Question 3.10: The lifetime of an automobile battery is described by a r.v.......
The lifetime of an automobile battery is described by a r.v. X having the Negative Exponential distribution with parameter \lambda={\frac{1}{3}} . Then:
(i) Determine the expected lifetime of the battery and the variation around this mean.
(ii) Calculate the probability that the lifetime will be between 2 and 4 time units.
(iii) If the battery has lasted for 3 time units, what is the (conditional) probability that it will last for at least an additional time unit?
(i) Since ~E X={\frac{1}{\lambda}}~and V\!\!ar(X)={\textstyle\frac{1}{\lambda^{2}}},we have here: E X=3,\ V\!\!ar(X)=9, and s.d.(X)=3.
(ii) Since, by (35),
F(x)=1-e^{-\lambda x},\qquad x\gt 0,\quad{\mathrm{so~that}}\,P(X\gt x)=e^{-\lambda x},\qquad x\gt 0.\quad(35)
F(x)=1-e^{-\frac x3}\mathrm{~for~}x\gt 0, we have P(2\lt X\lt 4)=P(2\lt {X\leq4})={P(X\leq4)}-{P(X\leq2)}={F(4)}-F(2)=(1-e^{-\frac{4}{3}})-{(1-e^{-\frac{2}{3}})}= e^{-{\frac{2}{3}}}-e^{-{\frac{4}{3}}}\simeq0.252.
(iii) The required probability is: P(X\gt 4\mid X\gt 3)=P(X\gt 1), by the memoryless property of this distribution, and P(X\ \gt 1)=1-P(X\leq1)= 1-F(1)=e^{-{\frac{1}{3}}}\simeq0.716.