# equation_確率分布

$$C_{ p } = \displaystyle\frac{ S_{ U } – S_{ L } }{ 6s }$$

$$k = \displaystyle\frac{ | ( S_{ U } + S_{ L } )/2 – \bar{ X } | }{ ( S_{ U } – S_{ L } )/2 }$$

$$k < 1 ならば C_{ pk } = ( 1 – k ) \times C_{ p }$$

$$k \geq 1 ならば C_{ pk } = 0$$

$$C_{ pk } = \displaystyle\min ( \displaystyle\frac{ S_{ U } – \bar{ X }}{ 3s} , \displaystyle\frac{ \bar{ X } – S_{ L } }{ 3s } )$$

$$C_{ p } = \displaystyle\frac{ S_{ U } – \bar{ X } }{ 3s }$$

$$C_{ p } = \displaystyle\frac{ \bar{ X } – S_{ L } }{ 3s }$$

$$y = f(x)$$　　　$$y = f(x)$$

$$f(x) \geq 0$$

$$Pr( a \lt x \leq b ) = \displaystyle \int_a^b f(x) dx$$

$$( a , b ] （ a \lt x \leq b ）$$　　　$$Pr( a \lt x \leq b )$$

$$f(x)$$　　　$$f(x)$$

$$Pr( – \infty \lt x \leq \infty ) = \displaystyle \int_{-\infty}^{ \infty } f(x) dx$$

$$Pr( – \infty \lt x \leq \infty ) = \displaystyle \int_{-\infty}^{ \infty } f(x) dx = 1$$

$$E(X) = \mu$$　　　$$V(X)$$　　　$$D(X)$$　　　$$\sigma$$

$$f(x)$$　　　$$f(x)$$

$$E(X) = \displaystyle \int_{}^{} x f(x) dx$$

$$V(X) = \displaystyle \int_{}^{} ( x – \mu )^2 f(x) dx$$

$$\bar{ X }$$

$$E( aX ) = a E(X)$$

$$E( a_{ 1 } X_{ 1 } + a_{ 2 } X_{ 2 } + \cdots + a_{ n } X_{ n } ) = a_{ 1 }E(X_{ 1 }) + a_{ 2 }E(X_{ 2 }) + \cdots + a_{ n }E(X_{ n })$$

$$E( X_{ 1 } + X_{ 2 } ) = E(X_{ 1 }) + E(X_{ 2 })$$

$$E( X_{ 1 } – X_{ 2 } ) = E(X_{ 1 }) – E(X_{ 2 })$$

$$V( aX ) = a^2 V(X)$$

$$V( a_{ 1 } X_{ 1 } + a_{ 2 } X_{ 2 } + \cdots + a_{ n } X_{ n } ) = a_{ 1 }^2 V(X_{ 1 }) + a_{ 2 }^2 V(X_{ 2 }) + \cdots + a_{ n }^2 V(X_{ n })$$

$$V( X_{ 1 } + X_{ 2 } ) = V(X_{ 1 }) + V(X_{ 2 })$$

$$V( X_{ 1 } – X_{ 2 } ) = V(X_{ 1 }) + V(X_{ 2 })$$

$$V( X_{ 1 } + X_{ 2 } ) = V(X_{ 1 }) + V(X_{ 2 }) + 2 Cov( X_{ 1 } , X_{ 2 })$$

$$V( X_{ 1 } – X_{ 2 } ) = V(X_{ 1 }) + V(X_{ 2 }) – 2 Cov( X_{ 1 } , X_{ 2 })$$

$$2 Cov( X_{ 1 } , X_{ 2 }) = E[ \{ X_{ 1 } – E( X_{ 1 } ) \} \{ X_{ 2 } – E( X_{ 2 } ) ]$$

$$Cov( X_{ 1 } , X_{ 2 }) = E[ \{ X_{ 1 } – E( X_{ 1 } ) \} \{ X_{ 2 } – E( X_{ 2 } ) ]$$

$$f(x) = \displaystyle\frac{ 1 }{ \sqrt{ 2 \pi \sigma } } \exp \left\{ – \frac{ 1 }{ 2 } \left( \displaystyle\frac{ x – \mu }{ \sigma } \right)^2 \right\}$$

$$N( \mu , \sigma^2 )$$　　　$$E(X) = \mu$$　　　$$D(X) = \sigma ( V(X) = \sigma^2 )$$

$$\mu = 0$$　　　$$\sigma^2 = 1$$　　　$$N( 0 , 1^2 )$$

$$Pr( X = x )$$　　　　$$Pr( X = x ) = {}_n \mathrm{ C }_x P^{ x } ( 1 – P)^{ n – x } = \displaystyle\frac{ n! }{ x! ( n – x )! } P^{ x } ( 1 – P )^{ n – x }$$

$${}_n \mathrm{ C }_x$$　　　$${}_n \mathrm{ C }_x = \displaystyle\frac{ n! }{ x! ( n – x )! }$$

$$E( X ) = nP$$　　 　$$D( X ) = \sqrt{ n P ( 1 – P ) }$$

$$p = X/n$$　　　$$E( p ) = P$$　　　$$D( p ) = \sqrt{ \displaystyle\frac{ P ( 1 -P ) }{ n } }$$

$$nP \geq 5 , n ( 1-P ) \geq 5$$

$$np = \lambda$$

$$Pr( X = k ) = \displaystyle\frac{ e^{ – \lambda } \lambda ^k }{ k! } ( k = 0, 1, 2, \cdots )$$

$$E( k ) = \lambda$$　　　$$D( k ) = \sqrt{ \lambda }$$

$$Pr( u \geq K_{ p } ) = P$$

$$U = \displaystyle\frac{ X – \mu }{ \sigma }$$

$$N( 20, 4^2 )$$

$$U = \displaystyle\frac{ 24 – 20 }{ 4 } = 1$$

$$N( 100, 30^2 )$$

$$U = \displaystyle\frac{ 53.5 – 100 }{ 30 } = 1.55$$

$$f(t) = \displaystyle\frac{ \Gamma ( \frac{ f + 1 }{ 2 }) }{ \sqrt{ f \pi } \Gamma ( \frac{ f }{ 2 } ) } ( 1 + \frac{ t^2 }{ f } ) ^ { -(f+1)/2 }$$

$$S(xy) = \displaystyle \sum_{ i=1 }^n ( x_{ i } – \bar{ x })( y_{ i } – \bar{ y })$$

$$S(xy) = \displaystyle \sum_{ i=1 }^n x_{ i }y_{ i } – n \sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i }$$

$$S(xy) = \displaystyle \sum_{ i=1 }^n x_{ i }y_{ i } – \displaystyle\frac{\sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } }{ n }$$

$$S(xy) = – \displaystyle\frac{\sum_{ i=1 }^n x_{ i } \sum_{ i=1 }^n y_{ i } }{ n }$$

t分布の確率密度関数

$$f(t) = \displaystyle\frac{ \Gamma ( \frac{ f + 1 }{ 2 }) }{ \sqrt{ f \pi } \Gamma ( \frac{ f }{ 2 } ) } ( 1 + \frac{ t^2 }{ f } ) ^ { -(f+1)/2 }$$

$$f(t) = \displaystyle\frac{ \Gamma ( ( f + 1 ) / 2 ) }{ \sqrt{ f \pi } \Gamma ( f / 2 ) } ( 1 + t^2 / f ) ^ { -(f+1)/2 }$$