During an experiment, some water was removed from each of 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?
(1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
(2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
Answer:A
这题有人解释过说平均数是原来的0.7,然后标准方差就不变。实在想不通。标准方差不是每项还要减去平均值再平方吗?标准方差应该变好多吧。
2.x的四次方加y的四次方=100.问the greatest possible value of x is between?
0~3 3~6 6~9 9~12 12~15
答案B,我选A。
选B的话,x取4,4的四次方都大于100了,这怎么可能呢?
请教啊!!!
第一题,平均数是原来的0.7没错,不过标准方差也是原来的0.7,也就是7 gallons,所以答案还是A
第二题,问的是greatest possible value of x,由于y的4次方的最小值为0,所以x的4次方的最大值为100,算出答案是3-4,也就是选B
第一题:设Xn为第n个蓄水池蓄的水,则
standard deviation = (X1-(X1+X2+X3+X4+X5+X6)/6)^2
+ (X2-(X1+X2+X3+X4+X5+X6)/6)^2
+ (X3-(X1+X2+X3+X4+X5+X6)/6)^2
+ (X4-(X1+X2+X3+X4+X5+X6)/6)^2
+ (X5-(X1+X2+X3+X4+X5+X6)/6)^2
+ (X6-(X1+X2+X3+X4+X5+X6)/6)^2
根据第一个条件,可以提出公共系数0.7,则
new standard deviation = 0.49*(X1-(X1+X2+X3+X4+X5+X6)/6)^2
+ 0.49*((X2-(X1+X2+X3+X4+X5+X6)/6)^2
+ 0.49*((X3-(X1+X2+X3+X4+X5+X6)/6)^2
+ 0.49*((X4-(X1+X2+X3+X4+X5+X6)/6)^2
+ 0.49*((X5-(X1+X2+X3+X4+X5+X6)/6)^2
+ 0.49*((X6-(X1+X2+X3+X4+X5+X6)/6)^2
第二个条件无法得出新的标准方差,选A。
第二题,3的四次方是81,4的四次方是256,可见X最大值在3和4之间,
那当然也就在3和6之间了,我是这么理解的。
啊!原来是说最大值在哪个范围内啊!郁闷,题目意思理解错了。谢谢了!
还有,哪个说数学简单的!!!我就有不少题不会做,尤其是考试时时间还那么紧张!!!
嗯~ 我也是做CXD的,不过大家都说他的简单
做PREP我也错很多
一战时数学没怎么准备就去了. 才拿了38分, 简直是惨不忍睹..
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