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The problem you're referring to involves calculating the probability that the difference in the sample means of two lines (checkout counters) exceeds a certain value. Let's break down the solution step-by-step.

### Problem Outline:
- **Service time**: Normally distributed with a mean (\(\mu\)) of 3 minutes and a standard deviation (\(\sigma\)) of 3.5 minutes.
- **Counters**: There are two lines:
  - Line 1: \(n_1 = 25\) customers.
  - Line 2: \(n_2 = 42\) customers.
  
The goal is to find the probability that the difference between the sample means for the shorter line (\(\overline{X_1}\)) and the longer line (\(\overline{X_2}\)) is more than 0.7 minutes.

### Approach:

1. **Difference in sample means**: We are interested in \(\overline{X_1} - \overline{X_2}\).

   The difference in sample means follows a normal distribution:
   \[
   \overline{X_1} - \overline{X_2} \sim N(\mu_1 - \mu_2, \sqrt{\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}})
   \]
   Where:
   - \(\mu_1 = \mu_2 = 3\) (since both lines have the same average service time),
   - \(\sigma = 3.5\).

   Thus, the mean of the difference is:
   \[
   \mu_{\overline{X_1} - \overline{X_2}} = \mu_1 - \mu_2 = 3 - 3 = 0
   \]
   And the standard deviation (standard error) of the difference is:
   \[
   \sigma_{\overline{X_1} - \overline{X_2}} = \sqrt{\frac{3.5^2}{25} + \frac{3.5^2}{42}} = \sqrt{\frac{12.25}{25} + \frac{12.25}{42}}
   \]
   Simplifying this:
   \[
   \sigma_{\overline{X_1} - \overline{X_2}} = \sqrt{0.49 + 0.2917} = \sqrt{0.7817} \approx 0.8842
   \]

2. **Standardize the difference**: We want to find the probability that the difference is greater than 0.7. To do this, we convert this difference into a standard normal variable (Z-score):
   \[
   Z = \frac{0.7 - 0}{\sigma_{\overline{X_1} - \overline{X_2}}} = \frac{0.7}{0.8842} \approx 0.7916
   \]

3. **Find the probability**: Now we use the standard normal distribution to find the probability that \(Z > 0.7916\). Using a Z-table or calculator, we find that the cumulative probability for \(Z = 0.7916\) is approximately 0.7852.

   Therefore, the probability that \(Z > 0.7916\) is:
   \[
   P(Z > 0.7916) = 1 - 0.7852 = 0.2148
   \]

So, the probability that the difference between the sample means is more than 0.7 minutes is approximately **0.2148**.

Would you like a more detailed explanation, or do you have any questions?

Here are five related questions you might want to explore:
1. How does the probability change if the standard deviation increases?
2. What is the interpretation of the Z-score in this context?
3. How would the result change if the sample sizes for the two lines were equal?
4. What role does the assumption of independence play in this calculation?
5. How can we compute confidence intervals for the difference in means?

**Tip**: When dealing with sample means, always consider the effect of sample size on the standard error. Larger samples reduce variability, which can affect the probability calculations.

Service time for a customer coming through a checkout counter in a retail store is a random variable with a mean of 3 minutes and standard deviation of 3.5 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n1 = 25 customers in the first line and n2 = 42 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X̄1 and the mean service time for the longer one X̄2 is more than 0.7 minutes. Assume that the service times for each customer can be regarded as independent random variables.

This problem involves calculating the probability that the difference in the sample means of service times for two lines in a retail store exceeds a given value. Key concepts include normal distribution, sample mean, and the central limit theorem. Suitable for undergraduate students.

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