CHAPTER 5 - Special Distributions

Section 5.2, page 250

Problem 4

We can do this without the table, since Maple can compute it for us:

>	P:=sum(binomial(15,k)*0.4^k*0.6^(15-k),k=6..9);

Problem 6

If EA, EB and EC are the expected number of times each man hits the target, then the expected total number of hits is EA+EB+EC. Now

>	EA:=sum(k*binomial(3,k)*(1/8)^k*(7/8)^(3-k),k=0..3);

>	EB:=sum(k*binomial(5,k)*(1/4)^k*(3/4)^(5-k),k=0..5);

>	EC:=sum(k*binomial(2,k)*(1/2)^k*(1/2)^(2-k),k=0..2);

>

EA+EB+EC;

We expect the target to be hit 21/8 times.

Section 5.4, page 262

Problem 6

We can assume that the occurrence of defects in the tape is a Poisson process. So in 1200 feet we expect 6/5(3) defects, or 3.6 defects. Use this as our Poisson mean and we get that the probability of no defects is simply

>	exp(-3.6);

Section 5.6, page 280

Problem 7

We look at one measurement first. We want the probability that the measurement is between one and two standard deviations to the left of the mean. To use Maple (instead of a table), we do the following:

>	stats[statevalf,cdf,normald](-1)-stats[statevalf,cdf,normald](-2);

That is the probability that one measurement is between 116 and 118 volts. Since the measurements are independent, the probability that all three lie in this interval is

>

%^3;

Problem 10

The sample mean is normally distributed with mean  and variance 4/n, so the standard deviation of the sample mean will be .  For 25 observations, this means 2/5. So "within one unit of  means within 2.5 standard deviations. We appeal to Maple again:

>	stats[statevalf,cdf,normald](2.5)-stats[statevalf,cdf,normald](-2.5);

The probability is nearly 99%.

Section 5.9, page 301

Problem 13

The amount of time (in minutes) that a student takes is exponentially distributed with mean 80, which means with parameter 1/80. We could use problem 8, which says that the minimum of the times taken by the five students is exponential with parameter 5/80, so that the probability that this minimum is less than 40 is

>	int(5/80*exp(-5/80*t),t=0..40);

>

evalf(%);

So it is pretty certain that at least one of them will finish in 40 minutes.

We can also ask what the probability is that none of them will finish within 40 minutes and subtract from 1. Since they're taking the test independently (we hope!), the probability that none finish by 40 minutes is

>	(int(1/80*exp(-1/80*t),t=40..infinity))^5;

and so we get the answer again:

>	simplify(1-%);

Problem 14

Starting at 9:25, the minimum of the rest is exponentially distributed with parameter 4/80. So the probability of 35 more minutes is:

>	int(4/80*exp(-4/80*t),t=0..35);

>

evalf(%);