为什么不是用x-80/24

NO.PZ2016062402000007

问题如下：

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?

选项：

4.56%

8.96%

13.36%

18.15%

解释：

First convert the cutoff points of 32 and 116 into standard normal deviates. The first is $z_1=\frac{(32-80)}{24}=\frac{48}{24}=-2$ , and the second is $z_1=\frac{116-80}{24}=\frac{36}{24}=1.5$ . From normal tables, P(Z > +1.5) = N(-1.5) = 0.0668 and P(Z < -2.0) = N(-2.0) = 0.0228. Summing gives 8.96%.

为什么不是用x-80/24看小于32和大于116的概率加总呢？

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1 个答案

DD仔_品职助教 · 2022年07月19日

嗨，从没放弃的小努力你好：

同学的你的思路和答案是一样的呀，已经给定了x是32或者116，带入x-80/24其实就是标准化的过程，我们得出z1=-2，z2=1.5，那么不在32~116之间的概率，也就是标准化之后不在-2~1.5之间的概率，那么如下图，通过查表得出两个阴影面积的概率，相加就是答案。

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NO.PZ2016062402000007问题如下 Assume tha ranm variable follows a normstribution with a meof 80 ana stanrviation of 24. Whpercentage of this stribution is not between 32 an116? 4.56% 8.96% 13.36% 18.15% First convert the cutoff points of 32 an116 into stanrnormviates. The first is z1=(32−80)24=4824=−2z_1=\frac{(32-80)}{24}=\frac{48}{24}=-2z1=24(32−80)=2448=−2, anthe seconis z1=116−8024=3624=1.5z_1=\frac{116-80}{24}=\frac{36}{24}=1.5z1=24116−80=2436=1.5. From normtables, P(Z +1.5) = N(-1.5) = 0.0668 anP(Z -2.0) = N(-2.0) = 0.0228. Summing gives 8.96%. 有快速判断的方法吗？

2023-10-18 08:35 1 · 回答

NO.PZ2016062402000007 Assume tha ranm variable follows a normstribution with a meof 80 ana stanrviation of 24. Whpercentage of this stribution is not between 32 an116? 4.56% 8.96% 13.36% 18.15% First convert the cutoff points of 32 an116 into stanrnormviates. The first is z1=(32−80)24=4824=−2z_1=\frac{(32-80)}{24}=\frac{48}{24}=-2z1=24(32−80)=2448=−2, anthe seconis z1=116−8024=3624=1.5z_1=\frac{116-80}{24}=\frac{36}{24}=1.5z1=24116−80=2436=1.5. From normtables, P(Z > +1.5) = N(-1.5) = 0.0668 anP(Z < -2.0) = N(-2.0) = 0.0228. Summing gives 8.96%. 技术上应该不是难事，这样体验感比较好

2021-08-02 16:10 1 · 回答

Assume tha ranm variable follows a normstribution with a meof 80 ana stanrviation of 24. Whpercentage of this stribution is not between 32 an116? 4.56% 8.96% 13.36% 18.15% First convert the cutoff points of 32 an116 into stanrnormviates. The first is z1=(32−80)24=4824=−2z_1=\frac{(32-80)}{24}=\frac{48}{24}=-2z1=24(32−80)=2448=−2, anthe seconis z1=116−8024=3624=1.5z_1=\frac{116-80}{24}=\frac{36}{24}=1.5z1=24116−80=2436=1.5. From normtables, P(Z > +1.5) = N(-1.5) = 0.0668 anP(Z < -2.0) = N(-2.0) = 0.0228. Summing gives 8.96%. 老师，这个考试的时候会给表吗

2020-09-18 07:28 1 · 回答

2020-09-07 08:57 1 · 回答

为什么不是用x-80/24

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