T-Distribution vs Standard Normal Distribution

The $t$-distribution is not one curve but a family, indexed by the degrees of freedom. It is the distribution of $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$ (x-bar minus mu, over s divided by root n) when the data come from a normal population. Swapping the fixed $\sigma$ (sigma) for the sample standard deviation $s$, which itself changes from sample to sample, is what puts the extra weight in the tails. Every member is symmetric about 0, and the family closes on the standard normal as the degrees of freedom grow.

The numbers make that convergence concrete. For a 95 percent interval the t-table gives $t^* = 2.145$ at 14 degrees of freedom, $2.042$ at 30, $1.984$ at 100 and $1.962$ at 1000, against $z^* = 1.960$ for the normal. The gap is 8.6 percent of the critical value at 14 degrees of freedom and about 0.1 percent at 1000.

"The sample is small so use $t$, and large so use $z$" is the wrong rule, and it is the one most students carry in. The trigger is whether $\sigma$ is known, not how big $n$ is. With 500 observations and a standard deviation estimated from them, the correct model is $t$ on 499 degrees of freedom, which happens to sit very close to the normal. Knowing $\sigma$ with $n = 8$ would put you back on $z$.

The heavier tails change verdicts, not just widths. A statistic of 2.00 read on $t$ with 14 degrees of freedom has a two-sided p-value of 0.0653, against the 0.0455 the normal returns for the same 2.00, so at $\alpha = 0.05$ one model rejects and the other does not.

Those tails cover the uncertainty in $s$ and nothing else. They do not repair a skewed population or a stray outlier, which is why a $t$ procedure still asks you to look at the shape of the sample first. The $t$-distribution enters the course at topic 4.2 of Unit 4, Inference for Quantitative Data: Means.

The standard normal distribution is the single normal curve with mean 0 and standard deviation 1, written $Z \sim N(0, 1)$. Any normal variable maps onto it through the z-score $z = \frac{x - \mu}{\sigma}$, where $\mu$ (mu) and $\sigma$ (sigma) are the mean and standard deviation of the original variable. Subtracting and dividing is a linear change of scale, so it moves no area: the proportion of $N(\mu, \sigma)$ below $x$ equals the proportion of $N(0, 1)$ below $z$. That equality is the reason one table can serve every normal distribution.

Suppose scores are $N(500, 100)$ and you want the share below 650. Standardize: $z = \frac{650 - 500}{100} = 1.5$. The area to the left of $z = 1.5$ is 0.9332, so about 93.3 percent of scores fall below 650 and 6.7 percent above. Symmetry hands you the mirror image for free, since the area to the left of $z = -1.5$ is that same 0.0668.

The sentence to unlearn is "I converted to z-scores, so the distribution is normal now." Standardizing relocates the center to 0 and rescales the spread to 1. It changes nothing about shape. A right-skewed set of values becomes a right-skewed set of z-scores with exactly the same skew, and reading a standard normal table on it returns an answer that is simply wrong. The table is valid because the distribution was normal to begin with, not because you standardized it.

Two smaller habits cost points. A table entry is the area to the left of $z$, so a "greater than" question needs 1 minus the entry. And a negative $z$ is not a mistake; it only means the value sits below the mean.

The z-table on this site runs from $z = -3.49$ to $z = 3.49$, matching the AP Statistics Table A layout. Past the ends there is very little left to account for: the area below $z = -3.49$ is about 0.00024.