[1]
\(X \sim N(\mu,\sigma^2)\)の製品を1つ検査した時に良品である確率は、
\(P(S_L \leq X \leq S_U)\)\(\displaystyle = \int ^{\frac{S_U-\mu}{\sigma}}_{\frac{S_L-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz\)
不良率を最小にするには、これを最大にすれば良い。
\(\displaystyle \frac{d}{d\mu}P(S_L \leq X \leq S_U)\)\(\displaystyle =\frac{1}{\sqrt{2\pi}}e^{-\left(\frac{S_U-\mu}{\sigma}\right)^2/2} \left(-\frac{1}{\sigma}\right)\)\(\displaystyle -\frac{1}{\sqrt{2\pi}}e^{-\left(\frac{S_L-\mu}{\sigma}\right)^2/2} \left(-\frac{1}{\sigma}\right)\)\(=0\)として、
\((S_U-\mu)^2=(S_L-\mu)^2\)
より、
\(\displaystyle \mu=\frac{S_U+S_L}{2}\)
[2]
\(\displaystyle\left \{\begin{array}\displaystyle C_p=\displaystyle \frac{S_U-S_L}{6\sigma}=\frac{2}{3} \\ \displaystyle \mu=\frac{S_U+S_L}{2} \end{array}\right.\)
を解くと、
\(\displaystyle\left \{\begin{array} \displaystyle S_U = \mu+2\sigma \\ S_L=\mu -2\sigma \end{array}\right.\)
なので、良品率は
\(P(S_L \leq X \leq S_U)\)\(\displaystyle = \int ^2_{-2} \frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz\)
\(=1-2\times 0.0228\)(標準正規分布表の利用)
よって、不良品率は\(0.0456\)
[3]
\(\displaystyle \frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)\)で、(1)(2)より、
\(\displaystyle \frac{S}{\sigma}=\frac{C_p}{\hat{C}_p}\)なので、
\(\displaystyle (n-1)\frac{C_p^2}{\hat{C}_p^2}\sim \chi^2(n-1)\)なので、95%信頼区間は与式の通り得られる。
[4]
\(\displaystyle \hat{C}_p=\frac{12.6-12.0}{6\times 0.05}=2.0\)
[3]で\(n=20\)として信頼区間を求めると、\((1.36,2.62)\)となる。
95%信頼区間が1.33以上なので、管理状態は良いと解釈される。
[5]
\(\displaystyle E[\hat{C}_p]=E\left[\frac{\sigma C_p}{S}\right]\)\(\displaystyle =\sigma C_pE\left[\frac{1}{S}\right]\)\(\displaystyle =\sigma C_pE[T]\)とすると、
\(a=\sigma E[T]\)
\(\displaystyle Y=\frac{(n-1)S^2}{\sigma^2}=\frac{n-1}{\sigma^2T^2}\)\(\sim \chi^2(n-1)\)と\(Y\)を\(T\)に変数変換を行うと、\(T\)の確率密度関数\(f(t)\)は、
\(\displaystyle f(t) =\frac{1}{2\Gamma\left(\frac{n-1}{2}\right)}\left(\frac{n-1}{2\sigma^2t^2}\right)^{\frac{n-1}{2}-1}e^{-\frac{n-1}{2\sigma^2t^2}}\)\(\displaystyle \times \frac{2(n-1)}{\sigma^2 t^3}\)
\(\displaystyle =\frac{2}{\Gamma\left(\frac{n-1}{2}\right)t}\left(\frac{n-1}{2\sigma^2t^2}\right)^{\frac{n-1}{2}}e^{-\frac{n-1}{2\sigma^2t^2}}\)
\(a\)\(\displaystyle = \sigma\int^\infty_0 tf(t) dt\)
\(\displaystyle =\sigma\int^\infty_0 \frac{2}{\Gamma\left(\frac{n-1}{2}\right)}\left(\frac{n-1}{2\sigma^2t^2}\right)^{\frac{n-1}{2}}e^{-\frac{n-1}{2\sigma^2t^2}}dt\)
ここで、\(\displaystyle \frac{n-1}{2\sigma^2t^2}=u\)と変数変換を行うと、
\(\displaystyle =\sigma\int^\infty_0 \frac{2}{\Gamma\left(\frac{n-1}{2}\right)}u^{\frac{n-1}{2}}e^{-u}\frac{\sqrt{n-1}}{2\sqrt{2}\sigma u^{3/2}}du\)
\(\displaystyle =\sqrt{\frac{n-1}{2}}\frac{1}{\Gamma\left(\frac{n-1}{2}\right)}\int^\infty_0 u^{\frac{n-4}{2}}e^{-u}du\)
\(\displaystyle =\sqrt{\frac{n-1}{2}}\frac{\Gamma\left(\frac{n-2}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}\)
別解
\(\displaystyle (n-1)\frac{C_p^2}{\hat{C}_p^2} \sim \chi^2(n-1)\)なので、\(\displaystyle (n-1)\frac{C_p^2}{\hat{C}_p^2}=Y\)と置くと、\(\displaystyle \hat{C}_p =\sqrt{\frac{n-1}{Y}}C_p\)となり、\(\displaystyle E[\hat{C}_p] =C_p \sqrt{n-1}E\left[\sqrt{\frac{1}{Y}}\right]\)
\(\displaystyle E\left[\sqrt{\frac{1}{Y}}\right]\)\(\displaystyle = \int^\infty _0 \sqrt{\frac{1}{Y}}f (y)dy\)
\(\displaystyle = \int^\infty _0 y^{-1/2}\frac{1}{2\Gamma \left(\frac{n-1}{2}\right)}\left(\frac{y}{2}\right)^{\frac{n-1}{2}-1}e^{-\frac{y}{2}}dy\)
\(\displaystyle = \frac{1}{\sqrt{2}}\int^\infty _0 \frac{1}{2\Gamma \left(\frac{n-1}{2}\right)}\left(\frac{y}{2}\right)^{\frac{n-2}{2}-1}e^{-\frac{y}{2}}dy\)
\(\displaystyle =\frac{\Gamma \left(\frac{n-2}{2}\right)}{\sqrt{2}\Gamma \left(\frac{n-1}{2}\right)} \int^\infty _0 \frac{1}{2\Gamma \left(\frac{n-2}{2}\right)}\left(\frac{y}{2}\right)^{\frac{n-2}{2}-1}e^{-\frac{y}{2}}dy\)
\(\displaystyle =\frac{\Gamma \left(\frac{n-2}{2}\right)}{\sqrt{2}\Gamma \left(\frac{n-1}{2}\right)} \)
よって、
\(\displaystyle E[\hat{C}_p] =\sqrt{\frac{n-1}{2}}\frac{\Gamma\left(\frac{n-2}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}C_p\)
コメント