The time required by workers to complete an assembly job has a mean of 50 minutes and a standard deviation of 8 minutes. To spot check the workers' progress on a particular day, their supervisor intends to record the times 60 workers take to complete one assembly job apiece. What is the probability that the sample mean will be more than 52 minutes?

Question

The time required by workers to complete an assembly job has a mean of 50 minutes and a standard deviation of 8 minutes. To spot check the workers' progress on a particular day, their supervisor intends to record the times 60 workers take to complete one assembly job apiece. What is the probability that the sample mean will be more than 52 minutes?

Answer 1

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

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

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 proportion related to the Z score.

Answer 2

To find the probability that the sample mean will be more than 52 minutes, we can use the Central Limit Theorem. According to the Central Limit Theorem, when taking a large enough sample from a population with a known mean and standard deviation, the distribution of the sample mean approaches a normal distribution regardless of the shape of the original population.

First, let's calculate the standard deviation of the sample mean. Since we are given the standard deviation of the population (8 minutes) and we are sampling 60 workers, we can use the formula for the standard deviation of the sample mean:

Standard deviation of the sample mean = (standard deviation of the population) / sqrt(sample size)

Standard deviation of the sample mean = 8 / sqrt(60)

Standard deviation of the sample mean ≈ 1.034

Next, we can use the Z-score formula to calculate the standardized score for a sample mean of 52 minutes:

Z = (x - μ) / (σ / sqrt(n))

where:
x = sample mean (52 minutes)
μ = population mean (50 minutes)
σ = standard deviation of the sample mean (1.034)
n = sample size (60 workers)

Z = (52 - 50) / (1.034 / sqrt(60))

Z ≈ 2.305

Now, we can find the probability that the sample mean will be more than 52 minutes by calculating the area under the standard normal curve to the right of the Z-score of 2.305 using a standard normal distribution table or a statistical software.

The probability can be found as P(Z > 2.305).

Using a standard normal distribution table or a statistical software, we find that P(Z > 2.305) ≈ 0.0104.

Therefore, the probability that the sample mean will be more than 52 minutes is approximately 0.0104 or 1.04%.