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

返回

小王爱学习 · 2022年01月11日

可以画图讲解下吗

  1. FRM I
  2. Quantitative

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 
添加评论
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

  • 418

    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