CHAPTER 3 - Random Variables and Distributions

Section 3.1, page 102

Problem 2

We must have

>	Sum(c*x,x=1..5)=1;

>	solve(%,c);

>

value(%);

So the probability distribution is P(x)=x/15.

Problem 8

This is about the binomial distribution. With Maple, we can do it the hard way:

>	sum(binomial(20,k)*(1/10)^k*(9/10)^(20-k),k=4..20);

>

evalf(%);

The probability is a little more than 13% that more than three of the balls will be red.

Section 3.2, page 109

Problem 4

First, the constant c:

>	solve(int(c*x^2,x=1..2)=1,c);

Now the sketch:

>	plot(3/7*x^2,x=1..2,color=blue,thickness=2,view=[0..3,0..2]);

Part (b) requires:

>	int(3/7*x^2,x=3/2..2);

Problem 8

This is like problem 4:

>	solve(int(c*exp(-2*x),x=0..infinity)=1,c);

>	plot(2*exp(-2*x),x=0..6,color=blue,thickness=2);

>	int(2*exp(-2*x),x=1..2);

>

evalf(%);

So the probability that we will find X between 1 and 2 is about 11.7 %

Section 3.3, page 116

Problem 8

We start with the cumulative distribution function of Z. First, note that the pdf on the disk is uniform with a value of , since the area of the whole disk is . So the probability that a point is within r units from the origin is , or simply . Since the pdf of the distribution of Z is simply the derivative of its cumulative distribution, we have that the df of Z is 2r. Here's a sketch:

>	plot(2*r,r=0..1,color=blue,thickness=2);