Extract the data and calculate the necessary sums.

Step 1: Extract the data and calculate the necessary sums.
The number of competitors is $n=10$.
Let Dancing marks be $X$ and Fashion Parade marks be $Y$.

$X$	$Y$	$XY$	$X^2$	$Y^2$
10	18	180	100	324
12	15	180	144	225
16	10	160	256	100
13	16	208	169	256
17	19	323	289	361
11	14	154	121	196
18	13	234	324	169
15	11	165	225	121
19	17	323	361	289
14	12	168	196	144
-----	-----	------	-------	-------
$\sum X = 145$	$\sum Y = 145$	$\sum XY = 2075$	$\sum X^2 = 2135$	$\sum Y^2 = 2185$

a) Calculate the product-moment correlation coefficient ($r$).
The formula for the product-moment correlation coefficient is:

$$r = \frac{n \sum XY - (\sum X)(\sum Y)}{\sqrt{[n \sum X^2 - (\sum X)^2][n \sum Y^2 - (\sum Y)^2]}}$$

Substitute the calculated sums into the formula:

$$r = \frac{10(2075) - (145)(145)}{\sqrt{[10(2135) - (145)^2][10(2185) - (145)^2]}}$$ $$r = \frac{20750 - 21025}{\sqrt{[21350 - 21025][21850 - 21025]}}$$ $$r = \frac{-275}{\sqrt{[325][825]}}$$ $$r = \frac{-275}{\sqrt{268125}}$$ $$r = \frac{-275}{517.807879...}$$ $$r \approx -0.53108$$

Rounding to 2 decimal places:

$$r \approx \boxed{-0.53}$$

b) Determine the least square regression line of the marks in fashion parade (Y) on the marks in dancing (X).
The regression line is of the form $Y = a + bX$.
The formula for the slope $b$ is:

$$b = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum X^2 - (\sum X)^2}$$

Using the values calculated in part a):

$$b = \frac{-275}{325}$$ $$b = -\frac{11}{13} \approx -0.8461538$$

The formula for the y-intercept $a$ is:

$$a = \bar{Y} - b\bar{X}$$

First, calculate the means $\bar{X}$ and $\bar{Y}$:

$$\bar{X} = \frac{\sum X}{n} = \frac{145}{10} = 14.5$$ $$\bar{Y} = \frac{\sum Y}{n} = \frac{145}{10} = 14.5$$

Now, calculate $a$:

$$a = 14.5 - \left(-\frac{11}{13}\right)(14.5)$$ $$a = 14.5 + \frac{11 \times 14.5}{13}$$ $$a = 14.5 + \frac{159.5}{13}$$ $$$$