(a) A continuous random variable $x$ has a probability density function defined
$f(x)=\left{\begin{array}{cc}
k\left(x-\frac{1}{a}\right), & 0 \leq x \leq 3 \\
0, & \text { elsewhere }
\end{array}\right.$
Given that $P ( x \geq 1)=0.8$, determine the:
(i) values of the constants a and k ;
(ii) mean of $x$.
Answer (2)
The values of the constants are a = -1 and k = 2/15. The mean of the continuous random variable x is calculated to be 1.8. These values were determined by evaluating the integrals of the given probability density function.
;
Find the values of constants a and k by using the properties of probability density functions and the given probability P ( x ≥ 1 ) = 0.8 .
Set up two equations based on the integrals of the pdf over the given ranges: ∫ 0 3 k ( x − a 1 ) d x = 1 and ∫ 1 3 k ( x − a 1 ) d x = 0.8 .
Solve the system of equations to find a = − 1 and k = 15 2 .
Calculate the mean of x using the formula μ = ∫ 0 3 x f ( x ) d x , which results in μ = 1.8 .
Explanation
Understand the problem and provided data We are given the probability density function (pdf) of a continuous random variable x as:
f ( x ) = { k ( x − a 1 ) , 0 , 0 ≤ x ≤ 3 elsewhere
Also, we know that P ( x ≥ 1 ) = 0.8 . Our goal is to find the values of the constants a and k , and then calculate the mean of x .
Set up the equations Since f ( x ) is a probability density function, the integral over its entire range must equal 1. This gives us our first equation:
∫ 0 3 k ( x − a 1 ) d x = 1
We are also given that P ( x ≥ 1 ) = 0.8 , which translates to:
∫ 1 3 k ( x − a 1 ) d x = 0.8
Evaluate the first integral Let's evaluate the first integral:
∫ 0 3 k ( x − a 1 ) d x = k ∫ 0 3 ( x − a 1 ) d x = k [ 2 x 2 − a x ] 0 3 = k ( 2 9 − a 3 ) = 1
So, we have:
k ( 2 9 − a 3 ) = 1
Evaluate the second integral Now, let's evaluate the second integral:
∫ 1 3 k ( x − a 1 ) d x = k ∫ 1 3 ( x − a 1 ) d x = k [ 2 x 2 − a x ] 1 3 = k [ ( 2 9 − a 3 ) − ( 2 1 − a 1 ) ] = k ( 2 8 − a 2 ) = k ( 4 − a 2 ) = 0.8
So, we have:
k ( 4 − a 2 ) = 0.8
Solve for a We now have a system of two equations with two unknowns, a and k :
k ( 2 9 − a 3 ) = 1
k ( 4 − a 2 ) = 0.8
Dividing the first equation by the second equation, we get:
4 − a 2 2 9 − a 3 = 0.8 1 = 4 5
a 4 a − 2 2 a 9 a − 6 = 4 5
2 ( 4 a − 2 ) 9 a − 6 = 4 5
4 ( 9 a − 6 ) = 10 ( 4 a − 2 )
36 a − 24 = 40 a − 20
− 4 a = 4
a = − 1
Solve for k Now that we have a = − 1 , we can substitute it back into either equation to solve for k . Let's use the first equation:
k ( 2 9 − a 3 ) = 1
k ( 2 9 − − 1 3 ) = 1
k ( 2 9 + 3 ) = 1
k ( 2 9 + 2 6 ) = 1
k ( 2 15 ) = 1
k = 15 2
So, a = − 1 and k = 15 2 .
Calculate the mean of x Now we need to find the mean of x . The formula for the mean is:
μ = ∫ 0 3 x f ( x ) d x = ∫ 0 3 x k ( x − a 1 ) d x
Substituting the values of a and k , we get:
μ = ∫ 0 3 x ⋅ 15 2 ( x − − 1 1 ) d x = 15 2 ∫ 0 3 x ( x + 1 ) d x = 15 2 ∫ 0 3 ( x 2 + x ) d x
μ = 15 2 [ 3 x 3 + 2 x 2 ] 0 3 = 15 2 ( 3 3 3 + 2 3 2 ) = 15 2 ( 9 + 2 9 ) = 15 2 ( 2 18 + 2 9 ) = 15 2 ( 2 27 ) = 15 27 = 5 9 = 1.8
Therefore, the mean of x is 1.8 .
Examples
Understanding probability density functions and calculating means is crucial in many real-world scenarios. For instance, in finance, it helps in modeling stock prices and predicting investment returns. In engineering, it aids in analyzing the reliability of systems and components. In environmental science, it's used to model pollution levels and predict environmental impacts. By determining the PDF and mean, we can make informed decisions and predictions in these fields.
