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In a uniform distribution, the probability density function (PDF) is a crucial concept. It describes how probability is distributed across a range of values. For a uniform distribution, the PDF is constant. This means that any value within the specified range is equally likely. In our example, the random variable \(x\) is uniformly distributed between 1.0 and 1.5.
The formula for the PDF of a uniform distribution on the interval \([a, b]\) is given by:\[f(x) = \frac{1}{b-a}\]This means that for each value between \(a\) and \(b\), the probability density is the same. Outside this interval, the probability density is zero.
For the given problem, \(x\) lies between 1.0 and 1.5, so the PDF is:\[f(x) = \frac{1}{1.5-1.0} = 2\]Thus, the graph of this PDF is a horizontal line at \(f(x) = 2\) between \(x = 1.0\) and \(x = 1.5\). The height of the line shows the constant density over this range.

• PDF provides a way to see where values are most concentrated.
• For uniform distributions, this concentration is even across the interval.

Continuous probability deals with events occurring within a certain range. Unlike discrete probability, where outcomes can be counted individually, in continuous probability, outcomes are part of a continuum.
For a continuous distribution like the uniform distribution, the probability of any single point is zero. This is because there are infinitely many points in any interval, and the probability is spread across these infinitely many possibilities. Instead, we calculate the probability over an interval.
In our example, finding \(P(x=1.25)\) involves a single point. So, in continuous distributions,

• Probability for any exact value, like \(P(x=1.25)\), is zero.
• Probabilities must be computed over intervals.

To find probabilities within a range for a uniform distribution, use the properties of the PDF. Specifically, probabilities are calculated using the length of the interval compared to the total range of the distribution.
For example, consider the problem of computing \(P(1.0 \leq x \leq 1.25)\). First, determine the interval length, which is \(1.25 - 1.0 = 0.25\). The total distribution range is \(1.5 - 1.0 = 0.5\).
Therefore, the probability is calculated as:\[P(1.0 \leq x \leq 1.25) = \frac{0.25}{0.5} = 0.5\]Do the same for computing \(P(1.20 < x < 1.5)\). Here, the interval length is \(1.5 - 1.20 = 0.3\). The probability becomes:\[P(1.20 < x < 1.5) = \frac{0.3}{0.5} = 0.6\]These steps show how to compute the probability for a continuous uniform distribution.

• Probability is the interval length divided by the total range.
• Helps in understanding how much of the distribution falls within a certain interval.