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?
What portion of the previous answer I gave did you not understand?
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.