What is the probability that a randomly selected battery will have a lifespan of at least 212 hours if the mean lifespan is 200 hours and the standard deviation is 7.6 hours, rounded to the nearest tenth of a percent?

To find the probability that a randomly selected battery will have a lifespan of at least 212 hours, we first need to understand the distribution of battery lifespans. Assuming the battery lifespan follows a normal distribution, we can use the mean (200 hours) and the standard deviation (7.6 hours) to calculate the z-score for 212 hours.

The z-score is calculated using the formula:

[ z = \frac{(X - \mu)}{\sigma} ]

where ( X ) is the value in question (212 hours), ( \mu ) is the mean (200 hours), and ( \sigma ) is the standard deviation (7.6 hours).

Substituting the values:

[ z = \frac{(212 - 200)}{7.6} = \frac{12}{7.6} \approx 1.5789 ]

Next, we need to find the probability that a z-score is greater than 1.5789. This is the area under the normal curve to the right of this z-score. You can look up the z-score in a standard normal distribution table or use a calculator to find:

[ P(Z > 1.5789) ]