標題:
M&S 問題 (stat)
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發問:
The passmark of an examination is set at 40. Below show the mark of 200 students who took the exam.mark range number of students mean standard deviationX>=40 160 64.0 6.0X=40 160 64.0 6.0 X=40, y=(5/4)X if x
最佳解答:
a) New mean of the 160 marks which are originally >= 40 will be 50 + 5(64 - 40)/6 = 70 New standard dev. of the 160 marks which are originally >= 40 will be 5 x 6/6 = 5 New mean of the 40 marks which are originally >= 40 will be 32 x 5/4 = 40 New standard dev. of the 40 marks which are originally >= 40 will be 4 x 5/4 = 5 So new mean of 200 students = (160 x 70 + 40 x 40)/200 = 64 New standard dev of 200 students: √{[160 x 25 + 40 x 25 + 160 x (70 - 64)2 + 40 x (40 - 64)2]/200} = 13 b) The combined set of the original marks has: Mean = 57.6 Standard dev = 14 So the median has a stanrard score of (52 - 57.6)/14 = -0.4 For the combined set of the new marks, median = 64 - 0.4 x 13 = 58.8 2010-05-30 15:52:44 補充： For lines 3 and 4, >= should be 2010-05-30 15:52:57 補充： should be less than "
其他解答:
Since >=40 is bigger than =40 For an original median of 52, the new median = 50+(5/6)(52-40)=60
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