From the table, we can form the line of best fit with the table below:
[tex]\begin{center}
\begin{tabular}
{|p{1.8cm}|c|c|c|c|}
&X&Y&X^2&XY\\[2ex]
Tyrone&6&9&36&54\\
Marcel&6&11&36&66\\
Patrick&7&15&49&105\\
Bobby&8&11&64&88\\
Dylan&9&15&81&135\\
Mike&10&16&100&160\\
Jonathan&12&17&144&204\\[2ex]
&\sum X=58&\sum Y=94&\sum X^2=510&\sum XY=812
\end{tabular}
\end{center}[/tex]
The line of best fit is given by [tex]y=a+bx[/tex], where: [tex]a= \frac{(\sum Y)(\sum X^2)-(\sum X)(\sum XY)}{n(\sum X^2)-(\sum X)^2} [/tex] and [tex]b= \frac{n(\sum XY)-(\sum X)(\sum Y)}{n(\sum X^2)-(\sum X)^2}[/tex].
Thus, we have:
[tex]a= \frac{(94)(510)-(58)(812)}{7(510)-(58)^2} \\ \\ = \frac{47,940-47,096}{3,570-3,364} = \frac{844}{206} \\ \\ =4.0971[/tex]
and
[tex]b= \frac{7(812)-(58)(94)}{7(510)-(58)^2} \\ \\ = \frac{5,684-5,452}{3,570-3,364} = \frac{232}{206} \\ \\ =1.1262[/tex]
Thus, the line of best fit is [tex]y=4.0971+1.1262x[/tex]
For Tyrone and Marcel with a shoe size of 6, their predicted age will be
[tex]y=4.0971+1.1262(6) \\ \\ =4.0971+6.7572=10.8543\approx11\ years[/tex]
Thus Tyrone with an age of 9 is more than 1 year different from his predicted age of 11.
For Patrick with a shoe size of 7, his predicted age will be
[tex]y=4.0971+1.1262(7) \\ \\ =4.0971+7.8834=11.9805\approx12\ years[/tex]
Thus Patrick with an age of 15 is more than 1 year different from his predicted age of 12.
For Bobby with a shoe size of 8, his predicted age will be
[tex]y=4.0971+1.1262(8) \\ \\ =4.0971+9.0096=13.1067\approx13\ years[/tex]
Thus Bobby with an age of 11 is more than 1 year different from his predicted age of 13.
For Dylan with a shoe size of 9, his predicted age will be
[tex]y=4.0971+1.1262(9) \\ \\ =4.0971+10.1358=14.2329\approx14\ years[/tex]
For Mike with a shoe size of 10, his predicted age will be
[tex]y=4.0971+1.1262(10) \\ \\ =4.0971+11.262=15.3591\approx15\ years[/tex]
For Jonathan with a shoe size of 12, his predicted age will be
[tex]y=4.0971+1.1262(12) \\ \\ =4.0971+13.5144=17.6115\approx18\ years[/tex]
Therefore, Tyrone, Patrick and Dylan which is [tex] \frac{3}{7} \times100\%=43\%[/tex] has their age 1 more than their predicted ages.
