In Chemistry, she was $\frac{102-90}{64}$ standard deviation units above the mean.
In Statistics, she was $\frac{77-70}{16}$ standard deviation units above the mean. Note that $\frac{7}{16}$ is quite a bit larger than $\frac{12}{64}$, so the performance in Statistics is (comparatively) better than in Chemistry.
Measuring in standard deviation units takes above (or below) the mean takes account of the different means and different amounts of variability in the class scores in the two subjects.
"Mound-shaped" presumably means more or less normally distributed. The probability that a normal random variable with mean $90$ and standard deviation $64$ is $\ge 102$ is equal to the probability that a standard normal is $\ge \frac{112-90}{64}$. If we look at a table of the standard normal, this is approximately $0.43$. The corresponding result for Statistics is about $0.33$. Informally, that means that about $43\%$ of the students were above her in Chemistry, and only about $33\%$ were above her in Statistics.
Remark: In my experience, unmanipulated marks from Mathematics exams have a very far from normal distribution. They are often consistent with being drawn from $2$ or more populations with radically different means.
If $X$ is a normal random variable, you can record an observation of it, $x$, and compare it to the mean. The usual way to do this is to standardize the variable, i.e.,
$$z = \frac{x - \mu}{\sigma}$$
Let's say that $X_1, X_2, \ldots X_n$ are random variables from the same distribution as $X$ above. If we record observations of each and calculate the mean, that's also a random variable. However, we can't expect the mean, $\overline{X}$, our new random variable to have the same distribution as our original distribution. It will have the same mean, but it won't have the same variance.
Think of it this way: Make $n$ larger and larger — record more and more observations. It seems that after a while, the mean of all those observations will be the same as the mean from the population. To make it a bit more concrete: Flip a coin a few times, letting $X = 1$ for heads and $X = 0$ for tails. Will your mean be $0.5$? Probably not. Flip it a few more times. Maybe you're a bit closer to $0.5$. By the time you flip the coin, say, a few thousand times, you'll probably be very close to $0.5$.
In other words, when we record a sample mean, making many observations, in the long term, restricts how far we can stray from the true mean. This is shown by the fact that
$$\text{Var}(\overline{X}) = \frac{\sigma^2}{n}$$
Note that as $n \rightarrow \infty$, $\text{Var}(\overline{X}) \rightarrow 0$. This is the gist of the Central Limit Theorem.
The Central Limit Theorem tells us that, regardless of the distribution of a random variable, as we take larger and larger samples, the distribution of the sampling mean (i.e., of $\overline{X}$) is normally distributed, and so we can use all the convenient properties of the normal distribution (like the standardized form). So when we write
$$z = \frac{\overline{x}-\mu}{\sigma / \sqrt{n}}$$
it's really the same thing as the earlier $z$: It's the difference of an observation and an expected value divided by the standard deviation of whatever distribution the observation came from. The new standard deviation (the standard error) is derived from the old one, but that's because the new distribution is derived from the old one.
Remember, $\sigma^2$ stands for population variance. So regardless of whether we're looking at a sample of size n or just one observation, it's always the population variance. $σ^2/n$ is the variance of the sample mean in terms of the population variance. (And, of course, $σ/\sqrt{n}$ is the square root of the variance of the sample mean.)
Best Answer
Standardization is the process of applying a linear transformation of a random variable so that its mean is zero and its variance is one.
Taking a non-standardized variable, if you subtract the mean, the new mean is obviously zero.