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標題:

mean median midhinge

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12 randomly sele cted students of ABC Secondary School are asked about the pocket money they received from parents every week, and the results (in $) :335 / 500 / 100 / 246 / 375 / 401 / 280 / 410 / 262 / 800 / 173 / 333Find:(a) the mean;(b) the median;(c) the standard deviation;(d) the... 顯示更多 12 randomly sele cted students of ABC Secondary School are asked about the pocket money they received from parents every week, and the results (in $) : 335 / 500 / 100 / 246 / 375 / 401 / 280 / 410 / 262 / 800 / 173 / 333 Find: (a) the mean; (b) the median; (c) the standard deviation; (d) the coefficient of skewness with comment; (e ) the midhinge; (f) the interquartile range; and (g) draw the box-and-whisker plot

最佳解答:

Ordered Data Set: { 100, 173, 246, 262, 280, 333, 335, 375, 401, 410, 500, 800 } Ans (a) Mean = (100+173+246+262+280+333+335+375+401+410+500+800) / 12 = 4215 /12 = 351.25 Ans (b) Median = 2nd Quartile = (333+335)/2 = 334 Ans (c) Standard Deviation = √{[(100-351.25)^2+(173-351.25)^2+(246-351.25)^2+(262-351.25)^2+(280-351.25)^2+(333-351.25)^2+(335-351.25)^2+(375-351.25)^2+(401-351.25)^2+(410-351.25)^2+(500-351.25)^2+(800-351.25)^2 ] / 12 } = 178.277 Ans (d) Coefficient ofSkewness = [3 x (Mean - Median)] / Standard Deviation = [ 3 x (351.25-334)] / 178.277 = 0.29 Comment: Sk = 0.29 >0, which means it is postively skewed, i.e. mean is larger than mode. Ans (e) Q1 = 1st Quartile = (246 + 262)/2 = 254 Q3 = 3rd Quartile = (401 + 410) /2 = 405.5 Midhinge = (Q1 + Q3) / 2 = (254 + 405.5) / 2 = 329.75 Ans (f) Interquartile range = Q3 - Q1 = 405.5 - 254 = 151.5 Ans (g) Based on following information: lowest observation = 100 1st Quartile = 254 Median = 334 3rd Quartile = 405.5 Largest observation = 800 then Box and Whisker plot = _______ I--------I__ I___ I-------------------I 100 254 334 405.5 800 +----+----+----+----+----+----+----+----+----+ 0 100 200 300 400 500 600 700 800 900

其他解答:6CC7293C79127CE5
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