# Probability Question

## Homework Statement

This question relates to a famous court case in the USA where for the first time results of
statistical analysis were presented as evidence of discrimination. In 1972 Mr Partida, a
Mexican-American, was convicted of burglary by a jury in a particular county in Texas.
Mr Partida’s lawyers, appealed the conviction on the grounds that the jury must have
been selected in a discriminatory (and thus non-random fashion) because it had a
disproportionately small number of Mexican-Americans on it. They based their argument
on the following two pieces of information. First, a recent census had shown Mexican-
Americans made up 79.1% of adults in the population in that county and that this fact
was widely accepted. Second, in a sample of court cases in the county involving a total of
870 jury members, only 39% were Mexican-Americans. Assume that we will conclude
that jury selection in the county was discriminatory if there is less than a 1% chance that
a sample proportion of 39% (or less) would result from random selection from a
population where the proportion was 79.1%. If you were a Justice of the US Supreme
Court, what would your finding be? Would you find that there probably was
discrimination in jury selection in the county?

## Homework Equations

lets put Pr[A] = 0.791
Pr = 0.39
number of Mexican American Jury is 870*0.39=339

## The Attempt at a Solution

i was thinking use Pr[A|B] to work it out, but since event A and B are dependent, so i can't keep going

also i try using 870C339 (0.791)^339 (0.209)^(870-339)
but this one is at zero..

so can anyone give me some hint? i am totally lost on "less than a 1% chance that
a sample proportion of 39% (or less) would result from random selection from a
population where the proportion was 79.1%"

HallsofIvy
Homework Helper
With large numbers like those, I would be inclined to use the Normal approximation to the binomal distribution. If Mexican-Americans make up 79.1% of the population, then p= 0.791 and q= 1-0.791= 0.209. Fair selection would be binomial distribution with mean p= 0.791 and standard deviation $\sqrt{pq}= \sqrt{(0.791)(0.209)}= \sqrt{0.165319}= 0.4066$.

The "standard variable" would be $z= \frac{x- \mu}{\sigma}= \frac{0.39- 0.791}{0.4066}$. Look that up in a table of the normal distribution to see if the probability of getting that, or less, is less than 1%.

A good table is at http://www.math.unb.ca/~knight/utility/NormTble.htm [Broken]

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