_{k}be the number of measurements y that fall in the interval y¯ − ks to y¯ + ks, or |y - y¯| < ks. We are trying to show that n

_{k}/n ≥1-1/k

^{2}. Before we can use their hint, we need to split the sum ∑(y - y¯)² that runs from 1 to n into two sums, the first from 1 to n

_{k}and the second from n

_{k}+1 to n. The first sum includes all those deviations within the interval, |y - y¯| < ks. The second sum includes all the other deviations, i.e., those outside the interval: |y - y¯| ≥ ks. Now the full sum is obviously greater than just the second sum:

_{i=1}

^{n}(y - y¯)² ≥ ∑

_{i=n}

_{k+1}

^{n}(y - y¯)²

_{i=1}

^{n}(y - y¯)² ≥ ∑

_{i=nk+1}

^{n}(ks)²

_{k})(ks)², so that

_{i=1}

^{n}(y - y¯)² ≥(n-n

_{k})(ks)²

^{2 }=(1/(n-1)) ∑

_{i=1}

^{n}(y - y¯)²

^{2}≥(1/(n-1)) (n-n

_{k})(ks)²

_{k}) k²

_{k})/(n-1)

_{k}-1)/(n-1)≈n

_{k}/n ≥1-1/k²

## Comments

_{k}of these measurements lie between y¯ - ks and y¯ + ks._{k}/n is at least 1-1/k².