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2 answers:
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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:
- 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:
- We will perform integration of each given pdf as follows:
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:
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:
- Evaluate the integral as follows:
... 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:
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 ) =
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