With $N=36$, $\sum fd^2=197500$ and $\sum fd=-750$, the variance of th…

Question

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

Accepted Answer

B. 5052.083

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_

Answer

5486.111

Answer

4618.056

Answer

434.028

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

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