问一道题：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%.

什么时候公式用(x-u)/standard deviation, 有得题用u+-a*standard deviation?

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1 个答案

orange品职答疑助手 · 2020年02月09日

同学你好，这边是求概率啊，之所以此处用(x-u)/standard deviation，是为了把X转化为服从标准正态分布后，可以直接查表得到分位点。

u+-a*standard deviation 是去计算置信区间的。

同学好好理一下这边，想清楚了做题就不会错了~

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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 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%. 为什么不是用x-80/24看小于32和大于116的概率加总呢？

2022-07-18 22:59 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

问一道题：NO.PZ2016062402000007

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