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In probability theory, a density function helps us determine how often an event will occur for a continuous random variable. For a continuous uniform distribution, like choosing a random number from the interval [2, 10], every number within that range is equally possible. The density function for a uniform distribution on this interval is constant, meaning it doesn't change value inside the interval. To find this function, you consider the length of the interval. Here, the interval is 8 units long (from 2 to 10). Thus, the density function is calculated as:

• \( f(x) = \frac{1}{8} \), for \( x \in [2, 10] \)

This means every point within [2, 10] has the same likelihood of being selected, making this function a fundamental tool in determining probabilities for subintervals in continuous uniform distributions.

The probability of an event in a continuous uniform distribution represents the size of the subinterval relative to the total interval. With a density function of \( f(x) = \frac{1}{8} \), the probability that the random variable falls between two points, say \( a \) and \( b \), is the area under the density function over that interval. This is calculated by integrating the density function from \( a \) to \( b \):

• \( P(a \leq X \leq b) = \int_{a}^{b} \frac{1}{8} \, dx = \frac{b-a}{8} \)

Integration here simplifies to just comparing the lengths of subintervals since 1 divided by the total interval length equals the consistent probability across the whole interval. This method is amazingly straightforward for finding probabilities in uniform distributions.

A continuous uniform distribution occurs when all outcomes in a range are equally likely. If you think of it like spinning a wheel or rolling a perfectly uniform die, every angle on the wheel or number on the die has the same chance. With our interval [2, 10], any number you point to has the same chance of being your random choice. The density function doesn't spike or dip here—it's flat. This distribution is characterized by:

• Constant density function: \( f(x) = \frac{1}{8} \), unchanged within [2, 10]
• Equal likelihood: Every subinterval probability depends solely on its length, \( \frac{b-a}{8} \).

The uniform nature makes it simple and intuitive for students when confirming they grasp these concepts. Even complex real-life systems often start with uniform distribution assumptions to simplify the math.

Solving a quadratic inequality involves finding out where a quadratic expression is greater or lesser than a given value. In our exercise, it was phrased as:\( X^2 - 12X + 35 > 0 \).First, find where the expression equals zero. Using the quadratic formula reveals it does this at the roots \( X = 5 \) and \( X = 7 \). Breaking this down further, the number line splits into segments:

• \( X < 5 \)
• \( 5 < X < 7 \)
• \( X > 7 \)

By testing values in each segment, you identify that the inequality holds true when \( X < 5 \) and \( X > 7 \). Thus, the entire probability is the sum of the chances that \( X \) falls outside [5, 7]. In a uniform distribution:

• \( P(X < 5) = \frac{3}{8} \)
• \( P(X > 7) = \frac{3}{8} \)

Combining, the total probability of \( X^2 - 12X + 35 > 0 \) is \( \frac{6}{8} = \frac{3}{4} \). Quadratic inequalities reveal how a distribution spreads beyond simply falling in a range.