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1(uniform distribution) , .

1) 2) X ~ U(a, b)

3) 4) x < a :x

a x < b :x b : :

5) 0 < p < 1 100p- xp : [a, b] p : 1-p xp=(1-p)a + pb1 Q1 = x0.25 = 0.75a + 0.25b2 Q2 = Me = x0.5 = 0.5a + 0.5b3 Q3= x0.75 = 0.25a + 0.75b I.Q.R = Q3 - Q1 = x0.75 - x0.25 = 0.5b 0.5a

X ~ U(0, 10) (1) X (2) X (m) (s2)(3) P(m - s < X < m + s ) (4) Q1 , Q2 , Q3 (5) X Mo = ? (1) X : X ~ U(0, 10)X :F(x) = P(X x) = 0 , x < 0, 0 x < 101 , x 100 , x < 0, 0 x < 101 , x 10=

s2 = 8.3333 s = = 2.89(2)(3)(m - s , m + s ) = (5 2.89, 5 + 2. 89) = (2.11, 7.89)(4)1 Q1 = (0.75)0 + (0.25) (10) = 2.52 Q2 = (0.5)0 + (0.5) (10) = 5.0 3 Q3= (0.25)0 + (0.75) (10) = 7.5(5) [0, 10] f(x) = f(x) . X .

X ~ U(0, 1) Y = a + (b a)X (a < b) , (1) Y (2) Y (3) Y (m) (s2) (4) Y Me = ? (1) X ~ U(0, 1)X :FX(x) = P(X x) = 0 , x < 0, 0 x < 11 , x 10 , x < 0x , 0 x < 11 , x 11 du=, y = a + (b a)x 0 x 1 a y b

a y < b Y :(2) Y :fY(y) = FY(y) = =b - a1, a y b

(3) Y ~ U(a, b)m = E(Y) =a+b2(4) Y ~ U(a, b)

, .2 (exponential distribution)

l l l .1) X ~ Exp(l)

2) m = E(X) = x f(x)dx = x le-lx dx 0u= lim - e-lx = u3) E(X2) = x2 f(x)dx = x2 le-lx dx 0= lim - e-lx = xu

X(:)

(1) (2) (3) 30 , ?f(x) = 3e-3x , x > 0 1(1) P(X > 1) = 3e-3x dx = (-1)e-3x = e-3 = 0.049802(2) P(X 2) = 3e-3x dx = (-1)e-3x = 1 - e-6 = 0.9975(3) l= 3 m= 1/3, 10.

4) x < 0 :x 0 :xxx

:5) (survival function) S(x) = P(X > x) = 1 F(x) = e-lx , x > 06) (hazard rate function), (failure rate function)

X ~ Exp(1/600) (1) X .(2) X .(3) X .(4) X X ~ Exp(1/600) X X F(x) = 1- e-x /600 , x 0X X S(x) = e-x/600 , x > 0 X X m = 1/ l = 600s2 = 1/ l2 = 360000

: X ~ Exp(1/100)

(1) 150 (2) 200 F(x) = 1- e-x/100 , x 0X S(x) = e-x/100 , x > 0 X (1) 150 :P(X < 150) = F(150) = 1- e-150/100 = 1 0.2231 = 0.7769(2) 200 :P(X 200) = S(200) = e-200/100 = e-2 = 0.1353

P(X > a+b) = le-lx dx = (-1) e-lx = e-l(a+b) a+bP(X > a) = le-lx dx = (-1) e-lx = e-la aP(X > b) = le-lx dx = (-1) e-lx = e-lb ba+bab

1,000 . (1) 500 , 100 .(2) (1) , x 0.3 x = ?(1) X m = 1000 X ~ Exp(1/1000)f(x) = e-x/1000 , x > 0 X :

(2) (1) , : 500 + xP(X 500 + x|X 500) = P(X x) = S(x) = e-x/1000 = 0.3

[T > t ] : : X ~ P(3)T : t [0, t] P(T > t) = P[X(t) = 0] = e-3t T :F(t) = P(T t) = 1 - P(T > t) = 1 - e-3t , t > 0 T :f(t) = 1 - F(t) = 3e-3t , t > 0 [T > t ] :

(1) l , T l . (2) , T l , Ti i.i.d. Exp(l).

T m=10 X(t) : t (1) 6 (2) 5 , 8 (3) X(t) (4) 5 2 (1) T m = 10 T ~ Exp(1/10)T :F(t) = 1 - e-t/10 , t > 0 :

(2)P(T 13|T > 5) = P(T > 8) = 1 F(8) = 1 - 1 - e-8/10 = e-0.8 = 0.4493(3) X(t) ~ P(t/10) X(t) :f(x) = e-t/10 , x = 0, 1, 2, (4) t=5 5 : X(t) ~ P(0.5) : ( )P(X = 2) = P(X 2) P(X 1) = 0.986 0.910 = 0.076

, .3 (gamma distribution)

n G(a) = ta-1 e-t dt , a > 0 :t = x/be-x/b dx = 1 p.d.f. ta-1 e-t dt = 1xa-1 e-x/b dx = 1

1) X ~ G(a, b)a: (shape parameter) b : (scale parameter)

2) m = E(X) = x f(x)dx = xa-1 e-x/b dx G(a)G(a+1) bG(a+1) ba+11=x(a+1)-1 e-x/b dx =G(a) ba1x(a+1)-1 e-x/b dx G(a)G(a+1) b=G(a)aG(a) b== a b

3) E(X2) = x2 f(x)dx = xa-1 e-x/b dx G(a)G(a+2)b2G(a+2) ba+21=x(a+2)-1 e-x/b dx =G(a) ba1x(a+2)-1 e-x/b dx G(a)G(a+2)b2=G(a)a(a+1)G(a) b2== a(a+1) b2

X1, X2, , Xn ~ i.i.d. Exp(l) S =X1 + X2 + + Xn ~ G(n, 1/l)(2) S : l n S ~ G(n, 1/l)

T m=2 X : 9:00 2 (1) X (2) 2 (3) 2 3 (1) T m=2 T ~ Exp(1/2)T1 : 9:00 T2 : T1 ~ Exp(1/2) , T2 ~ Exp(1/2)

(c2)(chi-squared distribution) a = r/2, b = 2 (degree of freedom; d.f.) r , X ~ c2(r) .1) X ~ c2(r)

d.f. = 7 a = 0.05 14.07

X ~ c2(5) P(X < x0) = 0.95 x0 = ?X ~ c2(2), Y ~ c2(4) ,P(X + Y > x0) = 0.01 x0 = ?

, .4 (Weibull distribution)

1) X ~ Wei(a, b)

2) 3) S(x) = 1 F(x) = e-(bx) , x > 0a4) h(x) = = = aba xa-1 , x > 0aaba xa-1e-(bx)e-(bx)aF(x) = aba ua-1e-(bu) du = (-1) e-(bu) = 1 - e-(bx) , x > 0aaa0x

X ~ Wei(2, 0.1) (1) X = ? P(X 4) = ?(2) X = ? P(X 10) = ?(3) X = ?(4) X Me = ?(1) a = 2, b = 0.1 F(x) = 1 exp[-(x/10)2 ] , x > 0 P(X 4) = F(4) = 1 exp[-(4/10)2 ]= 0.1479

2) 3) m = E(X) = x f(x)dx = abaxae-(bx) dx au = (bx)am = u [(1/a) +1]-1 e-u du = G 1 + E(X2) = x2 f(x)dx = aba xa+1e-(bx) dx = G 1 + a1b2s2 = Var(X) = E(X2) E(X)21b2G 1 + G 1 + 2-=

X ~ Wei(a, b)X : h(x) = cx, x > 0(1) P(X > 5) = e-1/4 = 0.7788 c = ?(2) X = ? 6 = ?(3) X = ? X = ?(1) h(x) = aba xa-1 = cx , x > 0a = 2, aba = cP(X > 5) = 2b2 xe-(bx) dx = - e-(bx) = e-25b = e-1/45222b = 0.1, c = aba = 2(0.1)2 = 0.0225 b2 = 1/4

(2) X :a = 2 , b = 0.16 0.5P(X < 0.5) = (0.02)x exp - dx(3) X = ? X = ?

, .5 (beta distribution)

, 0% 100% Beta(a, b) = xa-1 (1 - x)b-1 dx , a, b > 0 :p.d.f. xa-1 (1 - x)b-1 dx = 1

1) X ~ Beta(a, b)

(1) a b , b a .(2) a = b x = 0.5 , a b x = 0.5 .(3) a = b = 1, X Beta (1, 1) X U(0, 1)(4) X Beta (a, b) 1-X Beta (b, a)

2) 3) E(X2) = m = E(X) = x f(x)dx = x(a+1)-1 (1 x)b-1 dx Beta(a + 1, b) p.d.f.G(a + 1)G(b)G(a + b +1)G(a + 1)G(b)G(a + b +1)x(a+1)-1 (1 x)b-1 dx =aG(a)G(b)(a + b)G(a + b )==

: X ~ Beta(3, 4) (1) X = ?(2) X = ? X = ?(3) 70% (1) a = 3, b = 4X :(2) X :X :m = = 0.4286s2 = = 0.4286

, , , .6 (normal distribution)

1) A-4.2 e-z /2 dz =2e-z /2 dz = 12p.d.f. - < m < , s > 0exp - dx = 10e-z /2 dz = 2

2) :m = m3) :s2 = s2 m, s2 . :Z ~ N(0, 1)m = 0 s2 = 1 m s2 f(x) = , - < x< , - < m< , s > 0exp -1s( x - m )22s2 :X ~ N(m, s2)

m1 m2s1 = s2m1= m2s1 s2

(1) P(Z 0 ) = P(Z 0 ) = 0.5(2) P(Z -z0 ) = P(Z z0 ) = 1- P(Z < z0), z0 > 0(3) P(Z z0 ) = 0.5 + P(0 < Z < z0 ), P(Z z0) = 0.5 - P(0 < Z < z0), z0 > 0(4) P(|Z| z0 ) = P(-z0 < Z < z0 ) = 2P(0 < Z < z0), z0 > 0

(5) P(|Z| 1.645 ) = 0.9, P(|Z| 1.96 ) = 0.95, P(|Z| 2.58 ) = 0.99(6) P(|Z| 1 ) = 0.683, P(|Z| 2 ) = 0.954, P(|Z| 3 ) = 0.9980.050.0250.005

F(z) = f(u)du(7) 1 - F(z0 ) = P(Z z0 ) = P(Z -z0 ) = F(-z0 ) , z0 > 0

(11) P(m + as < X < m + bs) = P(a < Z < b) = F(b) F(a)

(12) P(m - s < X < m + s) = P(-1 < Z < 1) = 0.683 P(m - 2s < X < m + 2s) = P(-2 < Z < 2) = 0.954 P(m - 3s < X < m + 3s) = P(-3 < Z < 3) = 0.998

100(1-a)% : zaP(Z z0 ) = 1 a z0 100(1-a)% , za .

P(Z 1.36) = ? Z < 1.36 1.3 z , .06 z 0.9131 .

X ~ N(3, 4)P(1.5 X 5.5) = F - F = F(1.25) - F(-0.75)

(1) P(0 < Z < 1.54)(2) P(-1.10 < Z < 1.10)(3) P(Z < -1.78)(4) P(Z > -1.23)(1) P(0 < Z < 1.54) = P(Z < 1.54) 0.5 = 0.9382 0.5 = 0.4382

(2) P(-1.10 < Z < 1.10) = 2P(0 < Z < 1.10) = 2[P(Z < 1.10) 0.5)] = 2(0.8643 - 0.5) = 0.7286

(3) P(Z < -1.78) = P(Z > 1.78) = 1 - P(Z < 1.78) = 1 0.9625 = 0.0375

(4) P(Z > -1.23) = P(Z < 1.23) = 0.8907

X ~ N(5, 4) (1) P(X < 6.4) (2) P(X < x0) = 0.9750 x0 = ?(3) P(3 < X < x0) = 0.756 x0 =? (1) m = 5, s = 2 X P(X 6.4) = P Z < = F(0.70) = 0.7580 ( )(2) X P(Z < 1.96) = 0.9750P(X < x0) = P Z < ( )

P(Z < z0 ) = 0.9147 z0 = 1.37(3) P(3 < X < x0) = P <

128.4, 19.6 (1) 100 (2) 134 (3) 110 130 = P(Z -1.45) = 1 P(Z 1.45) = 1 0.9265 = 0.0735

(3) X ~ N(128.4, 19.62)

: X ~N(1.59, 0.582), : Y ~ N(2.18, 0.812)(1) 2.35(kg) (2) 5.04(kg) (3) 0.35(kg) = P(Z 1.31) = 0.9049(1) X ~ N(1.59, 0.582)P(X 2.35) = P ( )

P(D 0.35) = P ( )D 0.590.99620.35 0.59= P(Z -0.24) = 1 P(Z < 0.24) = 1 0.5948 = 0.4052 (3) D = Y - X ~ N(0.59, 0.99622)0.9962

(sample mean)

Xi ~ i.i.d. N(m , s2 ), i = 1, 2, , nY = (X1 + X2 + + Xn ) ~ N m , ( )(sample mean) : m, s2 i.i.d. Xi , i = 1, 2, , n .Xi ~ i.i.d. N(m , s2), i = 1, 2, , nX = (X1 + X2 + + Xn ) ~ N m , ( )

2 (central limit theorm)

X1 , X2 ~ i.i.d. f(x)= 1/6, x=1, 2, 3, 4, 5, 6X1 , X2

19,400(), 5,000() 1,000 (1) (2) 19,800,000() (3) (4) 19,600() (1) Xi , i = 1, 2, , 1000 : 19,400 5,000 (2)P[X (19.8)106 ] = P ( )

n m = np, s2 = np(1-p) , np 5, n(1-p) 5 B(n, p) N(np, np(1-p)) . (normal approximation)

X ~ B(15, 0.4) (1) P(7 X 9) = ?(2) P(7 X 9) = ?(3) P(6.5 X 9.5) = ?(1) P(7 X 9) = P(X 9) P(X 6) = 0.9662 -0.6098 = 0.3564(2) np = 6, npq = 3.6 X N(6, 3.6)~~..

X ~ B(30, 0.2) (1) P(X = 4) = ?(2) P(X = 4) = ?(1) X :

m , N(m, m) .X N(m, m)

9,500 1 2,000 228 ? , , ~

: 0.0099, 0.0045, 0.080, 0185 9,500 : 0.1129 2,000 : 2000(0.1129) = 226

1,5201152,3554424,32513451,3000.512

, t , F-, , .7

Y ~ N(m, s2)X = eY

1) X ~ LogN (m, s2)f(x) = , - < x< , - < m< , s > 0exp -( lnx - m )22s2 1sx

2) 4) 3) 5) 100(1-a)%

: X ~ LogN(6.95, 0.64)(1) (2) 1,750() (1)mX = exp m +( )m=6.95s2 = 0.64= 1436.55 ()(2)

1) 2) 3)

t (t-distribution)Z ~ N(0, 1), V ~ c2(r) : T =1) rt- , T ~ t(r) .2) m = 03) s2 =, r > 2

t (1) .(2) t = 0 , .(3) r .

5) t 4) 100(1-a)% : ta(r)P(T > t0) = 1- a ; t0 = ta(r)

P(T > t0.025(4)) = 0.025 t0.025(4) :d.f. = 4 a = .025 2.776 . t0.025(4) = 2.776, P(T > 2.776) = 0.025

T ~ t(3) (1) P(T > t0.025) = 0.025 t0.025 = ?(2) P(|T| < t0) = 0.99 t0 = ?(1) d.f. = 3 a = 0.025 t0.025 = 3.182(2) P(|T| < t0) = P(-t0 < T < t0) = 0.99 P(|T| t0) = 0.01P(T -t0) = P(T t0) = 0.005t0 = t0.005(3) = 5.841

F (F-distribution)1) 2) 3)

F

5) F 4) 100(1-a)% fa(m, n)P(F > f0) = 1- a ; f0 = fa(m, n))

F ~ F(5, 4) P(F > f0.05(5, 4)) = 0.05 f0.05(5, 4) : 4 a = 0.05 5 6.26 . ,f0.05(5, 4) = 6.26, P(F > 6.26) = 0.05

F ~ F(4, 5) (1) P(F > f0.025) = 0.025 f0.025 = ?(2) f0.95(4, 5) = ?(1) F- f0.025 = 7.39

(2) f0.95(4, 5) = = = 0.1597

(bivariate normal distribution) sX > 0, sY > 0, - < mX , mY < , -1 < r < 1 1) (X, Y ) ~ N(mX, mY, sX, sY, r) , r = Corr(X, Y )22

X Y :mX =0, mY = 0, sX =1, sY =1, r =0

(1) r > 0, X Y X Y y = x .(2) r < 0, X Y X Y y = -x .

2) 3)

4) 5)

(X) (Y) (X, Y ) ~ N(176, 160, 1.0, 1.52, 0.6) (1) 173cm , Y (2) P(154 < Y < 158|X = 173) = ?(1) (X, Y ) ~ N(176, 160, 1.0, 1.52, 0.6) Y :Y :Y : Y|X=173 ~ N(157.3, 1.44)= P(-2.75 < Z< 0.58) = F(0.58) F(-2.75)= 0.7190 + 0.9970 -1 = 0.7160(2) P(154 < Y < 158|X = 173) = P < Z