开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

Infinite · 2020年02月09日

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:

A.

4.56%

B.

8.96%

C.

13.36%

D.

18.15%

解释:

First convert the cutoff points of 32 and 116 into standard normal deviates. The first is z1=(3280)24=4824=2z_1=\frac{(32-80)}{24}=\frac{48}{24}=-2, and the second is z1=1168024=3624=1.5z_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%.

这里为什么是P(z>1.5) p(z<-2) 为什么不是p(z<1.5) p(z>-2)

1 个答案
已采纳答案

orange品职答疑助手 · 2020年02月09日

同学你好,因为题目问的是不在区间(32,116)里的呀,所以应该求区间以外的概率

  • 1

    回答
  • 0

    关注
  • 335

    浏览
相关问题

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

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%. 题目最后N(-1.5)和N(-2.0)怎么来的,应该怎么理解呢

2020-09-07 08:57 1 · 回答