Answer to Question #85925 in Statistics and Probability for Umalaxmi

Question #85925
For the random variable X with density function f(x)={2e^(-2x) ; x≥0
{0 ;x<0.
Find (i)the mean μ and variance σ^(2);
(ii)P [|X-μ|≥1].
Use Chebyshev's inequality to obtain an upper bound on P[|X-μ|]≥1] and compare with the result obtained in part (ii).
1
Expert's answer
2019-03-08T08:15:33-0500
μ=Rxf(x)dx=02xe2xdx=0.5\mu=\int\limits_{\mathbb{R}} xf(x) \, dx=\int\limits_0^{\infty} 2xe^{-2x}\,dx=0.5


σ2=R(xμ)2f(x)dx=02(x0.5)2e2xdx=0.25\sigma^2=\int\limits_{\mathbb{R}} (x-\mu)^2 f(x) \, dx=\int\limits_0^{\infty} 2(x-0.5)^2 e^{-2x}\,dx=0.25


P(Xμ1=2σ)=x0.51f(x)dx=1.52e2xdx=e3P(|X-\mu| \geqslant 1=2 \sigma)=\int\limits_{|x-0.5|\geqslant 1} f(x) \, dx=\int\limits_{1.5}^{\infty} 2e^{-2x}\,dx=e^{-3}


P(Xμ1=2σ)122=0.25P(|X-\mu| \geqslant 1=2 \sigma) \leqslant \dfrac{1}{2^2}=0.25


e30.05<0.25e^{-3}\approx 0.05<0.25


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