E[X^2] weights each time with the time, giving the square, and the E[X] in the denominator is the normalisation factor (also required to fix the dimensions).
Say that there are to different waiting times 1s and 3s, and they happen with probability 50% each. The average waiting time (1/2 1+1/2 3) is 2s. However, 75% of the time we are waiting on a 3s event and only 25% on a 1s event. The weighted average is 2.5s. E[X^2]=1/2 1+1/2 9=5(s^2) is not the right answer, it still has to be divided by E[X]=2(s) to get the correct answer.
Say that there are to different waiting times 1s and 3s, and they happen with probability 50% each. The average waiting time (1/2 1+1/2 3) is 2s. However, 75% of the time we are waiting on a 3s event and only 25% on a 1s event. The weighted average is 2.5s. E[X^2]=1/2 1+1/2 9=5(s^2) is not the right answer, it still has to be divided by E[X]=2(s) to get the correct answer.