问题如下图：
请问这道题题干是不是出错了，d.应该是Pr（b
NO.PZ2020010303000010 问题如下 Either using a Z table or the Excel function NORM.S.INV, computez so thPr(z Z) = .95 when Z ∼ N(0, 1)z so thPr(z Z) = .95 when Z ∼ N(0, 1)z so thPr(-z Z z) = .75 when Z ∼ N(0, 1) a anb so thPr( X = .75 anPr(X = 0.125 when X ∼ N(2, 4) 1.645. In Excel, the commanto compute this value is NORM.S.INV(.95).-1.645. In Excel, the commanto compute this value is NORM.S.INV(.05).1.15. Here the tail to the left shoulhave 12.5% anthe tail to the right shoulalso have 12.5%. In Excel, the commanto compute this value is –NORM.S.INV(.125). -0.3 an4.3. The area of the left anright shouleahave 12.5%. These cconstructeusing the answer to the previous problem re-centering on the meanscaling the stanrviation, so tha = 2 * -1.15 + 2 anb = 2 * 1.15 + 2. Note ththe formula is a=σ∗q+μa = \sigma * q + \mua=σ∗q+μ, where q is the quantile value. 没看懂问什么， 看解答是在求各种分位点 ？可以下excel的函数是怎么用的吗， 能用考试计算器求出来吗
NO.PZ2020010303000010 问题如下 Either using a Z table or the Excel function NORM.S.INV, computez so thPr(z Z) = .95 when Z ∼ N(0, 1)z so thPr(z Z) = .95 when Z ∼ N(0, 1)z so thPr(-z Z z) = .75 when Z ∼ N(0, 1) a anb so thPr( X = .75 anPr(X = 0.125 when X ∼ N(2, 4) 1.645. In Excel, the commanto compute this value is NORM.S.INV(.95).-1.645. In Excel, the commanto compute this value is NORM.S.INV(.05).1.15. Here the tail to the left shoulhave 12.5% anthe tail to the right shoulalso have 12.5%. In Excel, the commanto compute this value is –NORM.S.INV(.125). -0.3 an4.3. The area of the left anright shouleahave 12.5%. These cconstructeusing the answer to the previous problem re-centering on the meanscaling the stanrviation, so tha = 2 * -1.15 + 2 anb = 2 * 1.15 + 2. Note ththe formula is a=σ∗q+μa = \sigma * q + \mua=σ∗q+μ, where q is the quantile value. 题目Pr(z Z) = .95 Pr(z Z) = Pr(Z z) 求的是小z吧，z= -1.645？题目Pr(z Z) = .95 Pr(z Z) = Pr(Z z) ，z = 1.645？
NO.PZ2020010303000010 老师，第四问根据分位数反求分位点的公式要求掌握吗？讲义中只是提到了两者之间存在反函数关系
NO.PZ2020010303000010 个想请教一下为什么要加上2呀？
NO.PZ20200103030000101.645是怎么计算出的？用的是什么公式？
