Question
Calculate Karl Pearson's coefficient of correlation by actual mean method:

Answer

Let X denotes price and Y denotes quantity.
X $\text{X}-\overline{\text{X}}$ $(\text{X}-\overline{\text{X}})^2$ Y $\text{Y}-\overline{\text{Y}}$ $(\text{Y}-\overline{\text{Y}})^2$ $(\text{X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})$
10 -4 16 20 -4 16 16
12 -2 4 29 5 25 -10
14 0 0 21 -3 9 0
16 2 1 22 -2 4 -4
18 4 16 28 4 16 16
$\Sigma\text{X}=70\ \text{N}=5$   $\Sigma(\text{X}-\overline{\text{X}})^2=40$ $\Sigma\text{Y}=120$   $\Sigma(\text{Y}-\overline{\text{Y}})^2=70$ $\Sigma(\text{X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})=18$
$\overline{\text{X}}=\frac{\Sigma\text{X}}{\text{N}}=\frac{70}{5}=14$ $\overline{\text{Y}}=\frac{\Sigma\text{Y}}{\text{N}}=\frac{120}{5}=24$
$\text{r}=\frac{\Sigma\text{(X}-\overline{\text{X}})(\text{Y}-\overline{\text{Y}})}{\sqrt{\Sigma\text{(X}-\overline{\text{X}}) ^2}\sqrt{\Sigma(\text{Y}-\overline{\text{Y}})^2}}$
$\frac{180}{\sqrt{40}\sqrt{70}}=\frac{18}{6.32\times8.36}=\frac{18}{52.83}=0.34$

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