Should I be using normpdf to answer this question? If the mean of a sample is 30, and the standard deviation is 10, then how would I evaluate a question that asks me how likely it is that I would obtain a value of $34$? Also the size of the sample is 20.
I'm confused about this:  
Would I calculate $P(X=34)=P(Z=\frac{34-30}{8})=P(Z=0.5)$?
From this I'm confused because a cdf would give me $P(X\leq x )$ and I'm unclear on if I should use normpdf on a TI-84 calculator instead of normcdf. 
I'm an AP Statistics student and my teacher says never to use normpdf for this class. 
My second approach would be to see that $Z=\frac{34-30}{4} = 0.5$ so a value of 34 is only half a standard deviation above the mean. This would mean it's pretty likely. 
Also, does the sample size matter in this situation?
 A: There are a couple ways to interpret this question.
In many cases, we are interested in the question "how likely is it that we observe a value at least as extreme as X?" This is often the case in hypothesis testing, where we will reject the null hypothesis if a value at least as extreme as a specified threshold is observed. So in this case, we might respond with either $\mathbb{P}(Z \geq 0.5)$ or $\mathbb{P}(|Z| \geq 0.5)$, which both measure how likely a value at least this extreme is observed.
Another way might respond is to answer the question "how likely is it that we observe a value between $Z = 0.5 - \epsilon$ and $Z = 0.5 + \epsilon$?" Here $\epsilon$ is some constant that specifies the length of the interval we're interested in (you can think about this as the error of a measurement). In this case, our answer would be
$$\Phi(0.5 + \epsilon) - \Phi(0.5 - \epsilon)$$
where $\Phi$ is the standard normal CDF.
Of course, answering this question would require a reasonable choice of $\epsilon$.
