I need help in the setup of this problem. I'm sure that I'm making this far more complicated than what it actually needs to be.
"An anthropologist wishes to estimate the average height of men for a certain race of people. If the population standard deviation is assumed to be 2.5inches and if she randomly samples 100 men, find the probability that the difference between the sample mean and the true population mean will not exceed .5 inch."
Is this the correct setup? $ \frac{0.5-100\mu}{2.5/\sqrt(100)}$ How do i solve if I don't know the population mean?
I have a horrible feeling about this answer but...
Let the average height of be the random variable $X$. We know that $\sigma = 2.5$ inches.
By the central limit theorem for a sample of $100$ we have $\overline{X} \sim N(\mu,\frac{2.5^2}{100})$
The condition "find the probability that the difference between the sample mean and the true population will not exceed $0.5$ inch." can be written as $p(\mu -0.5 \le \overline{X} \le \mu+0.5)$
If we use the standard transformation $Z=\frac{\overline{X}-\mu}{\frac{2.5}{\sqrt{100}}}$ we have $$p \left( \frac{-0.5}{0.25} \le Z \le \frac{0.5}{0.25}\right) \\p( -2\le Z \le 2) $$
To calculate the probability that $Z$ lies in that interval we have $$p(Z\le2)-(1-p(Z\le2))$$ by looking at $\Phi(2)=p(Z\le2)=0.99725$ we finally have $$0.9975-0.00275=0.9945$$