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Davis and Lawrance (1989) present time-to-failure data for 171 automobile tires. Baltazar-Aban and Pena (1995) modeled these data with an exponential distribution with a failure rate of 0.004 tire per hour. a. Find the probability that a randomly selected tire fails during the first 100 hours of testing. b. Find the probability that a randomly selected tire fails between 50 and 150 hours of testing. c. Find the probability that a randomly selected tire survives more than 200 hours. d. Find the expected time to failure for a randomly selected tire. e. Find the variance and the standard deviation for the time to failure for a randomly selected tire.

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4 votes
This guy is correct
answered
User Saket Mehta
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2 votes

Answer:

So, to summarize:

a. Probability of failing during the first 100 hours ≈ 0.3220

b. Probability of failing between 50 and 150 hours ≈ 0.1318

c. Probability of surviving more than 200 hours ≈ 0.6780

d. Expected time to failure ≈ 250 hours

e. Variance ≈ 62500 hours^2, Standard Deviation ≈ 250 hours

Explanation:

To solve these problems, we'll use the exponential distribution with the given failure rate of 0.004 tire per hour. The probability density function (PDF) for the exponential distribution is given by:

f(x) = λ * e^(-λx)

Where:

λ (lambda) is the failure rate (0.004 tire per hour).

x is the time to failure (in hours).

e is the base of the natural logarithm, approximately 2.71828.

Let's address each part of the problem:

a. Probability of failing during the first 100 hours (0 ≤ x ≤ 100):

To find this probability, we need to calculate the integral of the PDF from 0 to 100:

P(0 ≤ x ≤ 100) = ∫[0 to 100] λ * e^(-λx) dx

P(0 ≤ x ≤ 100) = λ * ∫[0 to 100] e^(-λx) dx

P(0 ≤ x ≤ 100) = λ * [-e^(-λx)] [from 0 to 100]

P(0 ≤ x ≤ 100) = λ * [-e^(-100λ) + 1]

P(0 ≤ x ≤ 100) = 0.004 * [-e^(-0.4) + 1]

P(0 ≤ x ≤ 100) ≈ 0.3220

b. Probability of failing between 50 and 150 hours (50 ≤ x ≤ 150):

Similarly, we calculate the integral of the PDF from 50 to 150:

P(50 ≤ x ≤ 150) = ∫[50 to 150] λ * e^(-λx) dx

P(50 ≤ x ≤ 150) = λ * ∫[50 to 150] e^(-λx) dx

P(50 ≤ x ≤ 150) = λ * [-e^(-λx)] [from 50 to 150]

P(50 ≤ x ≤ 150) = λ * [-e^(-150λ) + e^(-50λ)]

P(50 ≤ x ≤ 150) = 0.004 * [-e^(-0.6) + e^(-0.2)]

P(50 ≤ x ≤ 150) ≈ 0.1318

c. Probability of surviving more than 200 hours (x > 200):

This is the complement of the probability of failing in the first 200 hours:

P(x > 200) = 1 - [P(0 ≤ x ≤ 200)]

P(x > 200) = 1 - [0.3220]

P(x > 200) ≈ 0.6780

d. Expected time to failure (mean or average):

The expected time to failure for a random tire is given by the inverse of the failure rate (lambda):

Expected Time to Failure = 1 / λ

Expected Time to Failure = 1 / 0.004

Expected Time to Failure = 250 hours

and. Variance and Standard Deviation:

The variance (σ^2) of the exponential distribution is given by:

Variance (σ^2) = 1 / λ^2

Variance (p^2)

Variance (σ^2) = 62500

The standard deviation (σ) is the square root of the variance:

Standard Deviation (σ) = √(62500)

Standard Deviation (σ) = 250 hours

answered
User RobinBrouwer
by
7.5k points

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