ANSWER
[tex]\begin{gathered} (a)\text{ }95\% \\ \\ (b)780.7hrs\text{ and }919.3hrs \end{gathered}[/tex]EXPLANATION
(a) Given that:
Mean, μ = 850 hours
Standard deviation, σ = 70 hours
To find the percentage of batteries that have lifetimes between 710 and 990 hours, we have to find:
[tex]P(710This can be simplified as follows:[tex]\begin{gathered} P(\frac{710-850}{70}Using the standard normal table, we have that:[tex]\begin{gathered} 0.97725-0.02275 \\ \\ 0.9545\Rightarrow95\% \end{gathered}[/tex]Hence, 95% of batteries have lifetimes between 710 hours and 990 hours.
(b) The middle 68% of the score represents:
[tex]P(-zWe can rewrite this as:[tex]\begin{gathered} P(ZNow, we have to find the z value corresponding to 0.84 from the standard normal table.We have that:
[tex]P(Z<0.99)=0.84[/tex]This implies that:
[tex]z=\pm0.99[/tex]Using the z-score formula, we have to find the scores corresponding to the z-scores of -0.99 and 0.99.
That is:
[tex]x=z*\sigma+\mu[/tex]For z = -0.99:
[tex]\begin{gathered} x=(-0.99*70)+850 \\ \\ x=780.7\text{ }hrs \end{gathered}[/tex]For z = 0.99:
[tex]\begin{gathered} x=(0.99*70)+850 \\ \\ x=919.3\text{ }hrs \end{gathered}[/tex]Therefore, approximately 68% of the batteries have lifetimes between 780.7 hours and 919.3 hours.