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GATE DA 2024 Q57: Exponential rate when 5E(X) equals Var(X)

Question

Let XX be a random variable exponentially distributed with parameter λ>0\lambda > 0. The probability density function of XX is given by:

fX(x)={λe−λx,x≥00,otherwisef_X(x) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0 \\ 0, & \text{otherwise} \end{cases}

If 5E(X)=Var(X)5E(X) = Var(X), where E(X)E(X) and Var(X)Var(X) indicate the expectation and variance of XX, respectively, the value of λ\lambda is ______ (rounded off to one decimal place).

Try it: numerical answer: type the value

Numerical answer. No negative marking.
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Answer: 0.2 (official GATE 2024 final answer key)

Concept used: Exponential(λ\lambda): E(X)=1/λE(X) = 1/\lambda and Var(X)=1/λ2Var(X) = 1/\lambda^2.

First step: 5λ=1λ2\dfrac{5}{\lambda} = \dfrac{1}{\lambda^2} gives λ=15\lambda = \dfrac{1}{5}.

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Source: official GATE 2024 DA question paper and final answer key, organised by IISc Bangalore. Question paper · Answer key