Question

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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

a) The probability that the car battery fails within 1 year is computed here as:

P(T <1) = f (1/2)e=/2 dt = 1 –e-12 0.39347

The expected number of batteries that were replaced last year is:

= np

= 10*0.39347

= 3.9347

b) Probability that Bob had to replace 4 batteries in 2019 is

Here we binomial probability function :

10 P(X = 4) = 0.393474 347*(1 – 0.39347)* = 0.2506

0.2506 is the required probability.

c) Exponential distribution follows a memoryless property, that is it does not depend on the past events. Therefore we just have to find the probability here that 2 batteries are to be replaced in 1 year

P(X = 2) = (2) 0.39347'(1 – 0.39347)* = 0.1276

0.1276 is the required probability.