The weight of men in the United States is normally distributed with mean $E(X) = 185 $ pounds and standard deviation, $\sigma = 10$. If a boat containing $n=25$ total passengers of U.S. men has a weight capacity of 6,000 pounds. what is the probability that these passengers will exceed the weight capacity?
I am not really sure how to go about this problem. I'm guessing that you have to convert it into a z-score somehow, but I'm not sure how to apply the 25 men sample. I know that $Z = x-\frac{185}{10}$ or maybe $Z = x-\frac{185\sqrt{25}}{10}$. So what would I chose as my $X$? Would I divide 6000 by 25 to find the average weight it has to be to exceed the capacity? Just trying a few things, I got the probability that they will exceed the weight capacity to be basically zero. Not sure if that is correct, and could use some help.
You are correct that the answer is very nearly 0, but (even after @Drew's nice edit) it does not seem you are at all clear on how to get the answer.
Here is an outline:
1) The total weight $T = X_1 + X_2 + \cdots + X_{25}$ is normally distributed. Figure out its mean $\mu_T$ and its SD $\sigma_T.$
2) Standardize: $P(T > 6000) = P\left(Z = \frac{T-\mu_T}{\sigma_T} > \frac{6000 - \mu_T}{\sigma_T}\right).$
3) Evaluate the quantity $\frac{6000 - \mu_T}{\sigma_T}$, and then the probability, using standard normal tables or software.
Below is a graph of the density function of the distribution of $T$, Your answer is the area beneath the curve to the right of the vertical red line.