题中的based on a z-statistic，不是查Z分布表的意思，对吗？
查Z分布表的1.96是97.5，而且Z分布表查出来的是不对称的，所以这个怎么理解呢？还是有两种Z分布表呢？在不考虑老师让记住这个数的情况下，百分之95的1.96是查表得出来的，还是通过查表算出来的呢？
问题如下图：
选项：
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B. 
C. 
D. 
解释：
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
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？
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
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
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？
