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When working with uniform distribution, the probability density function (PDF) provides a way to describe how the probability is distributed over an interval. For a uniform random variable between two points, say \(a\) and \(b\), each value within this interval is equally probable. That's why it's called "uniform". To find the PDF for a uniform distribution, we use the formula: \[ f(x) = \frac{1}{b-a} \quad \text{for} \quad a \leq x \leq b \]This formula tells us that the probability is distributed uniformly and equally across the interval \([a, b]\).In our case, with \(X\) distributed over \([0, 10]\), the PDF simplifies to:\[ f(x) = \frac{1}{10} \quad \text{for} \quad 0 \leq x \leq 10 \]This means that every number from 0 to 10 has an equal likelihood of occurring.

The expected value, often denoted \(E(X)\), serves as the "average" or "mean" outcome you'd expect from a random variable. For a uniform distribution, the expected value is simply the midpoint of the interval. The formula to calculate this is:\[ E(X) = \frac{a+b}{2} \]Here, \(a\) is the lower bound and \(b\) is the upper bound of the interval.For our exercise, where \(X\) is between 0 and 10, the expected value is:\[ E(X) = \frac{0+10}{2} = 5 \]This number, 5, is the balance point or the center of the interval, signifying where we expect \(X\) to land on average.

Variance, represented as \(\operatorname{Var}(X)\), measures the spread of a set of data points. In simpler terms, it tells us how much the values of a random variable differ from the expected value. For a uniform distribution, the variance is calculated by:\[ \operatorname{Var}(X) = \frac{(b-a)^2}{12} \]This formula accounts for the difference between the highest and lowest points, \(b\) and \(a\), respectively.Applying this to our uniform distribution on \([0, 10]\), we find:\[ \operatorname{Var}(X) = \frac{(10-0)^2}{12} = \frac{100}{12} = \frac{25}{3} \]This value, \(\frac{25}{3}\), indicates how the numbers in the distribution vary around the expected value.

Standard deviation, denoted by \(\sigma(X)\), is a measure derived from variance. It provides insight into the spread of a distribution: how much variation exists from the mean value.To find the standard deviation for a uniform distribution, we simply take the square root of the variance:\[ \sigma(X) = \sqrt{\operatorname{Var}(X)} \]Substituting the variance we computed earlier:\[ \sigma(X) = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}} \approx 2.89 \]This tells us that, on average, each value in the distribution is about 2.89 units away from the expected value (mean). It offers a more tangible grasp of deviation than variance, as it is in the same unit as the original data.

Calculating probabilities within a uniform distribution framework can be intuitive. Since all intervals of the same size have equal probabilities, finding the probability of a specific interval is straightforward.Suppose you need to find the probability that \(X\) falls within a sub-interval \([c, d]\) of \([a, b]\). You use:\[ P(c\leq X \leq d) = \frac{d-c}{b-a} \]For our example, the probability that \(X\) lies between 8 and 10 is:\[ P(8 \leq X \leq 10) = \frac{10-8}{10-0} = \frac{2}{10} = 0.2 \]So, there's a 20% chance that a value drawn from \(X\) will be between 8 and 10. This simple proportion reflects the equal likelihood of all sub-intervals in a uniform distribution.