Respuesta :
Answer:
f1 ( x ) valid pdf . f2 ( x ) is invalid pdf
k = 1 / 18 , i ) 0.6133 , ii ) 0.84792
Step-by-step explanation:
Solution:-
A) The two pdfs ( f1 ( x ) and f2 ( x ) ) are given as follows:
[tex]f_1(x) = \left \{ {{0.5(3x-x^3) } .. 0 < x < 2 \atop {0} } \right. \\\\f_2(x) = \left \{ {{0.3(3x-x^2) } .. 0 < x < 2 \atop {0} } \right. \\[/tex]
- To check the legitimacy of a continuous probability density function the area under the curve over the domain must be equal to 1. In other words the following:
[tex]\int\limits^a_b {f_1( x )} \, dx = 1\\\\ \int\limits^a_b {f_2( x )} \, dx = 1\\[/tex]
- We will perform integration of each given pdf as follows:
[tex]\int\limits^a_b {f_1(x)} \, dx = \int\limits^2_0 {0.5(3x - x^3 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - 0.125x^4 ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - 0.125*(16) ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 2 ] = 1\\[/tex]
[tex]\int\limits^a_b {f_2(x)} \, dx = \int\limits^2_0 {0.5(3x - x^2 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - \frac{x^3}{6} ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - \frac{(8)}{6} ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 1.3333 ] = 1.67 \neq 1 \\[/tex]
Answer: f1 ( x ) is a valid pdf; however, f2 ( x ) is not a valid pdf.
B)
- A random variable ( X ) denotes the resistance of a randomly chosen resistor, and the pdf is given as follows:
[tex]f ( x ) = kx[/tex] if 8 ≤ x ≤ 10
0 otherwise.
- To determine the value of ( k ) we will impose the condition of validity of a probability function as follows:
[tex]\int\limits^a_b {f(x)} \, dx = 1\\[/tex]
- Evaluate the integral as follows:
[tex]\int\limits^1_8 {kx} \, dx = 1\\\\\frac{kx^2}{2} ]\limits^1^0_8 = 1\\\\k* [ 10^2 - 8^2 ] = 2\\\\k = \frac{2}{36} = \frac{1}{18}[/tex]... Answer
- To determine the CDF of the given probability distribution we will integrate the pdf from the initial point ( 8 ) to a respective value ( x ) as follows:
[tex]cdf = F ( x ) = \int\limits^x_8 {f(x)} \, dx\\\\F ( x ) = \int\limits^x_8 {\frac{x}{18} } \, dx\\\\ F ( x ) = [ \frac{x^2}{36} ] \limits^x_8\\\\F ( x ) = \frac{x^2 - 64}{36}[/tex]
To determine the probability p ( 8.6 ≤ x ≤ 9.8 ) we will utilize the cdf as follows:
p ( 8.6 ≤ x ≤ 9.8 ) = F ( 9.8 ) - F ( 8.6 )
p ( 8.6 ≤ x ≤ 9.8 ) = [tex]\frac{(9.8)^2 - 64}{36} - \frac{(8.6)^2 - 64}{36} = 0.61333[/tex]
ii) To determine the conditional probability we will utilize the basic formula as follows:
p ( x ≤ 9.8 | x ≥ 8.6 ) = p ( 8.6 ≤ x ≤ 9.8 ) / p ( x ≥ 8.6 )
p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - p ( x ≤ 8.6 ) ]
p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 1 - 0.27666 ]
p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.61333 / [ 0.72333 ]
p ( x ≤ 9.8 | x ≥ 8.6 ) = 0.84792 ... answer
