# How many people should be on a jury to be representative?

Assuming a jury is like a sample of a population.

Assume the following:

The population is a million people. There are X percent who think the defendant is guilty. Choose N random people to be a jury.

Given X, how many jurors (what sample size) would you need to give 95% accuracy of the juror decision being the same majority decision as the entire population.

I would guess that if X is closer to 50% like 50.1% then you would need a very large jury to make sure you got the accurate result. Whereas if X is closer to 0% or 100% you might need just one juror.

So assume the everyone in the population chooses guilty or innocent at random 50/50. (So they might do this and 60% have randomly chosen guilty.) On average how many jurors would you need for them to give the same result as a census 95% of the time? [I added this assumption because we need to know how often the population is split and how often the population is in agreement which will affect the results. But thinking about it this condition would mean the jury would only agree with the population 50% of the time! Right??? So perhaps the condition should be "Provided above 60% of the population is in agreement the jury should agree with the population 95% of the time. How many jurors are needed in the worst case scenario?" ].

i.e. is there some way to calculate the best number of people for a jury such that the population as a whole will be happy most (95%) of the time? And justice is "seen" to be done?

• If everything before "I would guess..." is quoted verbatim from your problem, my interpretation of the problem is different than yours. You say So assume the everyone in the population chooses guilty or innocent at random 50/50. I dont think that is a correct assumption. Since ...X percent who think the defendant is guilty, anyone among them would definitely choose guilty (and the others definitely will choose innocent). – Just_to_Answer Aug 6 '17 at 20:37
• My thought would be to split the cases $X$% $<50$% and $X$% $>50$%. Then essentially it is a binomial distribution with parameters $N$ and $p=X/100$. If $Y_N$ denotes the proportion of the sample whose verdict is guilty, then you want $P(Y_N < 0.5) \geq 0.95$ if $X$% $<50$% and $P(Y_N > 0.5) \geq 0.95$ if $X$% $>50$%. Then my interpretation is that the question is to find the minimum such $N$. – Just_to_Answer Aug 6 '17 at 20:44
• A US jury is supposed to listen to the testimony presented in court during a trial, discuss it among themselves, and then arrive at a verdict based only on the information presented in court. If one accepts your premise that jurors are either PRO or CON at the start, and that is what matters, then one could just skip the trial. // Maybe you can restructure the problem so that the population consists of Young and Old members. (Say 'X percent' Young. )Then how large a sample is needed to be 95% sure that the majority 'age' in the sample matches the majority age in the population. – BruceET Aug 6 '17 at 21:45
• @BruceET Yes, I think that's pretty much the same problem. The tricky bit comes in because I think it would depend on how close to evens the split is. I was thinking that the whole population could watch the trial at the same time. And it just so happens N of the population are jurors. So it might be impossible to select the best number of jurors without knowing how split the opinion is. – zooby Aug 6 '17 at 21:52
• Perhaps a context or origin of the problem would be helpful. Is this from a textbook/exam, or are you making this up on your own? If someone-else made the question, they surely made up the context without thinking about the reality. The question mathematically (after all this is math.stackexchange, not criminaljustice.stackexchange) boils down to given a fixed $p$, finding the minimum $n$ needed such that $P(\hat p \text{ and } p \text{ are both either } < 0.5 \text{ or } > 0.5 ) \geq 0.95$. Of course, the minimum $n$ will depend on $p$. Normal approx to $\hat p$ might be a way to go. – Just_to_Answer Aug 7 '17 at 4:16