3. Distributions in Minitab

3.1 Binomial Distribution

In an attempt to assess the effect of attitude towards capital punishment on the voting behaviour of jurors in criminal cases, jurors who had voted in split ballots were interviewed.

In this exercise we will consider only the jurors who had voted in ballots where the jury was split in 6 for ``guilty'' vs 6 for ``not guilty''. Of these jurors, 19 stated that they had no conscientious scruples against the death penalty. Of these 19 jurors 12 voted ``guilty'' on the first ballot, the remaining 7 voted ``not guilty'' on the first ballot.

Are jurors without scruples against the death penalty more likely to vote ``guilty''? Let X denote the number of ``guilty'' votes among these n=19 jurors and let p be the - unknown - probability of a ``scrupleless'' juror voting ``guilty'' on the first ballot. We will try to settle the question about ''scrupleless jurors'' by looking at the probability distribution of X.

Question 1: Find the probability distribution for X if p=0.5.

Use the pdf command to answer Question 1:

	pdf;
	binomial n=19, p=0.5.

The pdf command will print out P(X=k) for k=1,...,19. All you have left to do is copy the numbers from the screen.

Click here for an answer.

Question 2: Interpret the computed probabilities. How likely is it that 12 or even more jurors without scruples vote ``guilty'' if the value for p was really equal 1/2?

You might find the cdf command helpful:

	cdf;
	binomial n=19, p=0.5.

The output gives probabilities of the type P(X < k), k=0,...,19.

Click here for an answer.

3.2 The Normal Distribution

Assume that the number of miles a driver gets on a set of radial tires is normally distributed with a mean of 30,000 miles and a standard deviation of 5000 miles.

Question 3: Would the manufacturer of these tires be justified in claiming that 90% of all drivers will get at least 25,000 miles?

The question can be answered by finding the 10-th percentile of the normal distribution with mean $\mu =30,000$ and standard deviation s=5000. We could use the normal table in the book to find the 10-th percentile of the standard normal distribution and then transform to the 10-th percentile of the normal distribution with m=30000, s=5000. by using an inverse z-transformation. Alternatively use the invcdf command in Minitab:

     invcdf 0.10;
     normal mu=30000, sigma=5000.

Or we could have used the cdf command to find P(X<25,000).

     cdf 25000;
     normal mu=30000, sigma=5000.

Click here for an answer.