This is a phrase from the book "Introduction to Algorithms" 3rd edition page 28.
Suppose that we randomly choose $n$ numbers and apply insertion sort. How long does it take to determine where in subarray $A[1\dots j-1]$ to insert element $A[j]$? On average, half the elements in $A[1\dots j-1]$ are less than $A[j]$, and half the elements are greater. On average, therefore, we check half of the subarray $A[1\dots j-1]$, and so $t_j$ is about $\frac{j}{2}$.
I can't comprehend why on average, $A[j]$ is bigger than half of the elements of $A[1\dots j-1]$. For $A[j]$ to be bigger than half of elements of $A[1\dots j-1]$, shouldn't it be the average value of the elements of $A[1\dots j-1]$? How do we know that $A[j]$ is the average value for the average case? I might be confused on the English language level, not on the mathematical level but English is not my primary language. So please take that into account.