GATE Mechanical Engineering (ME) 2018 Shift 2 Solved Paper

Question : 49 of 65

Marks: +1, -0

Let

x_{1}

and

x_{2}

be two independent exponentially distributed random variables with means

0.5

and

0.25

, respectively. Then

Y=\min (X_{1}, x_{2})

Solution:

Mean

(x_{1}) \;\; =0.5

\;{1}/{λ_{1}} \;\; =0.5

λ_{1} \;\; =\;{1}/{0.5}=2

Mean

(x_{2}) \;\; =0.25

\;{1}/{λ_{2}} \;\; =0.25

λ_{2} \;\; =\;{1}/{0.25}=4

y \;\; =\mean(x_{1}, x_{2})

Mean

(y)=\;{1}/{λ_{1}+λ_{2}}=\;{1}/{2+4}=\;{1}/{6}

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