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 is appropriate because the plot of the data points (d, f) suggests a roughly linear relationship. The values of f seem to increase as d increases, indicating a direct correlation. Additionally, the variation from a straight line is small, which is ideal for linear regression.

To calculate the sums: $S_{dd} = \sum d^2 = 152.09$ $S_{ff} = \sum f^2 = 3686$ Thus, we have:

• $S_{dd} = 152.09$
• $S_{ff} = 3686$

Using the formulas for the slope and intercept:

• Calculate $S_{dd} = 152.09$, $S_{ff} = 3686$, and $\sum fd = 723.1$.
• The slope $b$ is calculated as: $b = \frac{S_{fd} - \frac{\sum f \sum d}{n}}{S_{dd} - \frac{(\sum d)^2}{n}}$ Substituting the sums, we find: $b \approx 2.03$ Then, the intercept a can be found using: $a = \bar{f} - b \bar{d}$ Using the averages calculated previously, we get the regression line: $f = 15.0 + 2.03d$

The value of $b \approx 2.03$ indicates that for every 100 km increase in distance $d$, the fare $f$ increases by approximately £2.03. This suggests a direct correlation between distance and cost.

The competing travel agent charges 5p per km or £0.05 per km. Thus, the cost for a distance $t$ km is: $\text{Cost} = 0.05t$ We need to find when: $15.0 + 2.03t < 0.05t$

Solving this inequality:

1. Rearrange to get: $15.0 < 0.05t - 2.03t$
2. Combine like terms: $15.0 < -1.98t$
3. Divide by -1.98 (reversing the inequality): $t < \frac{15.0}{1.98} \approx 7.58$ Therefore, the range of values of $t$ for which the first travel agent is cheaper is: $0 < t < 7.58$