The lifetime mileage of a car can be modeled as a Weibull variable with parameter λ = 1 per 150000 miles and parameter a=1.25.a) What is the probability that a car will last between 100000 and 150000 miles?b) A dealer sold 10 cars on a day with a guarantee that any car lasting less than 100000miles will be replaced by another car with a 30% discount. What is the probabilitythat the dealer has to provide the discount on 2 or fewer of these cars?

Question
Asked Jul 18, 2019

The lifetime mileage of a car can be modeled as a Weibull variable with parameter λ = 1 per 150000 miles and parameter a=1.25.

a) What is the probability that a car will last between 100000 and 150000 miles?
b) A dealer sold 10 cars on a day with a guarantee that any car lasting less than 100000
miles will be replaced by another car with a 30% discount. What is the probability
that the dealer has to provide the discount on 2 or fewer of these cars?
check_circleExpert Solution
Step 1

It is given that the lifetime mileage of a car can be modelled as a Weibull variable with parameter λ=1 per150,000 miles and parameter a=1.25.

The cdf of a Weibull distribution with λ = 1 per 150,000 miles and a = 1.25 is given as follows:

1-exp
2
x 20
F(x)
| 0
Here,
1.25
1 -exp
F(x)
x 2 0
150,000
[0
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1-exp 2 x 20 F(x) | 0 Here, 1.25 1 -exp F(x) x 2 0 150,000 [0

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Step 2

a.

Define the random variable X as the lasting time of a car.

The probability that a car will last between 100,000 and 150,000 miles is obtained as shown below:

P(100,000< X <150,000)- P(X 150,000) P(X 100,000)
= F (150,000)- F(100,000)
1.25
100, 000
150, 000
1-exp
1-еxp
150, 000
150,000
-1-e-1+e
-0.6024
=0.1796
help_outline

Image Transcriptionclose

P(100,000< X <150,000)- P(X 150,000) P(X 100,000) = F (150,000)- F(100,000) 1.25 100, 000 150, 000 1-exp 1-еxp 150, 000 150,000 -1-e-1+e -0.6024 =0.1796

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Step 3

b.

It is given that a dealer sold 10 cars on a day with a guarantee that any car lasting less than 1000,000 miles will be replaced by anothe...

P(X <100,000) P(X<100,000)
=F(100,000)
1.25
100,000
-1-exp
150,000
=1-e 06024
=0.4525
help_outline

Image Transcriptionclose

P(X <100,000) P(X<100,000) =F(100,000) 1.25 100,000 -1-exp 150,000 =1-e 06024 =0.4525

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