Question

The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter λ\lambda. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is __________

Official answer and solution
Accepted answers
0.35 to 0.39
From
Official GATE 2021 answer key

Solution

Determine the expected value of an exponential distribution and evaluate the survival function at that value.

Let XX be an exponentially distributed random variable with parameter λ>0\lambda > 0. Its probability density function is f(x)=λe−λxf(x) = \lambda e^{-\lambda x} for x≥0x \ge 0, and its cumulative distribution function is F(x)=1−e−λxF(x) = 1 - e^{-\lambda x}.

The expected value of XX is:

E[X]=1λE[X] = \frac{1}{\lambda}

The probability that XX exceeds its expected value is given by the survival function evaluated at x=1λx = \frac{1}{\lambda}:

P(X>E[X])=P(X>1λ)=1−F(1λ)=e−λ(1/λ)=e−1P(X > E[X]) = P\left(X > \frac{1}{\lambda}\right) = 1 - F\left(\frac{1}{\lambda}\right) = e^{-\lambda (1/\lambda)} = e^{-1}

Computing the numerical value:

e−1=12.71828≈0.367879...≈0.37e^{-1} = \frac{1}{2.71828} \approx 0.367879... \approx 0.37

Rounded to 2 decimal places, the value is 0.37 (falling comfortably within the official range of 0.35 to 0.39).

Final answer

0.37

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question
28 of 65, printed as CS Q.18
Section
Computer Science and Information Technology

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