1. ## probability distribution

Thirty percent of the population in a southwestern community area are Spanish speaking Americans. A spanish speaking person is accused of killing a non-Spanish speaking American. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans and 10 are not. The defendant's lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. What do you think??

I was using of using a Binomial Probabilility Distribution table to look at n = 12 and x = 2 and the probability of success to be .3 and the result is 0.168. so do I multiply this number with 12 so I get 2 as the result after its rounded?

2. Originally Posted by EquinoX
Thirty percent of the population in a southwestern community area are Spanish speaking Americans. A spanish speaking person is accused of killing a non-Spanish speaking American. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans and 10 are not. The defendant's lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. What do you think??

I was using of using a Binomial Probabilility Distribution table to look at n = 12 and x = 2 and the probability of success to be .3 and the result is 0.168. so do I multiply this number with 12 so I get 2 as the result after its rounded?
Let X be the random variable number of Spanish speaking Americans in pool of potential jurors.

X ~ Binomial(n = 12, p = 0.3).

The probability of interest is $\displaystyle \Pr(X \leq 2)$ ..... I get 0.2528 (correct to four decimal places) as the answer ...... Is this small enough to represent bias, do you think ....?

3. I don't really understand why you mentioned it to be x<=2??

and then so as the probability of Spanish speaking American in pool of potential jurors is .2528 then the percentage of 2 Spanish speaking Americans and 10 are not is valid right? Therefore the government lawyer is correct?

4. Originally Posted by EquinoX
I don't really understand why you mentioned it to be x<=2??

and then so as the probability of Spanish speaking American in pool of potential jurors is .2528 then the percentage of 2 Spanish speaking Americans and 10 are not is valid right? Therefore the government lawyer is correct?
You need to think about why it's X <= 2 and not X = 2. What if there were 12 Spanish speaking Americans in the jury pool. You certainly wouldn't think there was a bias against them then, would you? And yet, what's Pr(X = 12) .......?

It's up to you to interpret the result 0.2528. It doesn't seem so low to me that X <= 2 could not have occured by chance ......

5. I still don't understand it...

the probability when x = 12 is 0..

so to make the questions more clear,.. the defender lawyer assumes that the spanish person is not wrong based on that probability but the government lawyer thinks that the killer is amongst those 12 people as the probability is common?

6. Originally Posted by mr fantastic
You need to think about why it's X <= 2 and not X = 2. What if there were 12 Spanish speaking Americans in the jury pool. You certainly wouldn't think there was a bias against them then, would you? And yet, what's Pr(X = 12) .......?

It's up to you to interpret the result 0.2528. It doesn't seem so low to me that X <= 2 could not have occured by chance ......
Originally Posted by EquinoX
I still don't understand it...

the probability when x = 12 is 0..
[snip]
My point was that the Pr(X = whatever) is always small and cannot be used to decide if there's bias or not. You have to use Pr(X <= whatever).

Originally Posted by EquinoX
[snip]
so to make the questions more clear,.. the defender lawyer assumes that the spanish person is not wrong based on that probability but the government lawyer thinks that the killer is amongst those 12 people as the probability is common?
?? I can't think of any legal system that has the defendant part of the 12 person jury .......!!!! I don't know what your're asking here.

The question you originally asked, as I understand it, is whether or not the defending lawyer is correct in claiming that the number of Spanish speaking people on the jury is so low that there's been a deliberate bias in the jury selection .......

7. ok so your answer on P(x<=2) shows the probability of the Spanish speakers in the jury?? am I right?

8. Originally Posted by EquinoX
ok so your answer on P(x<=2) shows the probability of the Spanish speakers in the jury?? am I right?
Yes, the probability that there are less than or equal to two Spanish speaking Americans in the jury. Recall my definition of X ........

9. Originally Posted by EquinoX
ok so your answer on P(x<=2) shows the probability of the Spanish speakers in the jury?? am I right?
It shows the probabilty of at most 2 of the 12 jurors being Spanish-speaking Americans. If you suspect there may be a bias when 2 are selected, then obviously you will think the same bias if 1 or 0 are selected because it shows the same trend in selection but stronger. So you find the probability of them picking 2 or less Spanish speaking americans and then see if it lies inside the critcal region.

10. Originally Posted by sean.1986
It shows the probabilty of at most 2 of the 12 jurors being Spanish-speaking Americans. If you suspect there may be a bias when 2 are selected, then obviously you will think the same bias if 1 or 0 are selected because it shows the same trend in selection but stronger. So you find the probability of them picking 2 or less Spanish speaking americans and then see if it lies inside the critcal region.
The critical region is a region of your choosing. It's a line in the sand - if the probability is below the critical value then you say there's evidence of bias ......

11. well if the probability of Spanish speaking person is 0.2528 out of that 12 then I think the proportion of 2 out of 12 makes sense right??

12. Originally Posted by TheRekz
well if the probability of Spanish speaking person is 0.2528 out of that 12 then I think the proportion of 2 out of 12 makes sense right??
Well, as I remarked earlier ....... I don't the probability is so low as to exclude the possibility that there's only two by pure chance.

13. so what your saying here is that the probability is low enough that we can reject that this proportion is not right?

14. Originally Posted by EquinoX
Thirty percent of the population in a southwestern community area are Spanish speaking Americans. A spanish speaking person is accused of killing a non-Spanish speaking American. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans and 10 are not. The defendant's lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. What do you think??

I was using of using a Binomial Probabilility Distribution table to look at n = 12 and x = 2 and the probability of success to be .3 and the result is 0.168. so do I multiply this number with 12 so I get 2 as the result after its rounded?
Originally Posted by EquinoX
so what your saying here is that the probability is low enough that we can reject that this proportion is not right?
It depends on the level of significance.