PART A
Calculating the z score for 670 and 950
We have;
[tex]\begin{gathered} z_1=\frac{670-810}{70} \\ z_1=-\frac{140}{70} \\ z_1=-2 \\ \text{and } \\ z_2=\frac{950-810}{70} \\ z_2=\frac{140}{70}_{} \\ z_2=2 \end{gathered}[/tex][tex]\begin{gathered} \text{from the graph } \\ \approx95\text{ \% of the batteries are within that range} \\ \end{gathered}[/tex]PART B
From the z score table we got z=0.47
[tex]\begin{gathered} 0.47=\frac{x-810}{70} \\ 70\times\text{ 0.47 = X-810} \\ 32.9=X-810 \\ X=842.9 \end{gathered}[/tex]Approximately 68% of the batteries have lifetimes between 0 hours and
hours is 842.9
