40. Figure [A] demonstrates a normal distribution of English midterm performance of Eagle class students. The statistical data shows that students’ mean score is 70 points, with a standard deviation of 10 points. There are 30 students in Eagle class. Based on the data and the graph, which of the following statement is     TRUE    ?phpMeWoyU
(A) w= 55, x=60, y=80, z=90
(B) Twenty-five students in Eagle Class scored above 60 and passed the English midterm exam.
(C) Students who scored within the 90-100 range were as many as students who scored within the 50-60 range.
(D) According to the graph, we are 95 percent certain that individual student’s actual score will be within one standard error of 70 points. W X M Y Z 

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統計： A(6), B(65), C(31), D(36), E(0) #1616601

詳解 (共 4 筆)

#3197692
above 60 60分以上依圖表了解6...
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#3300658

這是一個常態分配圖，所以正中間(M)是平均數、中數和眾數=60

而1個標準差10分，0.5個標準差就會是5分

因此

A) w=55 x=60 y=75 z=90

B) 正確

C) 90-100的學生數=1.7%+0.5%+0.1%=2.3%
      50-60的學生數=4.4%+9.2%+15%=28.6%
      應該是90-100分的學生數與40-50分的學生數相同
      或是50-60分的學生數與80-90分的學生數相同

D) 我看不太懂，但95%的常態分配隨機變數的值會落在離平均值±1.96個標準差內

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#2682495

 mean score is 70 points,  平均分數70分

standard deviation of 10 points  標準差10分


   

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#5449317

According to the graph, we are 95 percent certain that individual student’s actual score will be within one standard error of 70 points.

 

正負一個標準差為 68 % ，不是95 %

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