## Precalculus (6th Edition) Blitzer

a) The expression is $\underline{\frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}=3.9}$ which represents the mean hour spent on the digital media. b) The value of $\frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}$ using ${{a}_{n}}=0.5n+2$ is $4$ and it is overestimated by 0.1 hour with respect to part (a).
(a) From the given bar graph, ${{a}_{1}}=2.7$, ${{a}_{2}}=3.0$, ${{a}_{3}}=3.2$, ${{a}_{4}}=3.7$, ${{a}_{5}}=4.4$ ${{a}_{6}}=4.9$ and ${{a}_{7}}=5.3$ Therefore, \begin{align} & \frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}=\frac{1}{7}\left( 2.7+3.0+3.2+3.7+4.4+4.9+5.3 \right) \\ & =\frac{27.2}{7} \\ & =3.885 \\ & \approx 3.9\simeq \end{align} Hence, $\frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}=3.9$ represents the mean hours that U.S. adult users spent on the digital media. (b) We have the equation ${{a}_{n}}=0.5n+2$, For $n=1$, \begin{align} & {{a}_{1}}=0.5\times 1+2 \\ & =2.5 \end{align} For $n=2$ \begin{align} & {{a}_{2}}=0.5\times 2+2 \\ & =3.0 \end{align} For $n=3$ \begin{align} & {{a}_{3}}=0.5\times 3+2 \\ & =3.5 \end{align} For $n=4$ \begin{align} & {{a}_{4}}=0.5\times 4+2 \\ & =4.0 \end{align} For $n=5$ \begin{align} & {{a}_{5}}=0.5\times 5+2 \\ & =4.5 \end{align} For $n=6$ \begin{align} & {{a}_{6}}=0.5\times 6+2 \\ & =5.0 \end{align} For $n=7$ \begin{align} & {{a}_{7}}=0.5\times 7+2 \\ & =5.5 \end{align} Therefore, \begin{align} & \frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}=\frac{1}{7}\left( 2.5+3.0+3.5+4.0+4.5+5.0+5.5 \right) \\ & =\frac{28}{7} \\ & =4 \end{align} Hence, the relation will overestimate the value of $\frac{1}{7}\sum\limits_{i=1}^{7}{{{a}_{i}}}$ by 0.1 hour.