(a) A continuous random variable [tex]$x$[/tex] has a probability density function defined
[tex]f(x)=\left\{\begin{array}{cc}
k\left(x-\frac{1}{a}\right), & 0 \leq x \leq 3 \\
0, & \text { elsewhere }\\
\end{array}\right.[/tex]
Given that [tex]$P ( x \geq 1)=0.8$[/tex], determine the:
(i) values of the constants a and k ;
(ii) mean of [tex]$x$[/tex].
Answer (2)
We determined that the constants are a = − 1 and k = 15 2 , and the mean of the random variable x is 1.8.
;
Determine constants a and k by using given PDF f ( x ) and P ( x ≥ 1 ) = 0.8 .
Set up integral equations: ∫ 0 3 k ( x − a 1 ) d x = 1 and ∫ 1 3 k ( x − a 1 ) d x = 0.8 .
Solve for a and k : a = − 1 and k = 15 2 .
Calculate the mean: μ = ∫ 0 3 x f ( x ) d x = 1.8 .
Explanation
Analyze the problem and data We are given a probability density function (pdf) f ( x ) for a continuous random variable x :
f ( x ) = { k ( x − a 1 ) , 0 , 0 ≤ x ≤ 3 elsewhere We also 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 the following equation: ∫ 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 Now we have a system of two equations with two unknowns, a and k :
k ( 2 9 − a 3 ) = 1 k ( 4 − a 2 ) = 0.8 We can solve this system. Divide the first equation by the second equation: 4 − a 2 2 9 − a 3 = 0.8 1 = 4 5 2 9 − a 3 = 4 5 ( 4 − a 2 ) = 5 − 2 a 5 2 9 − 5 = a 3 − 2 a 5 − 2 1 = 2 a 6 − 5 = 2 a 1 − 2 1 = 2 a 1 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 Substitute the values of a and k :
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
State the final answer Therefore, the values of the constants are a = − 1 and k = 15 2 , and the mean of x is 1.8 .
Examples
Imagine you are analyzing the waiting time at a bus stop. The probability density function describes how likely people are to wait for different amounts of time. Knowing that people wait at least 1 minute 80% of the time, you can determine the specific shape of this distribution (the values of 'a' and 'k'). Furthermore, calculating the mean waiting time helps in scheduling and resource allocation to improve customer satisfaction.
