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The expected value, often denoted as \(E(X)\), is a fundamental concept in probability and statistics. It represents the average or mean value that a random variable is expected to take on. In the case of a uniform distribution, where every outcome is equally likely, calculating the expected value becomes quite straightforward.

For a uniform distribution, the expected value is simply the midpoint between the minimum \(a\) and the maximum \(b\) of the distribution. This is calculated using the formula:

• \( E(X) = \frac{a + b}{2} \)

Applying this formula to our problem of a highway patrol station with a response range from 0 to 30 miles, we find:

• \( E(X) = \frac{0 + 30}{2} = 15 \)

This tells us that, on average, the accidents occur 15 miles from the patrol station, which is right in the middle of the range.

Variance, denoted by \(\operatorname{Var}(X)\), measures the spread or dispersion of a set of data points. It shows how much the values differ from the expected value. In simple terms, it's a way to quantify how much the distances of accidents from the highway patrol station vary.

For a uniform distribution, the formula to find variance is:

• \( \operatorname{Var}(X) = \frac{(b-a)^2}{12} \)

Using our interval of 0 to 30 miles, the variance is calculated as:

• \( \operatorname{Var}(X) = \frac{(30-0)^2}{12} = \frac{900}{12} = 75 \)

The result, 75, indicates that the distance of accidents from the station tends to deviate quite a bit from the average 15 miles, when assessed over many emergencies.

The standard deviation provides another way to understand the variability in a data set. It is the square root of the variance and gives a measure of spread around the mean in the same units as the data. For our uniform distribution with a range from 0 to 30 miles, the standard deviation helps indicate how typical it is for an emergency to be a certain distance from the station.

The formula for standard deviation is:

• \( \sigma(X) = \sqrt{\operatorname{Var}(X)} \)

From the previously calculated variance (75), the standard deviation is:

• \( \sigma(X) = \sqrt{75} \approx 8.66025 \)

This result shows us that the actual distance between the accident and the station will typically vary by about 8.66 miles from the average of 15 miles.

Probability calculations allow us to determine the likelihood of certain outcomes. In the context of a uniform distribution, it's often about finding how likely it is for a random variable to fall within a particular range.

For the probability that a random variable is greater than a specific value \(c\), the formula is:

• \( P(X > c) = \frac{b-c}{b-a} \)

In our scenario, we're interested in finding the probability that an accident occurs more than 24 miles from the station:

• \( P(X > 24) = \frac{30-24}{30-0} = \frac{6}{30} = 0.2 \)

This means there is a 20% chance that any given accident will happen more than 24 miles from the station, making it a less likely, but still possible outcome given the uniform distribution.