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M&S 問題 (stat)
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 40 32.0 4.0The teacher wants to sacle the marks so that more students will pass the exam. The new mark corresponding... 顯示更多 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 deviation X>=40 160 64.0 6.0 X<40 40 32.0 4.0 The teacher wants to sacle the marks so that more students will pass the exam. The new mark corresponding to X is y= 50+ 5/6(x-40), if x >=40, y=(5/4)X if x <40. (a) Find the median and standard deviation of the new marks of the 200 students. (b) The median of the original 200marks is 52. Find the median of the new marks. 更新: the answer in b should be 60 更新 2: and i don't know how to calculate the combined standand dev also in line7 , how to obtain 25? and what is the formula? 更新 3: -if u can , plz explain line 11-13 , thz
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, the median belongs to the category >=40 For an original median of 52, the new median = 50+(5/6)(52-40)=60
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 40 32.0 4.0The teacher wants to sacle the marks so that more students will pass the exam. The new mark corresponding... 顯示更多 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 deviation X>=40 160 64.0 6.0 X<40 40 32.0 4.0 The teacher wants to sacle the marks so that more students will pass the exam. The new mark corresponding to X is y= 50+ 5/6(x-40), if x >=40, y=(5/4)X if x <40. (a) Find the median and standard deviation of the new marks of the 200 students. (b) The median of the original 200marks is 52. Find the median of the new marks. 更新: the answer in b should be 60 更新 2: and i don't know how to calculate the combined standand dev also in line7 , how to obtain 25? and what is the formula? 更新 3: -if u can , plz explain line 11-13 , thz
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, the median belongs to the category >=40 For an original median of 52, the new median = 50+(5/6)(52-40)=60
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