Finding the Cumulative Distribution Function (CDF) Value

The problem asks us to find the value of the Cumulative Distribution Function (CDF), denoted by \(F(x)\), at a specific point \(x=0.5\). We are given the Probability Density Function (PDF), \(f(x)\), for the length of time X an examinee needs to complete a 1-hour exam. The PDF is given by:

\(f(x) = \dfrac{6}{5}(x^2+x)\) for \(0 \le x \le 1.\)

Understanding the Cumulative Distribution Function (CDF)

For a continuous random variable X, the CDF \(F(x)\) is defined as the probability that X takes a value less than or equal to \(x\). Mathematically, it is given by the integral of the PDF \(f(t)\) from the lower limit of the distribution (or \(-\infty\)) up to \(x\):

\(F(x) = P(X \le x) = \int_{-\infty}^{x} f(t) dt\)

Since our PDF is defined for \(0 \le x \le 1\), for any \(x\) within this range, the integral starts from the lower bound of the non-zero PDF, which is 0. So, for \(0 \le x \le 1\), the CDF is:

\(F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} \dfrac{6}{5}(t^2+t) dt\)

Calculating F(0.5)

We need to find the value of \(F(x)\) when \(x=0.5\). We substitute \(x=0.5\) into the CDF formula:

\(F(0.5) = \int_{0}^{0.5} \dfrac{6}{5}(t^2+t) dt\)

Now, we evaluate this definite integral:

1. Bring the constant \(\dfrac{6}{5}\) outside the integral: \[ F(0.5) = \dfrac{6}{5} \int_{0}^{0.5} (t^2+t) dt \]
2. Integrate term by term. The integral of \(t^n\) is \(\dfrac{t^{n+1}}{n+1}\): \[ \int (t^2+t) dt = \dfrac{t^{2+1}}{2+1} + \dfrac{t^{1+1}}{1+1} + C = \dfrac{t^3}{3} + \dfrac{t^2}{2} + C \]
3. Evaluate the definite integral from 0 to 0.5: \[ F(0.5) = \dfrac{6}{5} \left[ \dfrac{t^3}{3} + \dfrac{t^2}{2} \right]_{0}^{0.5} \]
4. Apply the limits of integration \([G(t)]_a^b = G(b) - G(a)\): \[ F(0.5) = \dfrac{6}{5} \left[ \left( \dfrac{(0.5)^3}{3} + \dfrac{(0.5)^2}{2} \right) - \left( \dfrac{(0)^3}{3} + \dfrac{(0)^2}{2} \right) \right] \]
5. Calculate the values: \(0.5 = \dfrac{1}{2}\) \[ (0.5)^3 = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} \] \[ (0.5)^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4} \] \[ F(0.5) = \dfrac{6}{5} \left[ \left( \dfrac{1/8}{3} + \dfrac{1/4}{2} \right) - (0 + 0) \right] \] \[ F(0.5) = \dfrac{6}{5} \left[ \dfrac{1}{24} + \dfrac{1}{8} \right] \]
6. Find a common denominator for the terms inside the brackets. The common denominator for 24 and 8 is 24: \[ \dfrac{1}{8} = \dfrac{1 \times 3}{8 \times 3} = \dfrac{3}{24} \] \[ F(0.5) = \dfrac{6}{5} \left[ \dfrac{1}{24} + \dfrac{3}{24} \right] \] \[ F(0.5) = \dfrac{6}{5} \left[ \dfrac{1+3}{24} \right] \] \[ F(0.5) = \dfrac{6}{5} \left[ \dfrac{4}{24} \right] \]
7. Simplify the expression: \[ F(0.5) = \dfrac{6}{5} \times \dfrac{4}{24} \] We can simplify \(\dfrac{4}{24}\) to \(\dfrac{1}{6}\): \[ F(0.5) = \dfrac{6}{5} \times \dfrac{1}{6} \] Cancel out the 6 in the numerator and denominator: \[ F(0.5) = \dfrac{1}{5} \]

Thus, the value of \(F(0.5)\) is \(\dfrac{1}{5}\).

Comparing with Options

Let's compare our calculated value with the given options:

Our calculated value, \(\dfrac{1}{5}\), matches Option 4.

Revision Table: Key Concepts

Additional Information: Properties of PDF and CDF

It is important to understand the relationship and properties of PDF and CDF for continuous random variables.

• Properties of PDF \(f(x)\):
  • \(f(x) \ge 0\) for all \(x\). The given \(f(x) = \frac{6}{5}(x^2+x)\) is non-negative for \(x \in [0, 1]\) because \(x^2 \ge 0\) and \(x \ge 0\).
  • The total area under the PDF curve must be equal to 1. \(\int_{-\infty}^{\infty} f(x) dx = 1\). In this case, \(\int_{0}^{1} \frac{6}{5}(x^2+x) dx = \frac{6}{5} [\frac{x^3}{3} + \frac{x^2}{2}]_0^1 = \frac{6}{5} [(\frac{1}{3}+\frac{1}{2}) - (0+0)] = \frac{6}{5} [\frac{2+3}{6}] = \frac{6}{5} \times \frac{5}{6} = 1\). This confirms the PDF is valid.
• Properties of CDF \(F(x)\):
  • \(0 \le F(x) \le 1\) for all \(x\). Our result \(F(0.5) = 1/5\) is within this range.
  • \(F(x)\) is a non-decreasing function. This means if \(x_1 < x_2\), then \(F(x_1) \le F(x_2)\).
  • \(\lim_{x \to -\infty} F(x) = 0\).
  • \(\lim_{x \to \infty} F(x) = 1\). For our distribution defined on \([0, 1]\), \(F(x) = 0\) for \(x < 0\) and \(F(x) = 1\) for \(x \ge 1\).
  • The PDF is the derivative of the CDF: \(f(x) = \dfrac{d}{dx} F(x)\) where \(F(x)\) is differentiable.

Calculating the CDF at a specific point like \(F(0.5)\) tells us the probability that the random variable (exam completion time in this case) is less than or equal to 0.5 hours (or 30 minutes). A value of \(F(0.5) = \frac{1}{5}\) means there is a 20% chance that an examinee will complete the exam within the first 30 minutes.