= Solution
Put $z_i=x_i-\bar x$, so $\sum z_i=0$. The <ordinary least squares> criterion is $S(\alpha,\beta)=\sum_i(Y_i-\alpha-\beta z_i)^2$. Setting both partial derivatives to zero gives $\sum_i(Y_i-\alpha-\beta z_i)=0$ and $\sum_i z_i(Y_i-\alpha-\beta z_i)=0$. Therefore, provided $S_{xx}=\sum z_i^2>0$, the <centred simple linear regression> estimators are
$$
\boxed{\widehat\alpha=\bar Y,\qquad\widehat\beta=\frac{\sum_i(x_i-\bar x)Y_i}{\sum_i(x_i-\bar x)^2}.}
$$
The Hessian of $S$ is $2\operatorname{diag}(n,S_{xx})$, which is positive definite, so this stationary point is the unique minimum. From the mean-zero errors, both estimators are unbiased; independence and common error <variance> give $\operatorname{Var}\widehat\alpha=\sigma^2/n$, $\operatorname{Var}\widehat\beta=\sigma^2/S_{xx}$, and zero <covariance>. Normality is not needed for this derivation. If all $x_i$ coincide, the slope is not identifiable and only the intercept is determined.
Using the PDF data gives $\bar x=6.2$, $\bar y=42.4$, $S_{xx}=86.8$ and $S_{xy}=-257.4$. In particular, the fourth cost is 37 in the PDF; the TeX's 75 is a transcription error and would contradict the supplied covariance sum. The fitted <simple linear regression> is
$$
\boxed{\widehat y=42.4-\frac{1287}{434}(x-6.2).}
$$
Here the centred intercept estimate is 42.4; an intercept at $x=0$ would instead be $\widehat\alpha-\widehat\beta\bar x$. For eight units the predicted unit cost is
$$
\boxed{\widehat y(8)=42.4-\frac{1287}{434}(1.8)\approx37.06\ \text{pounds}.}
$$
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