The lifetime of a certain smart phone battery has an

unknown distribution with mean value of 8 hours and
standard deviation of 2 hours. What is the approximate
probability that the average battery lifetime of a sample
of 36 batteries will exceed 8.1 hours?

Assuming a normal distribution,

Z = (score-mean)/SEm

SEm = SD/√n

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

To find the probability that the average battery lifetime of a sample of 36 batteries will exceed 8.1 hours, we can use the Central Limit Theorem.

The Central Limit Theorem states that for a large enough sample size, the distribution of the sample means will approach a normal distribution, regardless of the shape of the original population distribution.

In this case, since we have a sample size of 36, we can assume that the sample means will be approximately normally distributed.

To solve this problem, we need to standardize the sample mean using the formula for standardizing a sample mean:

z = (x - μ) / (σ / √n)

Where:
- z is the standardized value (z-score) representing how many standard deviations the sample mean is from the population mean.
- x is the value we want to find the probability for (in this case, 8.1 hours).
- μ is the population mean (8 hours).
- σ is the population standard deviation (2 hours).
- n is the sample size (36 batteries).

We can plug in the values into the formula:

z = (8.1 - 8) / (2 / √36)

Simplifying, we get:

z = 0.1 / (2 / 6)
z = 0.1 * 6 / 2
z = 0.3

Now, we need to find the probability that the standardized value is greater than 0.3. We can look up this probability in the standard normal distribution table or use a statistical calculator.

Using the standard normal distribution table, we find that the probability corresponding to a z-score of 0.3 is approximately 0.6179.

Therefore, the approximate probability that the average battery lifetime of a sample of 36 batteries will exceed 8.1 hours is 0.6179, or 61.79%.