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### Problem 1
The time for a customer to be served is normally distributed with:
- Mean \( \mu = 176 \) seconds
- Variance \( \sigma^2 = 256 \), so standard deviation \( \sigma = \sqrt{256} = 16 \) seconds.

Since we are dealing with a sample of 100 customers, we apply the Central Limit Theorem (CLT) for the sampling distribution of the sample mean, which will be normally distributed with:
- Mean \( \mu_{\bar{X}} = 176 \)
- Standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{100}} = 1.6 \) seconds.

#### (a) Probability that the sample mean is between 175 and 178 seconds:
1. Convert 175 and 178 to z-scores:
   \[
   z_{175} = \frac{175 - 176}{1.6} = -0.625
   \]
   \[
   z_{178} = \frac{178 - 176}{1.6} = 1.25
   \]

2. Using the standard normal distribution table:
   - \( P(Z < -0.625) \approx 0.2660 \)
   - \( P(Z < 1.25) \approx 0.8944 \)

3. The probability that the sample mean is between 175 and 178 seconds is:
   \[
   P(175 < \bar{X} < 178) = P(Z < 1.25) - P(Z < -0.625) = 0.8944 - 0.2660 = 0.6284
   \]

#### (b) Probability that the sample mean will be at most 176 seconds:
1. Convert 176 to a z-score:
   \[
   z_{176} = \frac{176 - 176}{1.6} = 0
   \]

2. The probability for \( Z \leq 0 \) is 0.5, so:
   \[
   P(\bar{X} \leq 176) = 0.5
   \]

### Problem 2
The weight of an individual apple is normally distributed with:
- Mean \( \mu = 0.2 \) pounds
- Standard deviation \( \sigma = 0.03 \) pounds.

Since we are dealing with a sample of 40 apples, the total weight \( W \) is normally distributed with:
- Mean \( \mu_W = 40 \times 0.2 = 8 \) pounds
- Standard deviation \( \sigma_W = \sqrt{40} \times 0.03 = 0.19 \) pounds.

#### (a) Probability that the total weight of the bag exceeds 6.2 pounds:
1. Convert 6.2 to a z-score:
   \[
   z_{6.2} = \frac{6.2 - 8}{0.19} = -9.47
   \]

2. For \( Z = -9.47 \), the probability is almost 1, so:
   \[
   P(W > 6.2) \approx 1
   \]

#### (b) Probability that the total weight of the bag is less than 6 pounds:
1. Convert 6 to a z-score:
   \[
   z_{6} = \frac{6 - 8}{0.19} = -10.53
   \]

2. For \( Z = -10.53 \), the probability is essentially 0, so:
   \[
   P(W < 6) \approx 0
   \]

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

### Related Questions:
1. How does the Central Limit Theorem apply to these types of problems?
2. Why is the z-score used to find probabilities in normal distributions?
3. How would the probabilities change if the sample sizes were different?
4. What if the individual variance of times or weights were higher? How would that affect the solution?
5. How can these concepts apply in real-world business or quality control scenarios?

### Tip:
When working with sampling distributions, always check if the sample size is large enough for the Central Limit Theorem to apply, as it makes analyzing the data much simpler.

1. The time at the counter for a customer to be served at a post office can be modeled as a random variable having mean 176 sec., and variance 256. The sample mean will be obtained from the times for a random sample of 100 customers. What is the probability that:
(a) the sample mean will be between 175 and 178 seconds?
(b) the sample mean will be at most 176 seconds?

2. Several apple pickers are each asked to gather 40 ripe apples and put them in a bag. If the weight of an individual apple has a mean 0.2 pound and standard deviation 0.03 pound, find the probability that:
(a) the total weight of the bag will exceed 6.2 pounds.
(b) the total weight of the bag will be less than 6 pounds.

Solve probability questions involving sample means of customer service time and total weight of apples. This detailed solution covers key statistical concepts, including normal distribution and the Central Limit Theorem, to find probabilities for specific ranges in sample means and total weights.

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