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A probability density function (PDF) describes the likelihood of a continuous random variable falling within a certain range of values. For a uniform distribution like the one for Delta's flight time, the PDF is particularly straightforward. The key characteristic of a uniform distribution is that each outcome in the interval is equally likely. This is depicted by a constant horizontal line in the graph of its PDF.

To calculate the PDF for a uniform distribution between two points, say \([a, b]\), the formula used is:

• \( f(x) = \frac{1}{b-a} \) for \( a \leq x \leq b \).

In our exercise, since the possible flight times range from 120 to 140 minutes, the PDF is \( f(x) = \frac{1}{20} = 0.05 \). The graph features this horizontal line from 120 to 140, making a rectangle shape. This indicates that every flight time within this band is equally probable.

The expected value for a random variable provides a measure of the center of the distribution, often thought of as the 'average' value we'd anticipate. For a uniform distribution, calculating the expected value is straightforward: it is the midpoint of the distribution.

For the uniform distribution covering Delta's flight times, we use:

• \( E(X) = \frac{a+b}{2} \)

Here, \( a = 120 \) minutes and \( b = 140 \) minutes, leading to:

• \[ E(X) = \frac{120 + 140}{2} = 130 \text{ minutes} \]

Thus, the expected flight time is 130 minutes. This means, on average, flights are expected to last this long, showing the center of the probability distribution for flight durations.

Calculating probabilities with a uniform distribution is intuitive, as it involves computing areas under the PDF. Each interval on the horizontal axis of the graph holds equal likelihood in a uniform distribution.

To find the probability that a flight will be 'no more than 5 minutes late', indicating times up to 125 minutes, we calculate:

• \[ P(X \leq 125) = \frac{125 - 120}{140 - 120} = \frac{5}{20} = 0.25 \]

This result means there is a 25% chance the flight will take 125 minutes or less.
For being 'more than 10 minutes late', implying flight times over 130 minutes, the probability is:

• \[ P(X > 130) = \frac{140 - 130}{140 - 120} = \frac{10}{20} = 0.5 \]

Therefore, there's a 50% probability a flight will exceed 130 minutes, highlighting the balance of likelihoods in the interval of a uniform distribution.