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

小王爱学习 · 2022年01月11日

可以画图讲解下吗

NO.PZ2019040801000025

问题如下:

Assuming the height of a tree in a forest follows a normal distribution, there are more than 10,000 trees. The sample size of test is 200, the 95% confidence interval of the sample mean of the height is 11 to 35 meters based on a z-statistic, the standard error of mean height is closest to:

选项:

A.

1.96.

B.

3.58.

C.

6.12.

D.

10.50.

解释:

C is correct.

考点:The central limit theorem (CLT)

解析:95%的置信区间是11到 35,z值是1.96.

平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米

11-u/standard deviation =-1.96 35–u/sd =1.96 
1 个答案

DD仔_品职助教 · 2022年01月12日

嗨,从没放弃的小努力你好:


同学你好,具体请看下图:

----------------------------------------------
努力的时光都是限量版,加油!

  • 1

    回答
  • 0

    关注
  • 417

    浏览
相关问题

NO.PZ2019040801000025 问题如下 Assuming the height of a tree in a forest follows a normstribution, there are more th10,000 trees. The sample size of test is 200, the 95% confinintervof the sample meof the height is 11 to 35 meters baseon a z-statistithe stanrerror of meheight is closest to: A.1.96. B.3.58. C.6.12. 10.50. C is correct.考点The centrlimit theorem (CLT)解析95%的置信区间是11到 35,z值是1.96.平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米 如题,请老师赐教,盼复,谢谢

2024-09-11 14:38 1 · 回答

NO.PZ2019040801000025 问题如下 Assuming the height of a tree in a forest follows a normstribution, there are more th10,000 trees. The sample size of test is 200, the 95% confinintervof the sample meof the height is 11 to 35 meters baseon a z-statistithe stanrerror of meheight is closest to: A.1.96. B.3.58. C.6.12. 10.50. C is correct.考点The centrlimit theorem (CLT)解析95%的置信区间是11到 35,z值是1.96.平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米 题目没有提到ranmly select,另外sample mean是在11和35之间,求出来的为什么不是sample s直接就是标准误?

2024-04-18 11:00 1 · 回答

NO.PZ2019040801000025问题如下Assuming the height of a tree in a forest follows a normstribution, there are more th10,000 trees. The sample size of test is 200, the 95% confinintervof the sample meof the height is 11 to 35 meters baseon a z-statistithe stanrerror of meheight is closest to:A.1.96.B.3.58.C.6.12.10.50.C is correct.考点The centrlimit theorem (CLT)解析95%的置信区间是11到 35,z值是1.96.平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米看过之前类似提问,感觉没说清楚

2024-03-25 23:29 1 · 回答

NO.PZ2019040801000025 问题如下 Assuming the height of a tree in a forest follows a normstribution, there are more th10,000 trees. The sample size of test is 200, the 95% confinintervof the sample meof the height is 11 to 35 meters baseon a z-statistithe stanrerror of meheight is closest to: A.1.96. B.3.58. C.6.12. 10.50. C is correct.考点The centrlimit theorem (CLT)解析95%的置信区间是11到 35,z值是1.96.平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米 如题

2024-03-19 10:12 1 · 回答

NO.PZ2019040801000025问题如下Assuming the height of a tree in a forest follows a normstribution, there are more th10,000 trees. The sample size of test is 200, the 95% confinintervof the sample meof the height is 11 to 35 meters baseon a z-statistithe stanrerror of meheight is closest to:A.1.96.B.3.58.C.6.12.10.50.C is correct.考点The centrlimit theorem (CLT)解析95%的置信区间是11到 35,z值是1.96.平均数为0.5*(11+35)=23.所以置信区间是23 ± 1.96sx, sx是标准误。23+1.96sx等于35,sx等于6.12米老师,这个样本(容量200)的均值为什么是(11+35)/2呢?如果这么算不就相当样本容量2棵树,一颗高度11,一颗高度35?

2024-02-25 21:08 3 · 回答