Question #129652

A continuous r.U.X has the following probability density function
F(X) = 20e to the power -20x ; X>0
= 0 ; 0 otherwise
Find i) Mean of X, ii) P(X < 1000), iii) P(X>3000)

Expert's answer

X∈Γ(λ) (exponential random variable).We have λ=20. So (i) mean of X M(X)=1λ=120.(ii) First we find the distribution function of X F(x).F(x)=∫−∞xf(t)dt∫0x20e−20tdt=(20⋅(−120)e−20t)0x=1−e−20xF(x)=1−e−20x, x>0F(x)=0, otherwise.P(X<1000)=F(1000)=1−e−20⋅1000≈1.(iii)P(X>3000)=1−P(X≤3000)=1−(1−e20⋅3000)≈0.X\in \Gamma(\lambda)\text{ (exponential random variable)}.\\ \text{We have } \lambda=20.\text{ So (i) mean of } X\ M(X)=\frac{1}{\lambda}=\frac{1}{20}.\\ (ii)\text{ First we find the distribution function of } X\ F(x).\\ F(x)=\int_{-\infty}^xf(t)dt\\ \int_0^x20e^{-20t}dt=(20\cdot (-\frac{1}{20})e^{-20t})_0^x=1-e^{-20x}\\ F(x)=1-e^{-20x},\ x>0\\ F(x)=0,\text{ otherwise}.\\ P(X<1000)=F(1000)=1-e^{-20\cdot 1000}\approx 1.\\ (iii) P(X>3000)=1-P(X\leq 3000)=1-(1-e^{20\cdot 3000})\approx 0.


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