Final answer:
Using a z-score of approximately 0.84 for 80% of a Normal distribution, we solve for the battery lifetime that this percentage of batteries exceed, resulting in more than 532.83 hours.
Step-by-step explanation:
The question asks us to find the lifetime in hours for which 80% of all 2-volt non-rechargeable batteries last longer. This is a problem that involves the Normal distribution, mean (\(\mu\)), and standard deviation (\(\sigma\)). Given a Normal distribution with \(\mu = 516\) hours and \(\sigma = 20\) hours, we want to identify the value above which 80% of the battery lifetimes occur.
To find this, we use the concept of z-scores and lookup the z-score that corresponds to 80% in a standard Normal distribution table, or use an appropriate statistical calculator. A z-score table indicates that 80% corresponds to a z-score of approximately 0.84. Applying the z-score formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Solving for \(X\) gives us the battery lifetime that 80% of batteries exceed:
\[ X = Z \times \sigma + \mu \]
Substituting the z-score and the given values:
\[ X = 0.84 \times 20 + 516 \]
\[ X = 16.8 + 516 \]
\[ X = 532.8 \]
Therefore, 80% of all batteries have a lifetime of more than 532.83 hours, which corresponds to option b.