The yield per hectare, $y$ kg, of a crop is believed to depend on the average rainfall, $x$ mm, in the month of June. For 10 regions, records are kept of the values of $x$ and $y$, and these are shown in the table below. The yield from the tenth region was accidentally deleted from the records after the data was analysed, and this is indicated by the value $p$.

Average rainfall ($x$ mm)	149	110	188	135	156	140	168	118	122	174
Yield of crop ($y$ kg)	13.8	6.5	15.2	12.2	14.4	12.2	14.7	9.5	9.9	$p$

Given that the equation of the regression line of $y$ on $x$ is $y = -2.5652 + 0.10168x$, show that $p = 14.4$.

1. Calculate the sample mean of the average rainfall $x$, denoted as $\bar{x}$.

$\sum x = 149 + 110 + 188 + 135 + 156 + 140 + 168 + 118 + 122 + 174 = 1460$
$\bar{x} = \frac{1460}{10} = 146$

2. Find the sample mean of the yield $y$, $\bar{y}$, using the property that the regression line passes through the point $(\bar{x}, \bar{y})$.

Substitute $\bar{x} = 146$ into the given regression line equation $y = -2.5652 + 0.10168x$:
$\bar{y} = -2.5652 + 0.10168(146)$
$\bar{y} = -2.5652 + 14.84528 = 12.28008$

3. Set up an equation for $\bar{y}$ using the given dataset to solve for the missing value $p$.

Calculate the sum of the known $y$ values:
$\sum y_{\text{known}} = 13.8 + 6.5 + 15.2 + 12.2 + 14.4 + 12.2 + 14.7 + 9.5 + 9.9 = 108.4$
Using the formula for the mean $\bar{y} = \frac{\sum y}{n}$:
$\frac{108.4 + p}{10} = 12.28008$
$108.4 + p = 122.8008$
$p = 14.4008$
Rounding to 1 decimal place (consistent with the provided data), we obtain:
$p \approx 14.4 \text{ (shown)}$