A certain type of thermal battery for an airplane navigation device backup power has a mean life of 300 hours with a standard deviation of 15 hours. Assume a normal distribution of backup power device lives. What proportion of these batteries can be expected to have lives of 322 hours or less?

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To find the proportion of these batteries that can be expected to have lives of 322 hours or less, we need to calculate the z-score for 322 hours using the given information.

The z-score measures the number of standard deviations that a particular value is away from the mean. It is calculated using the formula:

z = (x - μ) / σ

Where:
- x is the value we are interested in (322 hours in this case)
- μ is the mean (300 hours)
- σ is the standard deviation (15 hours)

Let's calculate the z-score for 322 hours:

z = (322 - 300) / 15
z = 22 / 15
z ≈ 1.47

Once we have the z-score, we can use a standard normal distribution table or a calculator to find the proportion associated with that z-score.

Looking up the z-score of 1.47 in a standard normal distribution table, we find that the proportion of batteries that can be expected to have lives of 322 hours or less is approximately 0.9292.

Therefore, about 92.92% (or 92.92 out of 100) of these batteries can be expected to have lives of 322 hours or less.