A manufacturer of automobile batteries claims that the distribution of battery life is

54 months with a standard deviation of 6 months. We take a random sample of 50 batteries.
a. Find the probability that their mean life is less than 52 months.
b. Find the probability that their mean life is more than 53 months.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

If only one SD is provided, you can use just that to determine SEdiff.

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

To find the probabilities in this scenario, we can use the Central Limit Theorem. According to the Central Limit Theorem, if we take a random sample of a large enough size from any population with a finite mean (μ) and a finite standard deviation (σ), then the distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution.

In this case, we are given that the population mean (μ) is 54 months and the population standard deviation (σ) is 6 months. Since the sample size (n) is 50, we can assume that the sample mean will be approximately normally distributed.

a. To find the probability that the mean life of the batteries is less than 52 months, we need to calculate the z-score and use the standard normal distribution table.

The formula to calculate the z-score is:
z = (x - μ) / (σ / sqrt(n))

where:
x = 52 (the mean life we are interested in)
μ = 54 (population mean)
σ = 6 (population standard deviation)
n = 50 (sample size)

Substituting the values into the formula:
z = (52 - 54) / (6 / sqrt(50))
z = -2 / (6 / sqrt(50))
z = -2 / (6 / 7.071) ≈ -2.3548

Now, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score of -2.3548. The probability represents the area under the standard normal curve to the left of the z-score.

b. To find the probability that the mean life of the batteries is more than 53 months, we can use the same approach as in part a. Calculate the z-score using the formula:
z = (x - μ) / (σ / sqrt(n))

Substituting the values into the formula:
z = (53 - 54) / (6 / sqrt(50))
z = -1 / (6 / sqrt(50))
z = -1 / (6 / 7.071) ≈ -1.1785

Again, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score of -1.1785. The probability represents the area under the standard normal curve to the right of the z-score.

Remember to check the table or use a calculator to find the exact probabilities associated with the calculated z-scores.