To illustrate finding a $p$-value, consider a sample of $25$ individuals (which provides $24$ degrees of freedom) resulting in a computed $t$ score of $1.50$. The probability of obtaining a $t$ score at least this extreme depends on the proportion of expected scores in the corresponding distribution that are far from zero. Defining extreme in both directions, the $p$-value is the proportion of expected $t$ scores that are $1.50$ or greater, combined with those that are $-1.50$ or lower. In this scenario, that combined proportion is $.14$, meaning the final $p$-value is $.14$.