# random variables and distributions

Final exam review question #4

I think this is drawing from random variables and distributions. I don't remember how to approach this problem. Anyone have any advice? -_-

• edited July 15

Hi Jordan,
I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)

we are given a P = 0.20 and a P-hat = $(20/60)$ = 0.33

calculate your test statistic using $(P hat - P)/ SE$

SE = $sqrt((P(1-P)/n)$
Take this value to the Z table and find the p-value.
Since the question states "more than 20 of the sixty" , you will take the value of
(1-pvalue).
I hope I did this correctly.

• @Beau said:
Hi Jordan,
I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)

Yes, this is exactly right!

The estimate $\hat{p}$ is a random variable whose mean and standard deviation are based on the actual proportion $p=0.2$ and the sample size $n=60$. More precisely, the mean of $\hat{p}$ is $np$ and the standard deviation of $\hat{p}$ is
$\sqrt{p(1-p)/n}$. Thus, you use a normal distribution with those values for mean and standard deviation to do your estimate.