Validating PDF & Probabilities: Find C And Solve!
Hey guys! Today, we're diving into the exciting world of probability density functions (PDFs) and tackling a problem that involves finding a constant to make a PDF valid and then calculating some probabilities. So, buckle up, and let's get started!
Finding the Constant C
So, our mission, should we choose to accept it, is to find the constant C such that the probability density function (p.d.f.)
f(x) = \begin{cases} Cx^2 & \text{for } 0 < x < 3 \\ 0 & \text{elsewhere} \end{cases}
is valid. For a PDF to be valid, it must satisfy two conditions:
- f(x) ≥ 0 for all x: This means the function must always be non-negative.
- ∫-∞∞ f(x) dx = 1: The integral of the function over its entire domain must equal 1. This ensures that the total probability is 1.
In our case, f(x) = Cx² for 0 < x < 3, and 0 elsewhere. Since x² is always non-negative, the first condition is satisfied if C ≥ 0. Now, let's focus on the second condition.
We need to find C such that:
∫-∞∞ f(x) dx = 1
Since f(x) = 0 outside the interval (0, 3), we can rewrite the integral as:
∫03 Cx² dx = 1
Let's evaluate the integral:
C ∫03 x² dx = C [x³/3]03 = C [(3³/3) - (0³/3)] = C [27/3] = 9C
So, we have:
9C = 1
Solving for C, we get:
C = 1/9
Therefore, the valid probability density function is:
f(x) = \begin{cases} (1/9)x^2 & \text{for } 0 < x < 3 \\ 0 & \text{elsewhere} \end{cases}
Justification and Elaboration
The key to finding the constant C lies in understanding the fundamental properties of a probability density function. A PDF describes the relative likelihood of a continuous random variable taking on a given value. It's not the probability itself (as it is for discrete random variables), but rather the density of probability at a specific point. That's why we need to integrate the PDF over an interval to find the probability of the variable falling within that interval.
The condition that the integral of the PDF over its entire domain equals 1 is crucial. It reflects the certainty that the random variable must take on some value within its possible range. In simpler terms, the total probability of all possible outcomes must be 100%.
When we evaluated the integral ∫03 Cx² dx, we were essentially finding the area under the curve of the function f(x) = Cx² between the limits x = 0 and x = 3. This area represents the total probability of X falling within that interval. By setting this area equal to 1 and solving for C, we ensured that the function satisfies the requirement of a valid PDF.
Now that we've found the value of C, we can confidently use this PDF to calculate various probabilities related to the random variable X.
Calculating Probabilities
Now that we have our valid PDF, let's calculate the probabilities you asked about. Remember, for continuous random variables, the probability at a single point is always zero. This is because probability is represented by the area under the curve, and the area at a single point is infinitesimally small (zero).
(i) P(X = 1)
Since X is a continuous random variable,
P(X = 1) = 0
(ii) P(1 ≤ X ≤ 2)
To find the probability that X lies between 1 and 2, we need to integrate the PDF over this interval:
P(1 ≤ X ≤ 2) = ∫12 (1/9)x² dx
Evaluating the integral:
(1/9) ∫12 x² dx = (1/9) [x³/3]12 = (1/9) [(2³/3) - (1³/3)] = (1/9) [(8/3) - (1/3)] = (1/9) [7/3] = 7/27
So,
P(1 ≤ X ≤ 2) = 7/27
(iii) P(X ≤ 2)
To find the probability that X is less than or equal to 2, we integrate the PDF from 0 to 2 (since the PDF is 0 for x < 0):
P(X ≤ 2) = ∫02 (1/9)x² dx
Evaluating the integral:
(1/9) ∫02 x² dx = (1/9) [x³/3]02 = (1/9) [(2³/3) - (0³/3)] = (1/9) [8/3] = 8/27
So,
P(X ≤ 2) = 8/27
(iv) P(X > 1)
To find the probability that X is greater than 1, we can either integrate the PDF from 1 to 3, or use the complement rule:
P(X > 1) = 1 - P(X ≤ 1)
Let's use the complement rule. First, we need to find P(X ≤ 1):
P(X ≤ 1) = ∫01 (1/9)x² dx
Evaluating the integral:
(1/9) ∫01 x² dx = (1/9) [x³/3]01 = (1/9) [(1³/3) - (0³/3)] = (1/9) [1/3] = 1/27
Now, using the complement rule:
P(X > 1) = 1 - P(X ≤ 1) = 1 - (1/27) = 26/27
So,
P(X > 1) = 26/27
(v) P(X > 2)
To find the probability that X is greater than 2, we integrate the PDF from 2 to 3:
P(X > 2) = ∫23 (1/9)x² dx
Evaluating the integral:
(1/9) ∫23 x² dx = (1/9) [x³/3]23 = (1/9) [(3³/3) - (2³/3)] = (1/9) [(27/3) - (8/3)] = (1/9) [19/3] = 19/27
So,
P(X > 2) = 19/27
Elaboration on Probability Calculations
Understanding how to calculate probabilities from a PDF is fundamental in probability theory and statistics. The key concept is that the probability of a continuous random variable falling within a specific range is represented by the area under the PDF curve over that range. This area is calculated using integration.
For example, when we calculated P(1 ≤ X ≤ 2), we were finding the area under the curve of f(x) = (1/9)x² between x = 1 and x = 2. This area represents the probability that the random variable X will take on a value between 1 and 2.
The complement rule, which we used to calculate P(X > 1), is a powerful tool. It states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring. In this case, the event is X > 1, and the complement is X ≤ 1. Using the complement rule can sometimes simplify calculations, especially when it's easier to calculate the probability of the complement.
It's also important to remember that for continuous random variables, the probability of the variable taking on a single, specific value is always zero. This is because the area under the curve at a single point is infinitesimally small. That's why P(X = 1) = 0.
Conclusion
Alright, we've successfully found the constant C that makes our PDF valid and calculated various probabilities. Remember, the key is to understand the properties of PDFs and how to use integration to find probabilities. Keep practicing, and you'll become a pro at this in no time! You nailed it! Understanding PDFs and their applications is super useful in many fields, so great job sticking with it. Now you're ready to tackle more probability challenges. Keep up the awesome work!