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The lifetime of a 2-volt non-rechargeable battery in constant use has a Normal distribution, with a mean of 516 hours and a standard deviation of 20 hours. Eighty percent of all batteries have a lifetime of more than:

a. 499.16 hours
b. 532.83 hours
c. 516 hours
d. 20 hours

1 Answer

3 votes

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.

User Dzezzz
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