With $N=36$, $\sum fd^2=197500$ and $\sum fd=-750$, the variance of the distribution to three decimal places is:

1. Use $\sigma^2=\dfrac{\sum fd^2}{N}-\left(\dfrac{\sum fd}{N}\right)^2$. 2. First term: $\dfrac{197500}{36}=5486.111$. 3. Second term: $\left(\dfrac{-750}{36}\right)^2=(-20.8333)^2=434.028$. 4. Subtract: $5486.111-434.028=5052.083$. 5. $5486.111$ stops after step 2; $434.028$ reports only the correction; $4618.056$ subtracts twice. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 4: Statistics_