De Guzman Co. Claims that the average life of their battery product last of 26 hours with a standard deviation of 5 hours. What is the probability that 35 random pieces of their batteries have an average life span of less than 24.3 hours?

Question

Liam Burke · Accepted Answer

<p>

Ezekiel Anderson · Answer

undefined To find the probability that 35 random pieces of batteries have an average life span of less than 24.3 hours, we can use the Central Limit Theorem and the properties of the normal distribution.

First, we need to calculate the standard error of the mean (SEM) using the formula:

$$ SEM = \frac{\sigma}{\sqrt{n}} $$

Where:
- $ \sigma $ is the standard deviation of the population, which is 5 hours.
- $ n $ is the sample size, which is 35.

$$ SEM = \frac{5}{\sqrt{35}} $$

Next, we calculate the z-score using the formula:

$$ z = \frac{x - \mu}{SEM} $$

Where:
- $ x $ is the sample mean we're interested in, which is 24.3 hours.
- $ \mu $ is the population mean, which is 26 hours.
- $ SEM $ is the standard error of the mean.

$$ z = \frac{24.3 - 26}{\frac{5}{\sqrt{35}}} $$

Now, we use the z-table or statistical software to find the probability corresponding to this z-score. This probability represents the likelihood that the average life span of 35 random batteries is less than 24.3 hours.