The problem describes a flight's arrival time following a continuous uniform distribution. To calculate the mean and standard deviation for this uniform distribution, we can use the following standard formulas:

• For a uniform distribution $U(a, b)$, where $a$ is the minimum value and $b$ is the maximum value:
  • Mean (μ): $\mu = \frac{a + b}{2}$
  • Standard Deviation (σ): $\sigma = \frac{b - a}{\sqrt{12}}$

Step 1: Convert time to minutes.

• The early arrival time $a$ is 9:18 am.
• The late arrival time $b$ is 9:41 am.

Convert these times to the number of minutes after 9:00 am:

• 9:18 am = 18 minutes after 9:00 am, so $a = 18$.
• 9:41 am = 41 minutes after 9:00 am, so $b = 41$.

Step 2: Apply the formulas.

1. Mean (μ): $\mu = \frac{18 + 41}{2} = \frac{59}{2} = 29.5 \, \text{minutes}$ So, the mean is 29.5 minutes after 9:00 am, or 9:29.5 am.

2. Standard Deviation (σ): $\sigma = \frac{41 - 18}{\sqrt{12}} = \frac{23}{\sqrt{12}} \approx \frac{23}{3.4641} \approx 6.64 \, \text{minutes}$

Final Answers:

• Mean arrival time: $29.5$ minutes (which is 9:29.5 am)
• Standard deviation: $6.64$ minutes.

Would you like further details on the uniform distribution or the interpretation of these results?

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Relative questions:

1. How do you interpret a uniform distribution in real-world scenarios?
2. Can you explain why the standard deviation formula for a uniform distribution involves $\sqrt{12}$?
3. What happens if the flight's actual arrival time deviates from the uniform assumption?
4. How would the mean and standard deviation change if the flight window increased?
5. What are the implications of a small vs. large standard deviation in this context?

Tip: The uniform distribution is ideal for modeling scenarios where all outcomes in a range are equally likely.