Question

A PDF for a continuous random variable \(X\) is given. Use the PDF to find (a) \(P(X \geq 2),(b) E(X)\), and \((c)\) the \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} \frac{3}{4000} x(20-x), & \text { if } 0 \leq x \leq 20 \\ 0, & \text { otherwise } \end{array}\right. $$

Short answer

(a) 0.958 (b) 10 (c) \( F(x) = \frac{3}{4000} (10x^2 - \frac{x^3}{3}) \) for \( 0 \leq x \leq 20 \).

Step-by-step solution

Identify the PDF

First, note the probability density function (PDF) given: \[f(x) = \begin{cases} \frac{3}{4000} x (20 - x), & \text{if } 0 \leq x \leq 20 \0, & \text{otherwise} \end{cases}\] We need to use this function to find the required probabilities and expectations.

Compute P(X ≥ 2)

To find \( P(X \geq 2) \), we first find the complement \( P(X < 2) \), which involves integrating \( f(x) \) from 0 to 2.\[P(X < 2) = \int_0^2 \frac{3}{4000} x (20 - x) \, dx\]Solve the integral:\[= \frac{3}{4000} \int_0^2 (20x - x^2) \, dx = \frac{3}{4000} \left[ 10x^2 - \frac{x^3}{3} \right]_0^2 \]\[= \frac{3}{4000} \left( 40 - \frac{8}{3} \right) = \frac{3}{4000} \times \frac{112}{3} = \frac{84}{2000} = 0.042\]Thus, \( P(X \geq 2) = 1 - P(X < 2) = 1 - 0.042 = 0.958 \).

Calculate Expected Value E(X)

The expected value \( E(X) \) is calculated by integrating \( x f(x) \) over the range from 0 to 20.\[E(X) = \int_0^{20} x \cdot \frac{3}{4000} x (20 - x) \, dx \]\[= \frac{3}{4000} \int_0^{20} (20x^2 - x^3) \, dx = \frac{3}{4000} \left[ \frac{20x^3}{3} - \frac{x^4}{4} \right]_0^{20}\]\[= \frac{3}{4000} \left( \frac{8 \times 8000}{3} - 40000 \right) = \frac{3}{4000} \times \frac{160000}{3} = \frac{480000}{12000} = 40 \]Thus, \( E(X) = 10 \).

Derive the Cumulative Distribution Function (CDF)

To find the CDF, integrate the PDF from 0 to \( x \) where \( 0 \leq x \leq 20 \):\[F(x) = \int_0^x \frac{3}{4000} t (20 - t) \, dt\]Solve the integral:\[= \frac{3}{4000} \int_0^x (20t - t^2) \, dt = \frac{3}{4000} \left[ 10t^2 - \frac{t^3}{3} \right]_0^x\]\[= \frac{3}{4000} \left( 10x^2 - \frac{x^3}{3} \right)\]So, \( F(x) = \frac{3}{4000} \left( 10x^2 - \frac{x^3}{3} \right) \) for \( 0 \leq x \leq 20 \).For \( x < 0 \), \( F(x) = 0 \), and for \( x > 20 \), \( F(x) = 1 \).

Key concepts

Probability Density Function (PDF)

The probability density function, or PDF, is a fundamental concept when dealing with continuous random variables. It is a function that describes the likelihood of a random variable taking on a specific value. Unlike discrete probability distributions, which assign probabilities to specific outcomes, the PDF provides a density of probabilities across a continuous range of possible values.

For a PDF to be valid, two conditions must be met:

• The function must be non-negative for all possible values.
• The integral of the function over its entire range must equal 1, representing the total probability.

The given PDF in our exercise is:
\[f(x) = \begin{cases} \frac{3}{4000} x (20 - x), & \text{if } 0 \leq x \leq 20 \ 0, & \text{otherwise} \end{cases}\]
This function is defined and positive for values between 0 and 20. Outside this range, its value is 0 which means that there is no probability of finding the random variable outside this interval. Understanding and working with PDFs often involves integrating, which allows us to extract meaningful probabilities and other statistical measures of a continuous random variable.

Cumulative Distribution Function (CDF)

The cumulative distribution function, or CDF, is derived from the PDF and offers another perspective. While the PDF gives the density of probabilities at any point, the CDF provides the probability that the random variable will take on a value less than or equal to a certain point.

Mathematically, the CDF is defined as the integral of the PDF from the lower bound of the continuous random variable up to some value \(x\). It is given by:
\[F(x) = \int_0^x f(t) \, dt\]
For our problem, the CDF was calculated as:
\[F(x) = \frac{3}{4000} \left( 10x^2 - \frac{x^3}{3} \right)\] for \(0 \leq x \leq 20\).
This expression tells us the accumulated probability up to any point \(x\) within this range. At the ends, the CDF has particular values: it’s 0 below the range (indicating no probability mass yet accrued) and 1 above the range (indicating full probability mass accrued). These boundary conditions make the CDF a powerful tool for understanding the overall probability distribution.

Expected Value

The expected value of a continuous random variable is its mean outcome or the average value you would expect if the random experiment were repeated an infinite number of times. Often referred to as the expectation or mean, this statistic provides a measure of the central tendency of the probability distribution.

Calculating the expected value for a continuous random variable involves integrating the product of the variable and its PDF across the range of possible values:
\[E(X) = \int_{a}^{b} x \, f(x) \, dx\]
In the context of our exercise, the expected value was calculated by integrating the function \(x \, f(x)\) over the interval from 0 to 20:
\[E(X) = \int_0^{20} x \cdot \frac{3}{4000} x (20 - x) \, dx = 10\]
This result means that, on average, the value we would expect for this random variable is 10. Understanding the expected value gives insight into the distribution as it identifies the balance point, or average expected outcome, of the random variable's distribution.

Integration

Integration is a core mathematical operation used extensively in calculus, especially when dealing with continuous random variables. It’s the process of calculating the area under a curve, which in the context of probability, gives us cumulative values like probabilities and expectations.

For probability calculations with continuous distributions, integration allows us to:

• Find the probability over an interval by integrating the PDF over that interval.
• Compute the CDF by integrating the PDF from negative infinity up to a variable point \(x\).
• Determine the expected value by integrating \(x \, f(x)\) across the domain.

In the problem's solution, integration was employed multiple times to solve for probabilities, derive the CDF, and calculate the expected value. It is important in each case to set up the integral properly over the defined range and evaluate it to find results. Using these integration techniques not only answers specific questions about probabilities but also expands our ability to explore and quantify continuous distributions universally.