To answer (a), we will need to find the Z value for 710 hours and 990 hours.
[tex]\begin{gathered} For\text{ 710} \\ Z\text{ = }\frac{x-\mu}{\sigma} \\ \text{ = }\frac{710-850}{70} \\ \text{ = -2} \\ p\text{ =0.0228 } \\ For\text{ 990:} \\ Z\text{ = }\frac{990-850}{70} \\ \text{ =2} \\ p\text{ = 0.9772} \end{gathered}[/tex][tex]\begin{gathered} To\text{ find P \lparen–2 \le Z \le 2\rparen} \\ =\text{ 0.9772 -0.0228} \\ =\text{ 0.9544} \\ =95.44\% \end{gathered}[/tex]
B) 68% of data lies between one standard deviation of the mean.
850 +70 = 920
850 - 70 = 780
