Q6. A continuous random variable _I has the probability density function: f(x) 0 else (a) Find h that make f(x) a valid probability density function (PDF) and graph the PDF. (b) Find and graph the cumulative distribution function (CDF). (You can use Matlab for graphing the function, not for calculating it.) (c) Use the CDF to find i) Pr(YS 0) ii) Pr(X 5) iii) Pr(X56)
The Correct Answer and Explanation is:
To answer this question thoroughly, we’ll assume the missing part of the probability density function (PDF) was meant to be something like this:
Let f(x)=h(6−x)f(x) = h(6 – x) for 0≤x≤60 \le x \le 6, and f(x)=0f(x) = 0 otherwise.
This type of function is a triangular distribution that decreases linearly over the interval [0, 6]. We now go through each part:
(a) Find hh to make f(x)f(x) a valid PDF and graph the PDF
A probability density function must integrate to 1 over its domain: ∫06h(6−x) dx=1\int_{0}^{6} h(6 – x) \, dx = 1
Evaluate the integral: h∫06(6−x) dx=h[6x−x22]06=h(36−18)=h⋅18h \int_{0}^{6} (6 – x) \, dx = h \left[ 6x – \frac{x^2}{2} \right]_0^6 = h \left(36 – 18\right) = h \cdot 18
Set it equal to 1: h⋅18=1⇒h=118h \cdot 18 = 1 \Rightarrow h = \frac{1}{18}
✅ Answer: h=118h = \frac{1}{18}
(b) Find and graph the CDF
The CDF, F(x)F(x), is the integral of the PDF from 0 to xx: F(x)=∫0x118(6−t) dt=118[6t−t22]0x=118(6x−x22)F(x) = \int_0^x \frac{1}{18}(6 – t) \, dt = \frac{1}{18} \left[6t – \frac{t^2}{2} \right]_0^x = \frac{1}{18} \left(6x – \frac{x^2}{2} \right)
So the CDF is: F(x)={0x<0118(6x−x22)0≤x≤61x>6F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{18} \left(6x – \frac{x^2}{2} \right) & 0 \le x \le 6 \\ 1 & x > 6 \end{cases}
✅ CDF expression: F(x)=118(6x−x22)for 0≤x≤6F(x) = \frac{1}{18}(6x – \frac{x^2}{2}) \quad \text{for } 0 \le x \le 6
(c) Use the CDF to find:
i) Pr(X≤0)\Pr(X \le 0)
F(0)=0F(0) = 0
✅ Answer: 0
ii) Pr(X≤5)\Pr(X \le 5)
F(5)=118(6⋅5−252)=118(30−12.5)=17.518=3536F(5) = \frac{1}{18} (6 \cdot 5 – \frac{25}{2}) = \frac{1}{18} (30 – 12.5) = \frac{17.5}{18} = \frac{35}{36}
✅ Answer: 3536\frac{35}{36}
iii) Pr(X≤6)\Pr(X \le 6)
F(6)=118(36−18)=1F(6) = \frac{1}{18} (36 – 18) = 1
✅ Answer: 1
Explanation
This problem involves analyzing a continuous random variable XX with a given probability density function (PDF). A function qualifies as a valid PDF if it is non-negative and integrates to 1 over its domain. Here, the function f(x)=h(6−x)f(x) = h(6 – x) over the interval [0,6][0, 6] forms a triangle with height proportional to hh. To determine the constant hh, we integrate the function over its valid domain and equate the result to 1. Doing this gives h=118h = \frac{1}{18}, ensuring the total probability is normalized.
Next, we calculate the cumulative distribution function (CDF), which represents the probability that XX takes a value less than or equal to xx. The CDF is derived by integrating the PDF from the lower bound (0) to a point xx. This gives F(x)=118(6x−x22)F(x) = \frac{1}{18}(6x – \frac{x^2}{2}) for 0≤x≤60 \le x \le 6, with values of 0 for x<0x < 0 and 1 for x>6x > 6.
Finally, we use the CDF to compute specific probabilities. Since XX cannot take negative values, Pr(X≤0)=0\Pr(X \le 0) = 0. The value Pr(X≤5)\Pr(X \le 5) is found by substituting 5 into the CDF, yielding 3536\frac{35}{36}. Pr(X≤6)\Pr(X \le 6), the maximum of the domain, must equal 1.
These calculations illustrate the core concepts of probability density functions and cumulative distribution functions, emphasizing how integration and bounds define continuous probabilities.
