叶赫那拉坤坤 · 2020年06月01日
问题如下：
A variable obeying normal distribution has a mean of 0.9 and a standard deviation of 1.5. The probability for that variable to fall between 0.9 and 1.2 is:
选项：
A.7.93%
B.6.91%.
C.5.82%.
解释：
A is correct. Firstly, standardize the original normal distribution：Z= （x-0.9)/1.5, so P(0.9 ≤ x ≤ 1.2) = P( 0 ≤ Z ≤ 0.2).
(1.2-0.9)/1.5=0.2, (0.9-0.9)/1.5=0
Based on the given table, N(0) = 0.5000, N(0.2) = 0.5793. P( 0 ≤ Z ≤ 0.2) = 0.5793– 0.5000 = 0.0793 = 7.93%
不用除以根号n吗？
NO.PZ2018062016000087问题如下A variable obeying normstribution ha meof 0.9 ana stanrviation of 1.5. The probability for thvariable to fall between 0.9 an1.2 is:A.7.93%B.6.91%.C.5.82%.A is correct. Firstly, stanrze the originnormstributionZ= （x-0.9)/1.5, so P(0.9 ≤ x ≤ 1.2) = P( 0 ≤ Z ≤ 0.2).(1.2-0.9)/1.5=0.2, (0.9-0.9)/1.5=0Baseon the given table, N(0) = 0.5000, N(0.2) = 0.5793. P( 0 ≤ Z ≤ 0.2) = 0.5793– 0.5000 = 0.0793 = 7.93%为什么是用（0.9-0.9）/1.5求z，而不是z=0.9或1.2ne
NO.PZ2018062016000087 题目说“变量服从正态分布的变量均值为0.9，标准差为1.5。”这个均值0.9，是指实际总体的均值吗，还是只是我们通过n个样本测出来的期望值（理论值）？
NO.PZ2018062016000087 6.91%. 5.82%. A is correct. Firstly, stanrze the originnormstributionZ= （x-0.9)/1.5, so P(0.9 ≤ x ≤ 1.2) = P( 0 ≤ Z ≤ 0.2). (1.2-0.9)/1.5=0.2, (0.9-0.9)/1.5=0 Baseon the given table, N(0) = 0.5000, N(0.2) = 0.5793. P( 0 ≤ Z ≤ 0.2) = 0.5793– 0.5000 = 0.0793 = 7.93%表的横轴和纵轴都代表什么啊 z=0.2的时候 怎么查表呢
这步骤没看懂 P( 0 ≤ Z ≤ 0.2) = 0.5793– 0.5000 = 0.0793
