If the life X (in years) of a certain type of car has a Weibull distribution with the
parameter = 2, and then find the value of the parameter , given the probability that
the life of the car exceeds 5 years is e0:25. For these values of and , find the mean
and variance of X.
Answers
Answer:
Step-by-step explanation:
Definition 1: The Weibull distribution has the probability density function (pdf)
Weibull distribution pdf
for x ≥ 0. Here β > 0 is the shape parameter and α > 0 is the scale parameter.
The cumulative distribution function (cdf) is
Weibull distribution function
The inverse cumulative distribution function is I(p) =
image9188
Observation: There is also a three-parameter version of the Weibull distribution. Click here for more information about this version.
Observation: If x represents “time-to-failure”, the Weibull distribution is characterized by the fact that the failure rate is proportional to a power of time, namely β – 1. Thus β can be interpreted as follows:
A value of β < 1 indicates that the failure rate decreases over time. This happens if there is significant “infant mortality”, or where defective items fail early with a failure rate decreasing over time as the defective items are weeded out of the population.
A value of β = 1 indicates that the failure rate is constant over time. This might suggest random external events are causing mortality or failure.
A value of β > 1 indicates that the failure rate increases with time. This happens if there is an “aging” process; e.g. if parts are more likely to wear out and/or fail as time goes on.
1/α can be viewed as the failure rate. The mean of the Weibull distribution is the mean time to failure (MTTF) or mean time between failures (MTBF) = \alpha \Gamma \! \left( \! 1+\frac{1}{\beta} \! \right).
Key statistical properties of the Weibull distribution are:
Mean = \alpha \Gamma \! \left( \! 1+\frac{1}{\beta} \! \right)
Median = \alpha ({ln 2})^{1/\beta}
Mode (when β > 1) = \alpha \! \left( \! \frac{\beta-1}{\beta} \! \right)^{1/\beta}
Variance = \alpha^2 \Gamma \! \left( \! 1+\frac{2}{\beta} \! \right)- \mu^2
Step-by-step explanation:
see the working in the picture
