Single Variable Calculus, Chapter 1, 1.2, Section 1.2, Problem 23

The winning heights for the olympic pole vault competitions of various countries are shown in the table below.


$
\begin{array}{|c|c|c|c|}
\hline
\text{Year} & \text{Height(ft)} & \text{Year} & \text{Height(ft)}\\
\hline
1900 &10.83 &1956 &14.96\\
1904 &11.48 &1960 &15.42\\
1908 &12.17 &1964 &16.73\\
1912 &12.96 &1968 &17.71\\
1920 &13.42 &1972 &18.04\\
1924 &12.96 &1976 &18.04\\
1928 &13.77 &1980 &18.96\\
1932 &14.15 &1984 &18.85\\
1936 &14.27 &1988 &19.77\\
1948 &14.10 &1992 &19.02\\
1952 &14.92 &1996 &19.42\\
\hline
\end{array}
$

a.) Decide whether a linear model is appropriate by making a scatter plot of the given data



Linear model is appropriate because the graph resembles a straight line with positive slope.


b.) Find and graph the regression line


c.) Predict the height of the winning pole vault at the 2000 Olympics and compare with the actual winning height of 19.36 feet by using the linear model

The winning height for 2000 Olympics is...

$
\begin{equation}
\begin{aligned}
y &= 0.0891 x - 158.24\\
y &= 0.0891(2000) - 158.24\\
y &= 19.96 \text{ feet}
\end{aligned}
\end{equation}
$


To compare the winning height of 19.36 ft we can determine the year it was obtained by ...


$
\begin{equation}
\begin{aligned}
19.36 &= 0.891x - 158.24\\
x &= 1993.26 \text{ years}
\end{aligned}
\end{equation}
$


Therefore, we can say that the gap of the year is almost 7.