Trying to find the probability $b/w$ two scores. I know mapping to $z$-score is:
$$z_{11} = (11-10)/1.5 = .67$$
and
$$z_{14} = (14-10)/1.5 = 2.67$$
So,
$$\bar{x} = 10,\ σ = 1.5.$$
Would I simple subtract the two from each other? and so on the real number line it would range from $0$ to $100$ and that is how the distribution would be set up. So, $2.67 - 2 = 2$?
On the $z$-table $2$ is given a value of $0.9772$.
My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between $11$ and $14$ or if the question said what's the percentage of scores that fall between $11$ and $14$. How might be able to address both scenarios?
Just confused on the ideas of proportion, probability, and percentage and how they are relate.
Thank you.
Hint:
$P(11 < X < 14) = P(z_{11} < \frac{X-\mu}{\sigma} <z_{14}) = P(Z < z_{14}) - P(Z \leq z_{11})$