Re: A problem with solving some nonlinear system

*To*: mathgroup at smc.vnet.net*Subject*: [mg87298] Re: A problem with solving some nonlinear system*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Mon, 7 Apr 2008 05:15:31 -0400 (EDT)

On 4/6/08 at 6:44 AM, uvnarae at hotmail.com (Walkman) wrote: >The problem itself is in the book 428p. "Introduction to >Mathematical Statistics 6E" >In solving this problem, I've got stuck with this practical problem. >To find n and c such that >Integrate[f,{z,(c-75)/10/sqrt(n),inf}] = .05 >Integrate[f,{z,(c-78)/10/sqrt(n),inf}] = .9 >where f = N(0,1); N -> Normal Distribution of which mean = 0 and >variance = 1 What you have posted above isn't proper Mathematica code and leaves a lot for the reader to fill in. I am guessing f is meant to be density function for a normal distribution, i.e., in Mathematica f=PDF[NormalDistribution[0,1],z] and the integral you are expressing should be Integrate[f, {z, (c-75)/10/Sqrt[n],Infinity}] If I have what you want correctly, the integration limit you want can be found using the built in function Quantile. The integral is the upper tail for a normal distribution. So, the first limit can be found with In[5]:= y = Quantile[NormalDistribution[0, 1], .95] Out[5]= 1.64485 This will be (c-75)/10/Sqrt[n] Similary, In[8]:= x = Quantile[NormalDistribution[0, 1], .1] Out[8]= -1.28155 will be (c-78)/10/Sqrt[n] Now Solve can be used to find c and n In[9]:= Solve[{(c - 75)/10/Sqrt[n] == y, (c - 78)/10/Sqrt[n] == x}, {n, c}] Out[9]= {{n->0.0485576,c->75.176}}