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The average number of units that the factory workers at Benzo company assemble per week is μ = 45 with a standard deviation of σ = 12. Assume that the distribution of units assembled is normal. If a sample n = 25 were drawn from all the workers,

A. What range units manufactured by this sample would contain the sample mean 90% of the time? (Hint: Find the middle 90% of the distribution of sample means)

B. What is the probability that the mean will be between 40 and 50 units?

C. Is it reasonable for this sample to produce an average of 48 units per week, or is this mean very different from what would be normally expected?

D. How likely is it that this sample will produce more than an average of 55 units?

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A. To find the range of units that would contain the sample mean 90% of the time, we need to find the margin of error. According to the Central Limit Theorem, the distribution of sample means is approximately normally distributed with a mean equal to the population mean μ and a standard deviation equal to the population standard deviation σ divided by the square root of the sample size √n.

The formula for the margin of error is given by: Margin of Error = z * (σ/√n), where z is the z-score corresponding to the desired level of confidence. In this case, we want to find the range that contains 90% of the sample means, so the middle 90% of the distribution.

To find the z-score corresponding to a 90% confidence level, we calculate (1 - 0.90)/2 = 0.05. Looking up this value in a standard normal distribution table, we find that the z-score is approximately 1.645.

Therefore, the margin of error = 1.645 * (12/√25) = 1.645 * (12/5) = 3.95

So, the range of units that would contain the sample mean 90% of the time is μ ± margin of error = 45 ± 3.95 = (41.05, 48.95)

B. To find the probability that the mean will be between 40 and 50 units, we need to find the z-scores corresponding to these values.

The z-score for 40 units is calculated as (40 - 45) / (12/√25) = -2.083.
The z-score for 50 units is calculated as (50 - 45) / (12/√25) = 2.083.

Using a standard normal distribution table, we can find the cumulative probability up to these z-scores. The probability of the mean being between 40 and 50 units is the difference between these probabilities.

P(40 ≤ X ≤ 50) = P(-2.083 ≤ Z ≤ 2.083) = 2 * P(Z ≤ 2.083) - 1 ≈ 2 * 0.9817 - 1 ≈ 0.9634

Therefore, there is approximately a 96.34% probability that the mean will be between 40 and 50 units.

C. To determine if a mean of 48 units per week is reasonable, we need to compare it to the population mean μ.

To check if the mean of 48 units is significantly different from the population mean, we can perform a hypothesis test. We can set up the null hypothesis as H0: μ = 48 and an alternative hypothesis as H1: μ ≠ 48.

We can calculate the z-score for a sample mean of 48 units as (48 - 45) / (12/√25) = 1.25.

Using a standard normal distribution table, we can find the probability of obtaining a z-score of 1.25 or larger in either tail. If this probability is small (typically ≤ 0.05), we reject the null hypothesis and conclude that the sample mean of 48 units is significantly different from the population mean.

D. To find the probability that the sample produces more than an average of 55 units, we need to calculate the z-score for the value of 55 units.

The z-score for 55 units is calculated as (55 - 45) / (12/√25) = 2.083.

Using a standard normal distribution table, we can find the cumulative probability from the mean up to this z-score.

P(X > 55) = 1 - P(X ≤ 55) ≈ 1 - 0.9817 ≈ 0.0183

Therefore, there is approximately an 1.83% probability that the sample will produce more than an average of 55 units.