Question 29.2: A random variable, z, has a p.d.f. f (z) where f (z) = e^−z ......
A random variable, z, has a p.d.f. f (z) where
f(z)=\mathrm{e}^{-z} 0 ≤ z <∞
Calculate the probability that
(a) 0 ≤ z ≤ 2
(b) z is more than 1
(c) z is less than 0.5
Step-by-Step
Note that \textstyle{\int_{0}^{\infty}\mathrm{e}^{-z}\,\mathrm{d}z=1} so that f(z)=\mathrm{e}^{-z} is suitable as a p.d.f.
(a) P(0\leqslant z\leqslant2)={\int_{0}^{2}\mathrm{e}^{-z}\mathrm{d}z}=[-\mathrm{e}^{-z}]_{0}^{2}=0.865
(b) P(z\gt 1)=\int_{1}^{\infty}\mathrm{e}^{-z}\,\mathrm{d}z=\left[-\mathrm{e}^{-z}\right]_{1}^{\infty}=0.368
(c) P(z\lt 0.5)=[-\mathrm{e}^{-z}]_{0}^{0.5}=0.393
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