Answer:
Step-by-step explanation:
Let X denote the life span of a car battery and it follows and exponential distribution with average of 6 years.
Thus , the parameter of the exponential distribution is calculated as,
μ = 6
[tex]\frac{1}{\lambda} =6[/tex]
[tex]\lambda = \frac{1}{6}[/tex]
a) The required probability is
[tex]P(X>4)=1-P(X\leq 4)\\\\=1-F(4)\\\\1-(1-e^{- \lambda x})\\\\=e^{-\frac{4}{6}[/tex]
= 0.513
Hence, the probability that a randomly selected car battery will last more than four years is 0.513
b) The variance of the battery span is calculated as
[tex]\sigma ^2=\frac{1}{(\frac{1}{\lambda})^2 }\\\\\sigma ^2=\frac{1}{(\frac{1}{6})^2 } \\\\=6^2=36[/tex]
The 95% percentile [tex]x_{a=0.05}[/tex] (α = 5%) of the battery span is calculated
[tex]x_{0.05}=-\frac{log(\alpha) }{\lambda} \\\\=-\frac{log(0.05)}{1/6} \\\\=-6log(0.05)\\\\=17.97 \ years[/tex]
c)
Let [tex]X_r[/tex] denote the remaining life time of a car battery
i)the probability the battery will last an additional five years is calculated below
[tex]P(X_r>5)=e^{-5\lambda}\\\\=e^{-\frac{5}{6} }\\\\=0.4346[/tex]
ii) The average time that the battery is expected to last is calculated
[tex]E(X_r)=\frac{1}{\lambda} \\\\=6[/tex]
