Let π be a random variable exponentially distributed with parameter π > 0.β¦
2024
Let π be a random variable exponentially distributed with parameter π > 0. The probability density function of X is given by:
\(f_X(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{if } x \geq 0 \\ 0, & \text{otherwise} \end{cases} \)
If 5πΈ(π) = πππ(π), where πΈ(π) and πππ(π) indicate the expectation and variance of π, respectively, the value of π is ______ (rounded off to one decimal place).
Attempted by 2 students.
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Correct answer: 0.2
Answer: 0.2 (rounded to one decimal place)
For an exponential distribution with parameter Ξ»:
Expectation E(X) = 1/λ and variance Var(X) = 1/λ².
Use the given relation 5 E(X) = Var(X), so 5(1/λ) = 1/λ².
Solve the equation: 5/Ξ» = 1/λ² β multiply both sides by λ² to get 5Ξ» = 1.
Therefore Ξ» = 1/5 = 0.2, which is positive and satisfies the parameter constraint.