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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT0301 Elementary Statistical Methods Semester 1 2009/2010 Assignment 3 solution Q1 Cumulative frequency table: Class Class Boundaries Class Mark Frequency 10 – 19 9.5 – 19.5 14.5 7 20 – 29 19.5 – 29.5 24.5 9 30 – 39 29.5 – 39.5 34.5 14 40 – 49 39.5 – 49.5 44.5 12 80 – 99 49.5 – 59.5 54.5 9 100 – 119 59.5 – 69.5 64.5 8 120 – 139 69.5 – 79.5 74.5 6 (b) Mean = 42.9615 Standard deviation = 17.822 Variance = 317.633 Median ( ) m m m m f f f f N C L 1 2 1 2 − + + + − × + = L ( ) 583 . 41 12 14 9 7 2 65 10 5 . 39 = + + − × + = Mode ( ) 1 1 1 2 + − − − − − × + = M M M M M M M f f f f f C L ( ) 643 . 36 12 9 14 2 9 14 10 5 . 29 = − − × − × + = Coefficient of variation % 48 . 41 9615 . 42 17.822 = = = μ σ Skewness ( ) ( ) % 2 . 23 822 . 17 5833 . 41 9615 . 42 3 3 = − = − = σ μ Med Q2 (i) Let X be total score, then ( ) 90 , 426 ~ 2 = = σ μ N X ( ) ( ) 0057 . 53 . 2 426 450 450 = > = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − > = > Z P Z P X P σ (ii) Let Y be his average score, then ( ) 6 . 3 , 2 . 85 ~ 2 = = σ μ N Y ( ) ( ) 1 28 . 13 2 . 85 60 60 ≈ − > = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − > = > Z P Z P Y P σ Q3 (i) Let X be total marks, then ( ) 99 , 258 ~ 2 = = σ μ N X ( ) ( ) 01352 . 211 . 2 99 258 280 280 = > = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − > = > Z P Z P X P (ii) Let Y be his average marks, then ( ) 1875 . 6 , 5 .....
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 Spring '09
 Gabriell
 Statistics, Standard Deviation, Harshad number, σ, required probability

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