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 − 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 λk and a fixed expected value of ln(xk) equal to ln(λk) −  \(\E(X^n) = b^n \Gamma\left(1 + \frac{n}{k}\right)\) for \(n \ge 0\). k e [11] 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. ( [4][5] 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 \(b\). − = }, 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.[12]. ^ 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 − 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.