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}}

```

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