Suppose X is a normal random variable with mean 60 and standard deviation 2. A z-score was calculated for a number, and the z-score is 3.4. What is x?

Z-Score规范化可将数据归一化至0-1的范围内。

Z-score模型无法计算投资组合信用风险，因为Z-score模型主要是针对个别资产的信用风险进行评估，对整个投资组合的信用风险无法衡量。

下面代码的输出结果是()。 <?php $score=87; if($score>=90 && $score<=100){$grade="优秀";} if($score>=80){$grade="良好";} if($score>=70){$grade="一般";} if($score>=60){$grade="及格";} else{$grade="不及格";} echo $grade; ?>

以下参数传递正确的是： A: def score(mid_score, end_score): ……score(88, 79) B: def score( *，mid_score, end_score, mid_rate＝0.4): ……score(88, 79) C: def score( *，mid_score, end_score, mid_rate＝0.4): ……score(end_score = 79, mid_score = 88) D: def score( *，mid_score, end_score, mid_rate＝0.4): ……score(mid_score = 88, end_score = 79,0.5)

There are ________students playing basketball in the room. A: score of B: scores of C: two score D: two score of

若有定义int score&#91;10&#93;;，则对score数组中的元素的正确引用是（ ） A: score[10] B: score[6.0] C: score[0] D: score(7)

从学生成绩表SC中查询平均成绩大于60分的同学的学号和平均成绩的正确语句为： A: selectsno,avg(score)assscorefromscgroupbysnohavingavg(score)>60 B: selectsno,avg(score)assscorefromscgroupbysnowhereavg(score)>60 C: selectsno,avg(score)assscorefromscwhereavg(score)>60 D: selectsno,avg(score)assscorefromschavingavg(score)>60

查询成绩在60-80之间的学生信息，正确的语句是？ A: db.students.find({"score":{$gte:60,$lte:80}}) B: db.students.find({"score":{$gte:80},{"score":{$lte:60}}}) C: db.students.find({"score":{$gte:60},{"score":{$te:80}}}) D: db.students.find({"score":{$gt:80},{"score":{$lt:60}}})

下列程序片段用于产生成绩等第，已定义score%和mark$分别表示成绩和等第。即60分（含）以上及格、75分（含）以上良好、90分（含）以上优秀，能获得预期结果的是______。 A: If score>=90 Then Mark="优秀" ElseIf score>=75 Then Mark="良好" ElseIf score>=60 Then Mark="及格" End If B: If score>=60 Then Mark="及格" ElseIf score>=75 Then Mark="良好" ElseIf score>=90 Then Mark="优秀" C: If score>=90 Then Mark="优秀"If score>=75 Then Mark="良好"If score>=60 Then Mark="及格" D: If score>=60 Then Mark="及格" ElseIf score>=75 Then Mark="良好"ElseIf score>=90 Then Mark="优秀" End If