If $$Z$$ has the basic Weibull distribution with shape parameter $$k$$ then $$U = \exp\left(-Z^k\right)$$ has the standard uniform distribution. In the field of materials science, the shape parameter k of a distribution of strengths is known as the Weibull modulus. k â It will enhance any encyclopedic page you visit with the magic of the WIKI 2 technology. {\displaystyle N} But as we will see, every Weibull random variable can be obtained from a standard Weibull variable by a simple deterministic transformation, so the terminology is justified. Suppose again that $$X$$ has the Weibull distribution with shape parameter $$k \in (0, \infty)$$ and scale parameter $$b \in (0, \infty)$$. ( Suppose that $$Z$$ has the basic Weibull distribution with shape parameter $$k \in (0, \infty)$$. The probability density function $$g$$ is given by $g(t) = k t^{k - 1} \exp\left(-t^k\right), \quad t \in (0, \infty)$, These results follow from basic calculus. ) The Weibull distribution is the maximum entropy distribution for a non-negative real random variate with a fixed expected value of xk equal to Î»k and a fixed expected value of ln(xk) equal to ln(Î»k)Â âÂ  $$\E(X^n) = b^n \Gamma\left(1 + \frac{n}{k}\right)$$ for $$n \ge 0$$. k e  The Weibull plot is a plot of the empirical cumulative distribution function {\displaystyle \gamma } {\displaystyle x_{1}>x_{2}>\cdots >x_{N}} versus 0.4 It follows that $$U$$ has reliability function given by $\P(U \gt t) = \left\{\exp\left[-\left(\frac{t}{b}\right)^k\right]\right\}^n = \exp\left[-n \left(\frac{t}{b}\right)^k\right] = \exp\left[-\left(\frac{t}{b / n^{1/k}}\right)^k\right], \quad t \in [0, \infty)$ and so the result follows. Vary the shape parameter and note again the shape of the distribution and density functions. (  The shape parameter k is the same as above, while the scale parameter is , [ "article:topic", "showtoc:no", "Weibull distribution" ], $$\newcommand{\R}{\mathbb{R}}$$ $$\newcommand{\N}{\mathbb{N}}$$ $$\newcommand{\E}{\mathbb{E}}$$ $$\newcommand{\P}{\mathbb{P}}$$ $$\newcommand{\var}{\text{var}}$$ $$\newcommand{\sd}{\text{sd}}$$ $$\newcommand{\cov}{\text{cov}}$$ $$\newcommand{\cor}{\text{cor}}$$ $$\newcommand{\skw}{\text{skew}}$$ $$\newcommand{\kur}{\text{kurt}}$$. W If $$Y$$ has the Weibull distribution with shape parameter $$k$$ and scale parameter $$b$$ then $$Y / b$$ has the basic Weibull distribution with shape parameter $$k$$, and hence $$X = (Y / b)^k$$ has the standard exponential distributioon. Vary the parameters and note again the shape of the distribution and density functions. ) i x {\displaystyle f_{\rm {Frechet}}(x;k,\lambda )={\frac {k}{\lambda }}\left({\frac {x}{\lambda }}\right)^{-1-k}e^{-(x/\lambda )^{-k}}=-f_{\rm {Weibull}}(x;-k,\lambda ). It is a versatile distribution that can take on the characteristics of other types of distributions, based on the value of the shape parameter, [math] {\beta} \,\! If $$X$$ has the standard exponential distribution (parameter 1), then $$Y = b \, X^{1/k}$$ has the Weibull distribution with shape parameter $$k$$ and scale parameter $$b$$. â = }, f $$\E(X) = b \Gamma\left(1 + \frac{1}{k}\right)$$, $$\var(X) = b^2 \left[\Gamma\left(1 + \frac{2}{k}\right) - \Gamma^2\left(1 + \frac{1}{k}\right)\right]$$, The skewness of $$X$$ is $\skw(X) = \frac{\Gamma(1 + 3 / k) - 3 \Gamma(1 + 1 / k) \Gamma(1 + 2 / k) + 2 \Gamma^3(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^{3/2}}$, The kurtosis of $$X$$ is $\kur(X) = \frac{\Gamma(1 + 4 / k) - 4 \Gamma(1 + 1 / k) \Gamma(1 + 3 / k) + 6 \Gamma^2(1 + 1 / k) \Gamma(1 + 2 / k) - 3 \Gamma^4(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^2}$. , x In this section, we introduce the Weibull distributions, which are very useful in the field of actuarial science. x ) ^ {\displaystyle \lambda } F is the rank of the data point and For example, each of the following gives an application of the Weibull distribution. Approximate the mean and standard deviation of $$T$$. This follows trivially from the CDF $$F$$ given above, since $$F^c = 1 - F$$. is the number of data points.. ^ The axes are If $$Z$$ has the basic Weibull distribution with shape parameter $$k$$ then $$G(Z)$$ has the standard uniform distribution. k parameter given The characteristic function has also been obtained by Muraleedharan et al. only implicitly, one must generally solve for The Weibull distribution is related to a number of other probability distributions; in particular, it interpolates between the exponential distribution (k = 1) and the Rayleigh distribution (k = 2 and â 80 {\displaystyle \lambda } The reason for this change of variables is the cumulative distribution function can be linearized: which can be seen to be in the standard form of a straight line.