Sharron 是一名业余游泳爱好者，她一直在参加游泳课程。她记录了每个月游 50 米所花的时间。在前 7 个月中，她每个月 $x$ 记录的最佳成绩为 $t$ 秒，数据如下。

1. Set up axes with appropriate scales and labels.

Label the horizontal axis as $x$ (Month) covering at least the interval $[1, 7]$, and the vertical axis as $t$ (Time taken in seconds) covering at least the interval $[50, 120]$.

2. Plot the given data points.

Plot the coordinate pairs $(x, t)$: $(1, 115)$, $(2, 87)$, $(3, 75)$, $(4, 67)$, $(5, 62)$, $(6, 61)$, and $(7, 55)$.

1. Explain why a linear model is not appropriate.

The data shows a decreasing trend. A linear model with a negative gradient would decrease indefinitely and eventually predict negative timings $t$ for large values of $x$. This is physically impossible as swimming times must be strictly positive.

2. Explain why a quadratic model is not appropriate.

To fit the decreasing but flattening curve seen in the data, a quadratic model would be concave upwards (positive coefficient of $x^2$). This means the curve will eventually reach a minimum turning point and start to increase. This implies her swim times will eventually increase as she continues to practice, which is unrealistic. Her timings should instead plateau at a certain limit.

1. Determine the sign of $b$ by considering the relationship between $t$ and $\frac{1}{x}$.

From the given data, as the month $x$ increases, the time $t$ decreases.
However, as $x$ increases, the value of $\frac{1}{x}$ also decreases.
Since both $t$ and $\frac{1}{x}$ decrease together (they are positively correlated), the gradient $b$ in the linear relation $t = a + b\left(\frac{1}{x}\right)$ must be positive.

2. Determine the sign of $a$ by considering the long-term limits of the model in its real-world context.

Consider what happens after many months: as $x \to \infty$, $\frac{1}{x} \to 0$, which means $t \to a$.
In this context, $t$ represents the time taken to swim $50\text{ m}$. Sharron cannot swim the distance in $0$ or negative seconds (there is a minimum physical time limit).
Therefore, the theoretical lower limit $a$ must be positive.

1. Compute the product moment correlation coefficient ($r$) and the parameters of the regression line $t = a + b\left(\frac{1}{x}\right)$ using a graphing calculator.

Let $X = \frac{1}{x}$. We enter the paired data $(X, t)$ into a graphing calculator to perform linear regression:

From the graphing calculator:
$r = 0.991089 \dots$
$a = 49.500 \dots$
$b = 67.6858 \dots$

Rounding to 3 significant figures:
$r = 0.991$
$a = 49.5$
$b = 67.7$

1. Calculate the mean of the independent variable $X = \frac{1}{x}$ for the 8 months.

$\bar{X} = \frac{1}{8} \sum_{x=1}^{8} \frac{1}{x}$
$= \frac{1}{8} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right)$
$= \frac{761}{2240} \approx 0.33973$

2. Express the mean of $t$ in terms of the unknown timing $t_8$.

$\bar{t} = \frac{\sum t}{8}$
$= \frac{115 + 87 + 75 + 67 + 62 + 61 + 55 + t_8}{8}$
$= \frac{522 + t_8}{8}$

3. Apply the property that the regression line passes through the sample mean point $(\bar{X}, \bar{t})$ to solve for $t_8$.

Substitute $\bar{X}$ and $\bar{t}$ into the given regression equation $t = 48.28 + 69.45X$:

$\frac{522 + t_8}{8} = 48.28 + 69.45 \left( \frac{761}{2240} \right)$

Multiply the entire equation by 8:
$522 + t_8 = 8(48.28) + 69.45 \left( \frac{761}{280} \right)$
$522 + t_8 = 386.24 + 188.755...$
$t_8 = 574.995... - 522$
$t_8 = 52.995...$

Rounding to the nearest second:
$t_8 \approx 53$