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Let's go through each part of the question.

### Given Information:
- The time interval between Bob entering and being served is normally distributed.
- **Mean (μ) = 4.2 minutes**
- **Standard deviation (σ) = 1.3 minutes**

### Part (i)
We need to find the probability that the time until Bob is served will be at least 5 minutes.

1. Define \( X \) as the time until he is served, so \( X \sim N(4.2, 1.3^2) \).
2. We are asked to find \( P(X \geq 5) \).
3. To calculate this, we can standardize the value using the Z-score formula:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Here, \( X = 5 \), \( \mu = 4.2 \), and \( \sigma = 1.3 \).

   \[
   Z = \frac{5 - 4.2}{1.3} \approx 0.6154
   \]

4. Now, we look up the Z-score of 0.6154 in the standard normal distribution table or use a calculator. The probability for \( Z \leq 0.6154 \) is approximately 0.7301.
5. Therefore, \( P(X \geq 5) = 1 - P(Z \leq 0.6154) \approx 1 - 0.7301 = 0.2699 \).

So, the answer for part (i) is approximately **0.27**.

### Part (ii)
Now we need to find the probability that the average time in **eight visits** will be at least 5 minutes.

1. Let \( \bar{X} \) be the sample mean of the eight visits. According to the Central Limit Theorem, \( \bar{X} \) will also be normally distributed with:
   - Mean \( \mu_{\bar{X}} = \mu = 4.2 \)
   - Standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{1.3}{\sqrt{8}} \approx 0.4593 \)

2. We are looking for \( P(\bar{X} \geq 5) \).

3. Standardize this value using the Z-score:
   \[
   Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{5 - 4.2}{0.4593} \approx 1.742
   \]

4. Looking up \( Z = 1.742 \) in the standard normal distribution table, we find \( P(Z \leq 1.742) \approx 0.9591 \).
5. Thus, \( P(\bar{X} \geq 5) = 1 - 0.9591 = 0.0409 \).

So, the answer for part (ii) is approximately **0.041**.

### Summary of Answers
- **(i)** \( P(X \geq 5) \approx 0.27 \)
- **(ii)** \( P(\bar{X} \geq 5) \approx 0.041 \)

Would you like further details or have any questions?

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**Additional Related Questions:**

1. What would the probability be if we changed the standard deviation to 2 minutes?
2. How does increasing the number of visits in part (ii) affect the probability?
3. What is the probability that the time will be less than 3 minutes on a single visit?
4. If the mean waiting time increased to 5 minutes, how would that affect part (i)?
5. How is the Central Limit Theorem applied in part (ii)?

**Tip:** When calculating probabilities for averages, remember that the standard deviation of the average decreases as the sample size increases.

Bob dines at a consistent fast food restaurant every day. Suppose that the time interval between Bob entering the restaurant and being served his food follows a normal distribution with a mean of 4.2 minutes and a standard deviation of 1.3 minutes.

i. Find the probability that when he enters the restaurant today it will be at least 5 minutes until he is served.

ii. Find the probability that average time until he is served in eight randomly selected visits to the restaurant will be at least 5 minutes.

Solve a probability problem involving normal distribution where Bob's waiting time at a fast food restaurant is analyzed. This problem includes calculating probabilities for single and multiple visits with a mean wait of 4.2 minutes and standard deviation of 1.3 minutes. Suitable for high school students.

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