‘π (π) 1 4 3 5 6 3 2 52 − 54 55 − 57 58 − 60 61 − 63 64 − 66 67 − 69 70 − 72 Σf n = 24 π= ππππππππ‘ (π₯π ) π(π₯π ) π(π₯π )2 (π₯ − π) 53 56 59 62 65 68 71 53 224 177 310 390 204 142 Σπ(π₯π ) = 1500 2809 12544 10443 19220 25350 13872 10082 = 94320 −9.5 −6.5 −3.5 −0.5 2.5 5.5 8.5 (|π₯ − π|) (π₯ − π)2 (π₯ − π)3 (π₯ − π)4 90.25 42.25 12.25 0.25 6.25 30.25 72.25 −857.375 −274.625 −42.875 −0.125 15.625 166.375 614.125 8145.0625 1785.0625 150.0625 0.0625 39.0625 915.0625 5220.0625 9.5 6.5 3.5 0.5 2.5 5.5 8.5 Σx N 1500 24 = 62.5 = ππππβπ‘π (π) 52 − 54 55 − 57 58 − 60 61 − 63 64 − 66 67 − 69 70 − 72 ππππππππ π2 Σπ(π₯ − π)2] = n 570 24 π 2 = 23.75 π(π₯ − π)2 π(π₯ − π)3 π(π₯ − π)4 90.25 169 36.75 1.25 37.5 90.75 144.5 −857.375 −1098.5 −128.625 −0.625 93.75 499.125 1228.25 8145.0625 7140.25 450.1875 0.3125 234.375 2745.1875 10440.125 Σf(π₯ − π)2 = 570 Σf(π₯ − π)3 Σf(π₯ − π)4 = −264 = 29155.5 ππ‘ππππππ π·ππ£πππ‘πππ π=√ π2 = Σπ(π₯ − π)2] n 570 24 π = 4.87 π=√ ππππ€πππ π πΎπ’ππ‘ππ ππ Σπ(π₯ − π)3 π πππ€ = Σπ β π 3 ππ’ππ‘ = 264 π πππ€ = − 24 β (4.87)3 π πππ€ = −0.0952 Σπ(π₯ − π)4 Σπ β π 4 29155.5 24 β (4.87)4 ππ’ππ‘ = 2.16 ππ’ππ‘ =