quietvivian · 2020年09月18日
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 , and the second is . 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%.
老师，这个考试的时候会给表吗
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？
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？
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
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
