Question

For this question, answers should be decimal with 4 digits to right of the decimal point. ex. 0.1234 Bob owns a car rental company that has a fleet of 10 cars. Each car battery has an Exponentially distributed lifespan with a mean of 4 years. At the beginning of last year (Jan. 1, 2019), Bob replaced all the batteries. a) (5 points) What was the expected number of Batteries that Bob had to replace last year? b) (5 points) What was the probability that Bob had to replace 4 Batteries in 2019? c) (5 points) Bob replaced 1 battery in July and 3 batteries in September. It is now (Jan 1.2020) and all the batteries are still working. What is the probability that he will have to replace at least 2 batteries in the coming year?

Answer 1

As the lifespan of each battery is exponentially distribution with mean = 4 years, therefore the distribution of lifetime here is modelled as:

$T \sim exp(1/4)$ as parameter for exponential distribution is reciprocal of its mean.

a) The expected number of batteries that are to be replaced in a year here is computed as:

= 10*P(T < 1)

$= 10*(\int_{0}^{1}0.25e^{-0.25t} \ dt) = 10*(1 - e^{0.25*1}) = 2.2120$

Therefore 2.2120 is the expected number of batteries here.

b) The probability that 4 batteries would be replaced is computed here using binomial probability function as:

$= \binom{10}{4}0.2212^4(1 - 0.2212)^6 = 0.1122$

therefore 0.1122 is the required probability here.

c) Exponential distribution follows a memoryless property, therefore it does not matter what replacements happened in the last year, the probability that there will be at least 2 batteries replaced in the coming year is computed here as:

= 1 - Probability that 0 or 1 battery is replaced

= 1 - (1 - 0.2212)¹⁰ - 10*0.2212*(1 - 0.2212)⁹

= 0.6848

Therefore 0.6848 is the required probability here.