SL 4.10—X on y regression line

The regression line is given as $T=a+bf$, which is in the form x on y (here $T$ plays the role of $x$ and $f$ plays the role of $y$).

The x on y regression line predicts the variable on the left-hand side of the equation from the variable on the right-hand side. Here $T$ is on the left, so $T$ is being predicted from $f$.

The $T$ on $f$ regression line minimises horizontal distances (residuals in $T$), not vertical distances.

The line $T=a+bf$ is used to predict the water temperature $T$ for a given number of fish $f$.

We need to determine the correct use of the regression line $T=a+bf$.

Option A says it predicts $f$ from $T$. But $f$ is the input (right-hand side) and $T$ is the output (left-hand side), so this is backwards. Eliminated.

Option C is incorrect — regression lines predict one variable from another, not both simultaneously. Eliminated.

Option D describes the y on x regression line. The $T$ on $f$ line minimises horizontal (not vertical) distances. Eliminated.

Option B is correct: the line $T=a+bf$ predicts the water temperature $T$ for a given number of fish $f$.

The coach knows the resting heart rate $h=62$ bpm and wants to predict the training distance $d$.

Since $d$ is the variable to be predicted and $h$ is known, we use the $d$ on $h$ regression line: $d=95-1.1h$.

$d=95-1.1(62)=95-68.2=26.8\text{ km}$

The predicted training distance is 26.8 km, obtained using $d=95-1.1(62)$.

We need to predict $d$ from a known $h=62$, so we need the $d$ on $h$ regression line.

Options A and C use $h=85-0.9d$, which is the $h$ on $d$ line. This line predicts $h$, not $d$, so both are wrong. Eliminated.

Option D substitutes $d=85$ (the intercept of the other line) rather than $h=62$. This is a nonsensical substitution. Eliminated.

Option B correctly uses $d=95-1.1(62)=26.8$ km. This is the right equation with the right substitution.