A travel agent sells flights to different destinations from Beerow airport - Edexcel - A-Level Maths Statistics - Question 6 - 2010 - Paper 2

A linear regression model may be appropriate as the plotted points appear to lie reasonably close to a straight line, indicating a linear relationship between the distance $d$ and the fare $f$. Additionally, the trends observed suggest that as the distance increases, the fare tends to increase as well.

To calculate $S_{dd}$ and $S_{\mu}$, we use the formulas:

1. For $S_{dd}$:

   $S_{dd} = \sum d^2 - \frac{(\sum d)^2}{n}$

   Here, we first compute $\sum d^2 = 152.09$, $\sum d = 27.7$, and $n = 6$:

   $S_{dd} = 152.09 - \frac{(27.7)^2}{6} \approx 24.2$

2. For $S_{\mu}$:

   $S_{\mu} = \sum f - \frac{(\sum f)^2}{n}$

   Given $\sum f = 146$:

   $S_{\mu} = 723.1 - \frac{(146)^2}{6} \approx 49.1$

To calculate the regression line, use the formulas:

1. The slope $b$ is computed as:

   $b = \frac{S_{\mu}}{S_{dd}} = \frac{49.1}{24.2} \approx 2.03$

2. Then, find $a$:

   $a = \frac{\sum f - b\sum d}{n} = \frac{146 - 2.03 \times 27.7}{6} \approx 15.0$

Thus, the equation of the regression line is:

$f = 15.0 + 2.03d$

The value of $b = 2.03$ indicates that for every additional 100 km distance from the airport, the fare $f$ increases by approximately £2.03. This provides insight into how fare charges relate to distance.

The rival travel agent charges 5p per km. For a distance $t$:

The equation for the first agent's fare is:

$f = 15.0 + 2.03t$

The rival's fare is $0.05t$, thus we need:

$15.0 + 2.03t < 0.05t$

Rearranging gives:

$(15.0 < 0.05t - 2.03t)$

This simplifies to:

$15.0 < -1.98t$

Therefore:

$t > \frac{15.0}{1.98} \approx 7.58$

In conclusion, for $t$ values greater than approximately 7.58, the first travel agent is more expensive than the rival.