To solve this problem, we need to convert the given time data into a minute scale, and then calculate the mean and standard deviation for the continuous uniform distribution.

Step 1: Convert the time to minutes

Let's convert the given times into the number of minutes after 9:00 AM.

• Early arrival time (9:18 AM): $9:18 \, \text{AM} = 9 \times 60 + 18 = 558 \, \text{minutes}$
• Late arrival time (9:41 AM): $9:41 \, \text{AM} = 9 \times 60 + 41 = 581 \, \text{minutes}$

Thus, the time interval for the uniform distribution is from 558 minutes to 581 minutes.

Step 2: Calculate the mean

For a uniform distribution, the mean is given by the formula: $\mu = \frac{a + b}{2}$ where $a$ is the lower bound (early arrival time) and $b$ is the upper bound (late arrival time).

Substitute $a = 558$ and $b = 581$: $\mu = \frac{558 + 581}{2} = \frac{1139}{2} = 569.5 \, \text{minutes}$

Step 3: Calculate the standard deviation

For a uniform distribution, the standard deviation is given by the formula: $\sigma = \frac{b - a}{\sqrt{12}}$ Substitute $a = 558$ and $b = 581$: $\sigma = \frac{581 - 558}{\sqrt{12}} = \frac{23}{\sqrt{12}} \approx \frac{23}{3.4641} \approx 6.6390 \, \text{minutes}$

Final Answers:

• Mean arrival time: 569.5 minutes (which corresponds to 9:29.5 AM).
• Standard deviation: 6.64 minutes.

Would you like more details on these calculations, or do you have any questions?

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Here are 5 related questions to consider:

1. How would the mean change if the early and late arrival times were extended by 10 minutes each?
2. What is the probability that a flight arrives between 9:20 AM and 9:30 AM?
3. How would the standard deviation change if the time range was shortened?
4. How is the uniform distribution different from other continuous distributions like normal distribution?
5. What other statistical measures can be calculated for a uniform distribution?

Tip: For uniform distributions, all outcomes within the interval are equally likely, so the shape of the distribution is a rectangle.