问题及解答

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985:519--522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials(固体绝缘材料) subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens(样本) subjected to AC voltage(交流电压). The values of the parameters depend on the voltage and temperature; suppose $\alpha=2.5$ and $\beta=200$ (values suggested by data in the article).

• (a) What is the probability that a specimen's lifetime is at most $200$? Less than $200$? More than $300$?
• (b) What is the probability that a specimen's lifetime is between $100$ and $200$?
• (c) What value is such that exactly $50\%$ of all specimens have lifetimes exceeding the value?
   

The pdf of the rv $X$ which obeys the Weibull distribution is

\[
f(x;\alpha,\beta)=\begin{cases}
\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0.
\end{cases}
\]

The cdf is 

\[
F(x;\alpha,\beta)=\begin{cases}
0, & x < 0,\\
1-e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0.
\end{cases}
\]

(a)

The probability that a specimen's lifetime is at most $200$ is

\[
\begin{split}
P(X\leqslant 200)&=F(200;\alpha,\beta)=F(200;2.5,200)\\
&=1-e^{-(\frac{200}{200})^{2.5}}\\
&=1-e^{-1}\\
&\approx 0.63212056
\end{split}
\]

The probability that a specimen's lifetime is less than $200$ is equal to the probability that a specimen's lifetime is at most $200$.

\[
P(X < 200)=P(X\leqslant 200)\approx 0.63212056
\]

The probability that a specimen's lifetime is more than $300$ is

\[
\begin{split}
P(X > 300)&=1-P(X\leqslant 300)=1-(1-e^{-(\frac{300}{200})^{2.5}})\\
&=e^{-1.5^{2.5}}\\
&\approx e^{-2.75567596}\\
&\approx 0.06356604
\end{split}
\]

(b)

The probability that a specimen's lifetime is between $100$ and $200$ is

\[
\begin{split}
P(100\leqslant X\leqslant 200)&=F(200;\alpha,\beta)-F(100;\alpha,\beta)\\
&=F(200;2.5;200)-F(100;2.5,200)\\
&\approx 0.63212056-(1-e^{-(\frac{100}{200})^{2.5}})\\
&=0.63212056-1+e^{-0.5^{2.5}}\\
&\approx -0.36787944+e^{-0.17677670}\\
&\approx -0.36787944+0.83796689\\
&=0.47008745
\end{split}
\]

(c)

Suppose the value is $x$ that exactly $50\%$ of all specimens have lifetimes exceeding that value.

Then

\[P(X > x)=\frac{1}{2}\]

which infers that

\[
1-P(X\leqslant x)=\frac{1}{2}\quad\Rightarrow\quad P(X\leqslant x)=\frac{1}{2}
\]

That is,

\[
\begin{split}
\Rightarrow&F(x;\alpha,\beta)=\frac{1}{2}\\
\Rightarrow&1-e^{-(\frac{x}{\beta})^{\alpha}}=\frac{1}{2}\\
\Rightarrow&e^{-(\frac{x}{\beta})^{\alpha}}=\frac{1}{2}\\
\Rightarrow&-(\frac{x}{\beta})^{\alpha}=\ln\frac{1}{2}=-\ln 2\\
\Rightarrow&(\frac{x}{\beta})^{\alpha}=\ln 2\\
\Rightarrow&(\frac{x}{200})^{2.5}=\ln 2\\
\Rightarrow&\frac{x}{200}=(\ln 2)^{\frac{2}{5}}\\
\Rightarrow& x=200\times(\ln 2)^{\frac{2}{5}}\\
\Rightarrow& x\approx 200\times((0.69314717)^2)^{\frac{1}{5}}\\
\Rightarrow& x=200\times 0.4804529992790089^{\frac{1}{5}}\\
\Rightarrow& x=200\times 0.86363490\\
\Rightarrow& x=172.72698
\end{split}
\]