Battery Power Problem. 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. What proportion of these batteries can be
expected to have lives of 322 hours or less? Assume a normal
distribution of backup power device lives.

Z = (x - μ)/SD = (322-300)/15

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion higher than that Z score.

You want the proportion lower than that Z score, in the larger portion.

To solve this problem, we will use the concept of the standard normal distribution, also known as the Z-score.

Step 1: Calculate the Z-score
The Z-score is a measure of how many standard deviations a particular value is from the mean. It is calculated using the formula:

Z = (X - μ) / σ

Where:
- X is the given value (322 hours),
- μ is the mean (300 hours), and
- σ is the standard deviation (15 hours).

Substituting the values, we get:
Z = (322 - 300) / 15 = 22 / 15 = 1.47

Step 2: Find the proportion using a Z-table
A Z-table provides the proportion of values that fall below a given Z-score. We need to find the proportion below the Z-score of 1.47.

Using a Z-table, we find that the proportion corresponding to a Z-score of 1.47 is approximately 0.9292.

Step 3: Interpret the result
The proportion of these batteries that can be expected to have lives of 322 hours or less is approximately 0.9292, or 92.92%.

So, about 92.92% of these batteries can be expected to have lives of 322 hours or less.