问题及解答

A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength(延展强度) $X$ of a material has a lognormal distribution.

[基於某種材料失效機制的理論論證是以假定材料的延展強度 $X$ 具有對數正態分佈為基礎。]

Suppose the parameters are $\mu=5$ and $\sigma=.1$.

• (a) Compute $E(X)$ and $V(X)$.
• (b) Compute $P(X > 120)$.
• (c) Compute $P(110\leqslant X\leqslant 130)$.
• (d) What is the value of median ductile strength?
• (e) If ten different samples of an alloy steel(合金鋼) of this type were subjected to a strength test, how many would you expect to have strength at least $120$?
• (f) If the smallest $5\%$ of strength values were unacceptable, what would the minimum acceptable strength be?
   

(a)

\[
E(X)=e^{\mu+\frac{\sigma^2}{2}}=e^{5+\frac{0.1^2}{2}}=e^{5.005}\approx 149.15708315
\]

\[
\begin{split}
V(X)&=e^{2\mu+\sigma^2}\cdot(e^{\sigma^2}-1)=e^{2\cdot 5+0.1^2}\cdot(e^{0.1^2}-1)\\
&=e^{10.01}\cdot(e^{0.01}-1)\\
&\approx 22247.83545633\times(1.01005017-1)\\
&=223.5945284681440761
\end{split}
\]

(b)

\[
\begin{split}
P(X > 120)&=1-P(X\leqslant 120)=1-F(120;\mu,\sigma)\\
&=1-\Phi(\frac{\ln(120)-\mu}{\sigma})\\
&\approx 1-\Phi(\frac{4.78749174-5}{0.1})\\
&=1-\Phi(-2.1250826)\\
&\approx 1-\Phi(-2.13)\\
&=1-0.0166\\
&=0.9834
\end{split}
\]

(c)

\[
\begin{split}
P(110\leqslant X\leqslant 130)&=P(X\leqslant 130)-P(X\leqslant 110)\\
&=F(130;\mu,\sigma)-F(110;\mu,\sigma)\\
&=\Phi(\frac{\ln(130)-\mu}{\sigma})-\Phi(\frac{\ln(110)-\mu}{\sigma})\\
&\approx\Phi(\frac{4.86753445-5}{0.1})-\Phi(\frac{4.70048037-5}{0.1})\\
&=\Phi(-1.3246555)-\Phi(-2.9951963)\\
&\approx\Phi(-1.32)-\Phi(-3.00)\\
&=0.0934-0.0013\\
&=0.0921
\end{split}
\]

(d)

Suppose $x$ satisfies $P(X\leqslant x)=\frac{1}{2}$, then

\[
\begin{split}
0.5=P(X\leqslant x)&=F(x;\mu,\sigma)=\Phi(\frac{\ln x-\mu}{\sigma})\\
&=\Phi(\frac{\ln x-5}{0.1})
\end{split}
\]

It infers that $\frac{\ln x-5}{0.1}=0$, so $\ln x=5$. We have

\[
x=e^5\approx 148.41315910
\]

(e)

We have calculated the probability $P(X > 120)$ in (b). Thus $P(X\geqslant 120)=P(X > 120)=0.9834$.

Therefore, we expect there are $0.9834\time 10=9.834$ different samples of an alloy steel of this type have strength at least $120$.

(f)

Suppose the minimum acceptable strength is $x_0$, then

\[
P(X\leqslant x_0)=5\%=0.05
\]

That is

\[
\begin{split}
0.05&=P(X\leqslant x_0)=F(x_0;\mu,\sigma)=\Phi(\frac{\ln(x_0)-\mu}{\sigma})\\
&=\Phi(\frac{\ln(x_0)-5}{0.1})
\end{split}
\]

By Table A.3, we have $\Phi(-1.64)=0.0505$, and $\Phi(-1.65)=0.0495$. So it is reasonable to let $\Phi(-1.645)=0.05$. Then we get

\[
\begin{split}
&\frac{\ln(x_0)-5}{0.1}=-1.645\\
\Rightarrow&\ln(x_0)=5+0.1\times(-1.645)=4.8355\\
\Rightarrow& x_0=e^{4.8355}\\
\Rightarrow& x_0\approx 125.90151823
\end{split}
\]

So the minimum acceptable strength is $125.90151823$.