To determine the percentage of batteries that will need to be replaced, we calculate the probability that a battery fails within 36 months. Given a normally distributed lifespan, we can use the z-score formula:

$z = \frac{X - \mu}{\sigma}$

where:

• $X = 36$ months (the failure period),
• $\mu = 45$ months (the mean lifespan),
• $\sigma = 7$ months (the standard deviation).

1. Calculate the z-score:

   $z = \frac{36 - 45}{7} = \frac{-9}{7} \approx -1.29$

2. Find the probability associated with $z = -1.29$:

   Using standard normal distribution tables, the probability $P(Z \leq -1.29) \approx 0.0985$.

This value, 0.0985, indicates that approximately 9.85% of batteries will fail within 36 months, meaning the company would expect to replace about 9.85% of its batteries under the guarantee policy.

Would you like a more detailed explanation on z-scores and normal distribution?

Here are some related questions:

1. How would the replacement percentage change if the guarantee was extended to 40 months?
2. What would happen to the replacement percentage if the standard deviation increased to 9 months?
3. How is the replacement percentage impacted if the average lifespan increases?
4. How does altering the guarantee period affect company costs?
5. What is the likelihood of a battery lasting more than 50 months?

Tip: Familiarizing yourself with z-score tables can simplify calculating probabilities for normally distributed data.