To solve this, we’ll use the Central Limit Theorem, which states that for a large enough sample size, the sampling distribution of the sample mean will be approximately normal, even if the population distribution is not. Here’s the solution step-by-step.

Step 1: Identify the Sample Mean and Standard Deviation

Each worker assembles an average of $\mu = 4.5$ computers per hour with a standard deviation of $\sigma = 0.8$ computers.

Since we have 42 workers, we are looking at the distribution of the sample mean $\bar{X}$.

Step 2: Calculate the Standard Error of the Mean (SEM)

The SEM, which is the standard deviation of the sample mean, is given by: $\text{SEM} = \frac{\sigma}{\sqrt{n}}$ where:

• $\sigma = 0.8$
• $n = 42$

So: $\text{SEM} = \frac{0.8}{\sqrt{42}} \approx 0.1235$

Step 3: Convert the Range to Z-Scores

Now, we want to find the probability that the mean production per hour for a 42-person shift falls between 184.8 and 189 computers.

Step 3a: Calculate the Mean Total Production for 42 Workers

The expected total production for 42 workers per hour is: $42 \times 4.5 = 189$

Thus, the probability question converts to finding the probability that the total production is between 184.8 and 189.

Step 3b: Convert Total Production to Mean Production

To work with Z-scores, convert the total production range to the corresponding sample means:

1. Lower Bound: $\frac{184.8}{42} = 4.4$
2. Upper Bound: $\frac{189}{42} = 4.5$

So, we want to find the probability that the sample mean falls between 4.4 and 4.5.

Step 3c: Calculate Z-Scores

Using the formula for Z-scores: $Z = \frac{\bar{X} - \mu}{\text{SEM}}$

1. For $\bar{X} = 4.4$: $Z = \frac{4.4 - 4.5}{0.1235} \approx -0.8108$

2. For $\bar{X} = 4.5$: $Z = \frac{4.5 - 4.5}{0.1235} = 0$

Step 4: Find the Probability

Using a standard normal distribution table, find the probability for each Z-score:

1. Probability for $Z = -0.8108$: approximately 0.2090
2. Probability for $Z = 0$: 0.5000

The probability that the mean production per hour is between 4.4 and 4.5 is: $0.5000 - 0.2090 = 0.2910$

Final Answer

The probability that a 42-person shift will put together between 184.8 and 189 computers per hour is approximately: $\boxed{0.2910}$

Would you like further details on any part of this solution?

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1. How does the Central Limit Theorem apply to this problem?
2. What is the significance of the standard error in sampling distributions?
3. Why do we convert total production to mean production for this calculation?
4. How would the probability change with a different number of workers?
5. How can we calculate probabilities without assuming a normal distribution?

Tip: Always verify if the sample size is large enough to apply the Central Limit Theorem, especially when working with averages and sums.