Questions: Question 16 (1 point) An object is thrown from the top of a building. The following data measure the height of the object from the ground for a five second period. Calculate the correlation coefficient using technology (you can copy and paste the data into Excel). Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. Seconds Height 0.5 112.5 1 110.875 1.5 106.8 2 100.275 2.5 91.3 3 79.875 3.5 70.083 4 59.83 4.5 30.65 5

Solution Steps

The height of an object thrown from a building was measured at various time intervals. The data collected is as follows:

\[ \begin{array}{|c|c|} \hline \text{Seconds} & \text{Height} \\ \hline 0.5 & 112.5 \\ 1 & 110.875 \\ 1.5 & 106.8 \\ 2 & 100.275 \\ 2.5 & 91.3 \\ 3 & 79.875 \\ 3.5 & 70.083 \\ 4 & 59.83 \\ 4.5 & 30.65 \\ \hline \end{array} \]

The covariance \( \text{Cov}(X,Y) \) between the time (seconds) and height (meters) was calculated to be:

\[ \text{Cov}(X,Y) = -35.8981 \]

The standard deviations for the time and height were computed as follows:

\[ \sigma_X = 1.3693 \] \[ \sigma_Y = 27.4075 \]

Using the formula for the correlation coefficient \( r \):

\[ r = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y} \]

Substituting the values:

\[ r = \frac{-35.8981}{1.3693 \times 27.4075} = -0.9565 \]

Final Answer