Statistics and Probability - Scatter graphs and bivariate data

A student records the number of hours spent studying ($x$) and the test scores ($y$) for 5 students: $(2, 40), (4, 60), (6, 70), (8, 90), (10, 90)$. Calculate the mean point $(\bar{x}, \bar{y})$ for this data.

$\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$ $\bar{y} = \frac{40 + 60 + 70 + 90 + 90}{5} = \frac{350}{5} = 70$ The mean point is $(6, 70)$.

To find the mean point, calculate the arithmetic mean of all $x$-coordinates and all $y$-coordinates separately.

Given a line of best fit equation $y = 5x + 20$ where $x$ is the number of hours worked and $y$ is the total earnings. Predict the earnings for someone working $7$ hours.

Substitute $x = 7$ into the equation: $y = 5(7) + 20$ $y = 35 + 20$ $y = 55$ The predicted earnings are $55$.

Interpolation involves substituting a known independent variable value into the linear equation derived from the scatter graph.

Determine the type of correlation for the following data points: $(1, 10), (2, 8), (3, 5), (4, 3), (5, 1)$.

As $x$ increases ($1 \to 5$), the values of $y$ decrease ($10 \to 1$). Therefore, the data shows a strong negative correlation.

Correlation is identified by observing the direction in which $y$ moves as $x$ increases. Since $y$ decreases as $x$ increases, the gradient of the relationship is negative.

$m = \frac{0 - 60}{30 - 0} = \frac{-60}{30} = -2$ $c = 60$ $y = -2x + 60$

The points show a perfect negative linear correlation. Using the coordinates provided for the line of best fit, we calculate the gradient $m$ and the y-intercept $c$ to form the linear equation.

$\bar{x} = \frac{160+170+180+190}{4} = 175$ $\bar{y} = \frac{60+70+80+90}{4} = 75$ $\text{Mean Point} = (175, 75)$ $\text{Substitute into line: } 75 = 175 - 100 \implies 75 = 75$

The mean point is calculated by averaging the $x$ and $y$ values separately. Since the mean point satisfies the equation $y = x - 100$, it is a valid candidate for the line of best fit.