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Math  /  Data & Statistics

QuestionLet XX be a random variable with a CDF equal to F(x)=c⋅(11.5−xe−x)⋅1[1.8;∞)(x)F(x)=c \cdot\left(\frac{1}{1.5}-x e^{-x}\right) \cdot \mathbb{1}_{[1.8 ; \infty)}(x), where cc is a constant. Find cc :
Answer: 3.361 □\square
The correct answer is: 1.500
For the random variable defined aboveP (X∈[3.96;∞))(X \in[3.96 ; \infty)) amounts to
Answer: □\square 1

Studdy Solution

STEP 1

1. The cumulative distribution function (CDF) F(x) F(x) is defined for x≥1.8 x \geq 1.8 .
2. The CDF must satisfy lim⁡x→∞F(x)=1 \lim_{x \to \infty} F(x) = 1 .

STEP 2

1. Determine the value of the constant c c by ensuring the CDF approaches 1 as x x approaches infinity.
2. Calculate the probability P(X∈[3.96,∞)) P(X \in [3.96, \infty)) .

STEP 3

Evaluate the CDF F(x) F(x) as x→∞ x \to \infty :
F(x)=c⋅(11.5−xe−x) F(x) = c \cdot \left(\frac{1}{1.5} - x e^{-x}\right)
As x→∞ x \to \infty , the term xe−x→0 x e^{-x} \to 0 .
Thus,
lim⁡x→∞F(x)=c⋅11.5=1 \lim_{x \to \infty} F(x) = c \cdot \frac{1}{1.5} = 1

STEP 4

Solve for c c :
c⋅11.5=1 c \cdot \frac{1}{1.5} = 1
c=1.5 c = 1.5

STEP 5

Calculate P(X∈[3.96,∞)) P(X \in [3.96, \infty)) :
Since F(x) F(x) is a CDF, P(X∈[3.96,∞))=1−F(3.96) P(X \in [3.96, \infty)) = 1 - F(3.96) .
First, calculate F(3.96) F(3.96) :
F(3.96)=1.5⋅(11.5−3.96e−3.96) F(3.96) = 1.5 \cdot \left(\frac{1}{1.5} - 3.96 e^{-3.96}\right)
Compute 3.96e−3.96 3.96 e^{-3.96} and substitute back into the equation to find F(3.96) F(3.96) .

STEP 6

Compute P(X∈[3.96,∞))=1−F(3.96) P(X \in [3.96, \infty)) = 1 - F(3.96) .
Substitute the calculated value of F(3.96) F(3.96) to find the probability.
The probability P(X∈[3.96,∞)) P(X \in [3.96, \infty)) is:
1 \boxed{1}

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