[ ] =. =3.5 3 =.3 =. =0.30 : (f i ) u i u i f i u i f i
|
|
|
- あいぞう かくはり
- 5 years ago
- Views:
Transcription
1 [ ] ( x s, x + s) 5% x s = 3, x + s = 7 s = = 70 5 = 0 a = 3, b = = = % 5 A = 56.5, = 60 6 x = 70.7, s = 5. 7 :
2 [ ] =. =3.5 3 =.3 =. =0.30 : (f i ) u i u i f i u i f i ū = = 0.0 x = = { } 6.0 s u = 589 ( ) =. 0 0 s x = 7. = < < ( x s, x + s) = (7.39, 8.7) % ( x s, x + s) = (.85, 6.0) % ( x 3s, x + 3s) = ( 3.69, 79.55) 0 00%
3 [ ] 3 ( x s, x + s).. 3 ( x s, x + s) 3 3: () ( ) ( ) ( ) 0 = =3 3 == =0.3 x = 3.95 s = 5.56 =0.7 ( x s, x + s) = (7.39, 8.5) 78 ( x s, x + s) = (.83, 6.07) 8 ( x 3s, x + 3s) = ( 3.73, 79.63)
4 [ ] 8 : =3.8 =.5 3 =55.9 =. =0.8 5 ū = 36 = 0.36 x = = ( ) 3.0 s u = = s x =.83 = < < 5 ū = = 0. x = =.3 98 ( ) 3.0 s u = =.53 s x =.53 =
5 [ ] 5 5: (f i ) u i u i f i u i f i ( x s, x + s) = (7.8, 6.) ( x s, x + s) = (0.65, 77.97) ( x 3s, x + 3s) = ( 6.8, 9.80) x s % 6.5%
6 [ ] ( x s, x + s) = (5.65, 65.93) : 79 (7.6, 60.8) : 66 ( x s, x + s) = (5.5, 86.07) : 96 (0.77, 77.53) : 9 ( x 3s, x + 3s) = (.63, 06.) : 98 ( 5.9, 9.) : B A B A a: 9% b: 8% c: 78% d: % e: 83% f: 7% ψ = 3. Q = 0.5
7 分 ) [ ] 7 7 i 0 5(.8) 0(0.) (0.) 8(7.8) iii ψ =.05, Q = (35.6) 8(7.) (.) 6(6.6) ψ = 0.897, Q = 0.05 ii 0 35(35.8) 5(.) (7.) 8(8.8) iv ψ = 0.875, Q = (76.) 33(39.6) 6 0 9(35.6) 5(8.) ψ =.7, Q = i ii iii iv x = y = 7 (, ), (, ), (3, 5) 9 5 r = 0. 待ち時間 ( 予約時刻 - 到着時刻 ( 分 ) I ū = ( x a)c, v = (ȳ b)d s uv = n n (u i ū)(v i v) i=
8 [ ] 8 = = n cd n n {(x i a)c ( x a)c}{(y i b)d (ȳ b)d} i= n (x i x)(y i ȳ) = cds xy i= u v r uv x y r xy.5. I s u = cs x, s v = ds y r uv = s uv s u s v = cds xy cds x s y = r xy Ω = { AAAA, AAAD, AADA, AADD, ADAA, ADAD, ADDA, ADDD, DAAA, DAAD, DADA, DADD, DDAA, DDAD, DDDA, DDDD } { AAAA, AAAD, AADA, AADD }. { DDAA, DADA, DAAD, ADDA, ADAD, AADD } Ω = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA }. A B C D i ii iii iv = = 0.5=( ) + ( ) 0.5 = = = P (A G) = P (G) P (A G) !!! 5! = 3,560,!!!! =,30 6 = 3
9 [ ] C C = 55 9 C 3 3 C C = x p(x) x 3 p(x) 3 = 6 3 x 0 p(x) E(x) = 0 3, E(x) = 5, V (x) = 3 5 x 3 3 p(x) 3 = 3 3 = 6 µ =, σ = V (x) = 5 9 ( ) 3 x p(x) =, 7 90 x =,, 3,..., , ,000 8 = 3,50 00,500 0, B(, 0.9) m 5m 8 8 x ( B 8, ) x = 35 8
10 [ ] P o() p(x) = x x! e p(0) = e, 0.35 p() = e, 0.7 p() = e, (e + e + e ) = 5e P o() p(x) = x x! e p(0) + p() = 3e, e p(x) = 3x x! e 3 p(0) = e 3, p() p(0) = 3e e 3 ( ) x ( ) 00 x ( 00 9 B(00, 00 ) p(x) = 99 00C x 00 00, p(0) = 99 00) = , p() = ( ) 99 = , p() = ( 00 ) ( 99 00) 98 = 0.89 P o() : p(x) = x! e, p(0) = e = , p() = e = , p() = e = p(x) = x x! e 0 p(0) + 7,000 p() +,000 p() +,000 [ {p(0) + p() + p()}] 8,000 =,7,000,003 =.5, Q(.5) = % 998,003,00,003 =.5, Q(.5) = ( ) = % = 0.5, Q(0.5) = =.5, =.0, µ u(0.) = µ % u(0.) =.8, = 0.8, Q(0.8) = 0.9, a 65 5 =.5, Q(.5) = % =.0, {Q(.0) + Q(.0)} = % = µ = = 0.3, Q(0.3) = 0.38, = =.8 a = cm
11 [ ] B(50, 0.8) N(0, ( 8) ) = , Q(0.53) = 0.98, 0.98 = % = , Q(0.88) = 0.89, 9% 5 00, B(00, 0.) N(0, ).5 0 =.875, Q(.88) = % 6 B (,600, u(0.0) =.36, 7 ) N(86, 8.8 ) 5 =.33, Q(.3) = % a =.36 a = 93 P {x = 0, y = 0} = 0., P {x = 0} P {y = 0} = = 0.5 E(x) = 0.5, V (x) = 0.5, E(y) =, V (y) = 0.6. x + y E(x + y) =.5, V (x + y) = 0.65 E(x + y) = E(x) + E(y) V (x + y) = V (x) + V (y) P {x = y} = 0.5, P {y > x} = 0.30 E(x) = 0.88, V (x) = , E(y) = 0.95, V (y) = E(xy) = 0.90 Cov(x, y) = = 0.06, ρ xy = = N(360, (0 ) ) = , Q(0.35) = %
12 [ ] 30 = 0 5 = = 0.65, Q(0.65) = N(60, ) 00 N(60, ( 5) ) =.789, Q(.79) = = % σ u(0.0) =.36 + σ ( ) σ = + σ.0 g.36 = a 5 a 3 {( 5, 5), ( 5, 5), ( 5, 5), ( 5, 5), ( 5, 5), ( 5, 5), (5, 5), (5, 5), (5, 5)}. x \x x x E(x x ) = 0 V (x x ) = 7 V (x ) = V (x ) = 36 3 N(., (0.0 ) ) =., Q(.) = % = 3, = 0 N(3,000, 0 ) 3,00 3,000 0 =.0, Q(.0) = = = 97 N(6,97) =.0, Q(.0) =
13 [ ] ,500, ,000 5, ,000, , µ µ (0.,.8) 95% 95% µ,050 ± t(5, 0.05) 00 = (997,,03) n = = ( ( ) 3 6 x N µ, ) P { 0.3 < x µ < 0.3} = {0.5 Q(.)} = n 0. n 70.6 n 7. ± = (9.50,.90) s = 3.0, t(, 0.05) =.0. ± = (9.9, 3.) σ 3.0 s iii,000,000 B(, 000, p) p 6 n =.96 pq =,0 ˆpˆq 95% = 0.039
14 [ ] 7 n \ p , , , e \ p ,537,07,305, ,58 6,7 8,068 9,0 9,60 % (,050,050,500 ±.96 8 p : p : 990,500 ±.96,500 50,500 ( 990,500 50,500 ) = (0.677, 0.73),500 ) = (0.636, 0.68),500, ,500 ±.96,050,500 50,500, ,500 50,500 = (0.007, 0.073),500 95% 95% 0 0 N(0, ).65 9 =.355 Q(.36) = % u(0.0) = n.36 n n 6 3 µ 0 t(n, 0.05) s x µ 0 + t(n, 0.05) s µ 0 n n x t(n, 0.05) s n µ 0 x + t(n, 0.05) s n [ x t(n, 0.05) s, x + t(n, 0.05) s ] n n x ± t(n, 0.05) s n µ 95% µ 0
15 [ ] 5 N(µ, 3.5 ) 0 x = 5.9 H 0 : µ = 60. H : µ x = 5.9 < 56.0 = A 5 00 x x H 0 : µ = 0 6 x N(50, (/5) ) /5 =.5, /5 =.5, Q(.5) = µ = 9. N(9., (/5) ) [9.5, 50.5] /5 =.5, /5 = 6.5, Q(6.5) 0 Q(.5) Q(6.5) =. Q(.5) = , 0.07 µ = µ = Z = (5 ) + (39 0) 0 + (6 6) 6 = 9.65 χ (, 0.05) = 5.99, χ (, 0.0) = 9., χ (, 0.005) = n = n = 600, ˆp = = 8 5, ˆp = = , ˆp = = ( ˆp ˆp = > = ) 600 p 95% 8 ( 8 5 ± ) = (0.93, 0.573) p 95% 7 ( 7 5 ± ) = (0.7, 0.507) 5 600
16 [ ] 6 H 0 : p = p α c α x = 3 x = 7 α x 7 x 6.5 ˆp ˆp /(0 ) =.333, Q(.33) = , = x x y y nˆα + x i ˆβ = yi () xi ˆα + x ˆβ i = x i y i () () () xi n { x i ( x i ) } n ˆβ = x i y i xi yi n (xi x) ˆβ = (xi x)(y i ȳ) ˆβ = (xi x)(y i ȳ) (xi x) = s xy s x () ˆα = ȳ ˆβ x.
17 [ ] 製作時間 製作個数 6 x i y i x i x (x i x) y i ȳ (y i ȳ) (x i x)(y i ȳ) x i x i y i x = 3, ȳ = 36 5ˆα + 5 ˆβ = 80, 5ˆα + 55 ˆβ = 65 ˆα = 3.5, ˆβ = 7.5 ˆβ = s xy s x = 75 0 = 7.5, ˆα = ȳ ˆβ x = = 3.5. ŷ = x x = 6 ŷ = = y i ŷ i y i ŷ i (y i ŷ i ) S e = 7.5 S T = 570 r = S e = 7.5 S T 570 = 75 = σ s = 7.5 = ± t(3, 0.05) 0 = (5.9, 9.09)
18 [ ] 8 ˆβ = 7.5 >.59 = t(3, 0.05) 読書 音楽鑑賞の時間 テレビを視る時間 7 ŷ = x S e = 0.9, r = ( 0.9, 0.5) ˆβ = 0.35 > 0.77 = t(7, 0.05) 民間最終消費支出 ( 千億円 ) 国内総生産 ( 千億円 ) 8 ŷ = x GDP r = s = ± t(, 0.05) = (0.8, 0.536) ˆα =. > 73.8 = t(, 0.05) 5.6
グラフと組み合わせ 課題 7 ( 解答例 ) 2013/5/27 1 列挙 n 個の文字の集合 { } S = a, a,, an の全てからなる文字列 つまり同じ文字を含まない 長さ n の文字列を列挙する 方法を考える 1. 何通りの文字列があるかを答えなさい また そのことが正しい
グラフと組み合わせ 課題 7 ( 解答例 ) 2013/5/27 1 列挙 n 個の文字の集合 { S = a, a,, an 0 1 1 の全てからなる文字列 つまり同じ文字を含まない 長さ n の文字列を列挙する 方法を考える 1. 何通りの文字列があるかを答えなさい また そのことが正しいことを数学的帰納法で示しなさい 2. 文字列を列挙する再帰的アルゴリズムを構築しなさい 3. n = 4
renshumondai-kaito.dvi
3 1 13 14 1.1 1 44.5 39.5 49.5 2 0.10 2 0.10 54.5 49.5 59.5 5 0.25 7 0.35 64.5 59.5 69.5 8 0.40 15 0.75 74.5 69.5 79.5 3 0.15 18 0.90 84.5 79.5 89.5 2 0.10 20 1.00 20 1.00 2 1.2 1 16.5 20.5 12.5 2 0.10
6.1 (P (P (P (P (P (P (, P (, P.
(011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.
6.1 (P (P (P (P (P (P (, P (, P.101
(008 0 3 7 ( ( ( 00 1 (P.3 1 1.1 (P.3.................. 1 1. (P.4............... 1 (P.15.1 (P.15................. (P.18............3 (P.17......... 3.4 (P................ 4 3 (P.7 4 3.1 ( P.7...........
R R 16 ( 3 )
(017 ) 9 4 7 ( ) ( 3 ) ( 010 ) 1 (P3) 1 11 (P4) 1 1 (P4) 1 (P15) 1 (P16) (P0) 3 (P18) 3 4 (P3) 4 3 4 31 1 5 3 5 4 6 5 9 51 9 5 9 6 9 61 9 6 α β 9 63 û 11 64 R 1 65 13 66 14 7 14 71 15 7 R R 16 http://wwwecoosaka-uacjp/~tazak/class/017
ii 3.,. 4. F. (), ,,. 8.,. 1. (75% ) (25% ) =9 7, =9 8 (. ). 1.,, (). 3.,. 1. ( ).,.,.,.,.,. ( ) (1 2 )., ( ), 0. 2., 1., 0,.
23(2011) (1 C104) 5 11 (2 C206) 5 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 ( ). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5.. 6.. 7.,,. 8.,. 1. (75%
統計学のポイント整理
.. September 17, 2012 1 / 55 n! = n (n 1) (n 2) 1 0! = 1 10! = 10 9 8 1 = 3628800 n k np k np k = n! (n k)! (1) 5 3 5 P 3 = 5! = 5 4 3 = 60 (5 3)! n k n C k nc k = npk k! = n! k!(n k)! (2) 5 3 5C 3 = 5!
Microsoft Word - 表紙.docx
黒住英司 [ 著 ] サピエンティア 計量経済学 訂正および練習問題解答 (206/2/2 版 ) 訂正 練習問題解答 3 .69, 3.8 4 (X i X)U i i i (X i μ x )U i ( X μx ) U i. i E [ ] (X i μ x )U i i E[(X i μ x )]E[U i ]0. i V [ ] (X i μ x )U i i 2 i j E [(X i
ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.
(1 C205) 4 10 (2 C206) 4 11 (2 B202) 4 12 25(2013) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7. 8. 1., 2007 ( ).,. 2. P. G., 1995. 3. J. C., 1988. 1... 2.,,. ii 3.,. 4. F. ( ),..
<4D F736F F D B A836F838A F C83432E646F6378>
1 シーケンスカバリングアレイ 2010/11/26( 金 ) 秋山浩一 1. はじめに 2010 年 11 月 3 日に 辰巳さんから TEF のメーリングリストに一通のメールが発信されました それは NIST のウェブに掲載された D. Richard Kuhn, Raghu N. Kacker, Yu Lei の 3 名による 組合せテスト技法のチュートリアル資料 PRACTICAL COMBINATORIAL
研修コーナー
l l l l l l l l l l l α α β l µ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l
( )/2 hara/lectures/lectures-j.html 2, {H} {T } S = {H, T } {(H, H), (H, T )} {(H, T ), (T, T )} {(H, H), (T, T )} {1
( )/2 http://www2.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html 1 2011 ( )/2 2 2011 4 1 2 1.1 1 2 1 2 3 4 5 1.1.1 sample space S S = {H, T } H T T H S = {(H, H), (H, T ), (T, H), (T, T )} (T, H) S
ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.
24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)
..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A
![..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A](/thumbs/93/112175219.jpg "..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A")
.. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.
untitled
17 5 13 1 2 1.1... 2 1.2... 2 1.3... 3 2 3 2.1... 3 2.2... 5 3 6 3.1... 6 3.2... 7 3.3 t... 7 3.4 BC a... 9 3.5... 10 4 11 1 1 θ n ˆθ. ˆθ, ˆθ, ˆθ.,, ˆθ.,.,,,. 1.1 ˆθ σ 2 = E(ˆθ E ˆθ) 2 b = E(ˆθ θ). Y 1,,Y
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
i ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P
1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A
医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.
医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987
chap10.dvi
. q {y j } I( ( L y j =Δy j = u j = C l ε j l = C(L ε j, {ε j } i.i.d.(,i q ( l= y O p ( {u j } q {C l } A l C l
t χ 2 F Q t χ 2 F 1 2 µ, σ 2 N(µ, σ 2 ) f(x µ, σ 2 ) = 1 ( exp (x ) µ)2 2πσ 2 2σ 2 0, N(0, 1) (100 α) z(α) t χ 2 *1 2.1 t (i)x N(µ, σ 2 ) x µ σ N(0, 1
t χ F Q t χ F µ, σ N(µ, σ ) f(x µ, σ ) = ( exp (x ) µ) πσ σ 0, N(0, ) (00 α) z(α) t χ *. t (i)x N(µ, σ ) x µ σ N(0, ) (ii)x,, x N(µ, σ ) x = x+ +x N(µ, σ ) (iii) (i),(ii) z = x µ N(0, ) σ N(0, ) ( 9 97.
5 Armitage x 1,, x n y i = 10x i + 3 y i = log x i {x i } {y i } 1.2 n i i x ij i j y ij, z ij i j 2 1 y = a x + b ( cm) x ij (i j )
5 Armitage. x,, x n y i = 0x i + 3 y i = log x i x i y i.2 n i i x ij i j y ij, z ij i j 2 y = a x + b 2 2. ( cm) x ij (i j ) (i) x, x 2 σ 2 x,, σ 2 x,2 σ x,, σ x,2 t t x * (ii) (i) m y ij = x ij /00 y
L P y P y + ɛ, ɛ y P y I P y,, y P y + I P y, 3 ŷ β 0 β y β 0 β y β β 0, β y x x, x,, x, y y, y,, y x x y y x x, y y, x x y y {}}{,,, / / L P / / y, P
005 5 6 y β + ɛ {x, x,, x p } y, {x, x,, x p }, β, ɛ E ɛ 0 V ɛ σ I 3 rak p 4 ɛ i N 0, σ ɛ ɛ y β y β y y β y + β β, ɛ β y + β 0, β y β y ɛ ɛ β ɛ y β mi L y y ŷ β y β y β β L P y P y + ɛ, ɛ y P y I P y,,
201711grade1ouyou.pdf
2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2
![ESD-巻頭言[003-004].indd](/thumbs/41/22693718.jpg "ESD-巻頭言[003-004].indd")
分散分析・2次元正規分布
2 II L10(2016-06-30 Thu) : Time-stamp: 2016-06-30 Thu 13:55 JST hig F 2.. http://hig3.net ( ) L10 2 II(2016) 1 / 24 F 2 F L09-Q1 Quiz :F 1 α = 0.05, 2 F 3 H 0, : σ 2 1 /σ2 2 = 1., H 1, σ 2 1 /σ2 2 1. 4
chap9.dvi
9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =
α β *2 α α β β α = α 1 β = 1 β 2.2 α 0 β *3 2.3 * *2 *3 *4 (µ A ) (µ P ) (µ A > µ P ) 10 (µ A = µ P + 10) 15 (µ A = µ P +
Armitage 1 1.1 2 t *1 α β 1.2 µ x µ 2 2 2 α β 2.1 1 α β α ( ) β *1 t t 1 α β *2 α α β β α = α 1 β = 1 β 2.2 α 0 β 1 0 0 1 1 5 2.5 *3 2.3 *4 3 3.1 1 1 1 *2 *3 *4 (µ A ) (µ P ) (µ A > µ P ) 10 (µ A = µ P
1 911 9001030 9:00 A B C D E F G H I J K L M 1A0900 1B0900 1C0900 1D0900 1E0900 1F0900 1G0900 1H0900 1I0900 1J0900 1K0900 1L0900 1M0900 9:15 1A0915 1B0915 1C0915 1D0915 1E0915 1F0915 1G0915 1H0915 1I0915
i
14 i ii iii iv v vi 14 13 86 13 12 28 14 16 14 15 31 (1) 13 12 28 20 (2) (3) 2 (4) (5) 14 14 50 48 3 11 11 22 14 15 10 14 20 21 20 (1) 14 (2) 14 4 (3) (4) (5) 12 12 (6) 14 15 5 6 7 8 9 10 7
( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................
( ) g 900,000 2,000,000 5,000,000 2,200,000 1,000,000 1,500, ,000 2,500,000 1,000, , , , , , ,000 2,000,000
( ) 73 10,905,238 3,853,235 295,309 1,415,972 5,340,722 2,390,603 890,603 1,500,000 1,000,000 300,000 1,500,000 49 19. 3. 1 17,172,842 3,917,488 13,255,354 10,760,078 (550) 555,000 600,000 600,000 12,100,000
2 Part A B C A > B > C (0) 90, 69, 61, 68, 6, 77, 75, 20, 41, 34 (1) 8, 56, 16, 50, 43, 66, 44, 77, 55, 48 (2) 92, 74, 56, 81, 84, 86, 1, 27,
/ (1) (2) (3) ysawano@tmu.ac.jp (4) (0) (10) 11 (10) (a) (b) (c) (5) - - 11160939-11160939- - 1 2 Part 1. 1. 1. A B C A > B > C (0) 90, 69, 61, 68, 6, 77, 75, 20, 41, 34 (1) 8, 56, 16, 50, 43, 66, 44,
.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T
NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
: , 2.0, 3.0, 2.0, (%) ( 2.
2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................
第１部 一般的コメント
(( 2000 11 24 2003 12 31 3122 94 2332 508 26 a () () i ii iii iv (i) (ii) (i) (ii) (iii) (iv) (a) (b)(c)(d) a) / (i) (ii) (iii) (iv) 1996 7 1996 12
表1票4.qx4
iii iv v 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 22 23 10 11 24 25 26 27 10 56 28 11 29 30 12 13 14 15 16 17 18 19 2010 2111 22 23 2412 2513 14 31 17 32 18 33 19 34 20 35 21 36 24 37 25 38 2614
(iii) x, x N(µ, ) z = x µ () N(0, ) () 0 (y,, y 0 ) (σ = 6) *3 0 y y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 y ( ) *4 H 0 : µ
t 2 Armitage t t t χ 2 F χ 2 F 2 µ, N(µ, ) f(x µ, ) = ( ) exp (x µ)2 2πσ 2 2 0, N(0, ) (00 α) z(α) t * 2. t (i)x N(µ, ) x µ σ N(0, ) 2 (ii)x,, x N(µ, ) x = x + +x ( N µ, σ2 ) (iii) (i),(ii) x,, x N(µ,
I II III IV V
I II III IV V N/m 2 640 980 50 200 290 440 2m 50 4m 100 100 150 200 290 390 590 150 340 4m 6m 8m 100 170 250 µ = E FRVβ β N/mm 2 N/mm 2 1.1 F c t.1 3 1 1.1 1.1 2 2 2 2 F F b F s F c F t F b F s 3 3 3
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
第１章 国民年金における無年金
1 2 3 4 ILO ILO 5 i ii 6 7 8 9 10 ( ) 3 2 ( ) 3 2 2 2 11 20 60 12 1 2 3 4 5 6 7 8 9 10 11 12 13 13 14 15 16 17 14 15 8 16 2003 1 17 18 iii 19 iv 20 21 22 23 24 25 ,,, 26 27 28 29 30 (1) (2) (3) 31 1 20
基礎数学I
I & II ii ii........... 22................. 25 12............... 28.................. 28.................... 31............. 32.................. 34 3 1 9.................... 1....................... 1............
?
240-8501 79-2 Email: nakamoto@ynu.ac.jp 1 3 1.1...................................... 3 1.2?................................. 6 1.3..................................... 8 1.4.......................................
ii 2. F. ( ), ,,. 5. G., L., D. ( ) ( ), 2005.,. 6.,,. 7.,. 8. ( ), , (20 ). 1. (75% ) (25% ). 60.,. 2. =8 5, =8 4 (. 1.) 1.,,
(1 C205) 4 8 27(2015) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7.... 1., 2014... 2. P. G., 1995.,. 3.,. 4.. 5., 1996... 1., 2007,. ii 2. F. ( ),.. 3... 4.,,. 5. G., L., D. ( )
.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
I? 3 1 3 1.1?................................. 3 1.2?............................... 3 1.3!................................... 3 2 4 2.1........................................ 4 2.2.......................................
20 15 14.6 15.3 14.9 15.7 16.0 15.7 13.4 14.5 13.7 14.2 10 10 13 16 19 22 1 70,000 60,000 50,000 40,000 30,000 20,000 10,000 0 2,500 59,862 56,384 2,000 42,662 44,211 40,639 37,323 1,500 33,408 34,472
- 2 -
- 2 - - 3 - (1) (2) (3) (1) - 4 - ~ - 5 - (2) - 6 - (1) (1) - 7 - - 8 - (i) (ii) (iii) (ii) (iii) (ii) 10 - 9 - (3) - 10 - (3) - 11 - - 12 - (1) - 13 - - 14 - (2) - 15 - - 16 - (3) - 17 - - 18 - (4) -
2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4 4 4 2 5 5 2 4 4 4 0 3 3 0 9 10 10 9 1 1
1 1979 6 24 3 4 4 4 4 3 4 4 2 3 4 4 6 0 0 6 2 4 4 4 3 0 0 3 3 3 4 3 2 4 3? 4 3 4 3 4 4 4 4 3 3 4 4 4 4 2 1 1 2 15 4 4 15 0 1 2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4
1 (1) (2)
1 2 (1) (2) (3) 3-78 - 1 (1) (2) - 79 - i) ii) iii) (3) (4) (5) (6) - 80 - (7) (8) (9) (10) 2 (1) (2) (3) (4) i) - 81 - ii) (a) (b) 3 (1) (2) - 82 - - 83 - - 84 - - 85 - - 86 - (1) (2) (3) (4) (5) (6)
( 30 ) 30 4 5 1 4 1.1............................................... 4 1.............................................. 4 1..1.................................. 4 1.......................................
読めば必ずわかる 分散分析の基礎 第２版
2 2003 12 5 ( ) ( ) 2 I 3 1 3 2 2? 6 3 11 4? 12 II 14 5 15 6 16 7 17 8 19 9 21 10 22 11 F 25 12 : 1 26 3 I 1 17 11 x 1, x 2,, x n x( ) x = 1 n n i=1 x i 12 (SD ) x 1, x 2,, x n s 2 s 2 = 1 n n (x i x)
S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.
![S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.](/thumbs/91/105754396.jpg "S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.")
() 1.1.. 1. 1.1. (1) L K (i) 0 K 1 K (ii) x, y K x + y K, x y K (iii) x, y K xy K (iv) x K \ {0} x 1 K K L L K ( 0 L 1 L ) L K L/K (2) K M L M K L 1.1. C C 1.2. R K = {a + b 3 i a, b Q} Q( 2, 3) = Q( 2
* n x 11,, x 1n N(µ 1, σ 2 ) x 21,, x 2n N(µ 2, σ 2 ) H 0 µ 1 = µ 2 (= µ ) H 1 µ 1 µ 2 H 0, H 1 *2 σ 2 σ 2 0, σ 2 1 *1 *2 H 0 H
1 1 1.1 *1 1. 1.3.1 n x 11,, x 1n Nµ 1, σ x 1,, x n Nµ, σ H 0 µ 1 = µ = µ H 1 µ 1 µ H 0, H 1 * σ σ 0, σ 1 *1 * H 0 H 0, H 1 H 1 1 H 0 µ, σ 0 H 1 µ 1, µ, σ 1 L 0 µ, σ x L 1 µ 1, µ, σ x x H 0 L 0 µ, σ 0
³ÎΨÏÀ
2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p
untitled
( œ ) œ 2,000,000 20. 4. 1 25. 3.27 44,886,350 39,933,174 4,953,176 9,393,543 4,953,012 153,012 4,800,000 164 164 4,001,324 2,899,583 254,074 847,667 5,392,219 584,884 7,335 4,800,000 153,012 4,800,000
9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P
9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
I L01( Wed) : Time-stamp: Wed 07:38 JST hig e, ( ) L01 I(2017) 1 / 19
I L01(2017-09-20 Wed) : Time-stamp: 2017-09-20 Wed 07:38 JST hig e, http://hig3.net ( ) L01 I(2017) 1 / 19 ? 1? 2? ( ) L01 I(2017) 2 / 19 ?,,.,., 1..,. 1,2,.,.,. ( ) L01 I(2017) 3 / 19 ? I. M (3 ) II,
zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
エクセルカバー入稿用.indd
i 1 1 2 3 5 5 6 7 7 8 9 9 10 11 11 11 12 2 13 13 14 15 15 16 17 17 ii CONTENTS 18 18 21 22 22 24 25 26 27 27 28 29 30 31 32 36 37 40 40 42 43 44 44 46 47 48 iii 48 50 51 52 54 55 59 61 62 64 65 66 67 68
provider_020524_2.PDF
1 1 1 2 2 3 (1) 3 (2) 4 (3) 6 7 7 (1) 8 (2) 21 26 27 27 27 28 31 32 32 36 1 1 2 2 (1) 3 3 4 45 (2) 6 7 5 (3) 6 7 8 (1) ii iii iv 8 * 9 10 11 9 12 10 13 14 15 11 16 17 12 13 18 19 20 (2) 14 21 22 23 24
N cos s s cos ψ e e e e 3 3 e e 3 e 3 e
3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >
「産業上利用することができる発明」の審査の運用指針（案）
1 1.... 2 1.1... 2 2.... 4 2.1... 4 3.... 6 4.... 6 1 1 29 1 29 1 1 1. 2 1 1.1 (1) (2) (3) 1 (4) 2 4 1 2 2 3 4 31 12 5 7 2.2 (5) ( a ) ( b ) 1 3 2 ( c ) (6) 2. 2.1 2.1 (1) 4 ( i ) ( ii ) ( iii ) ( iv)
II (No.2) 2 4,.. (1) (cm) (2) (cm) , (
II (No.1) 1 x 1, x 2,..., x µ = 1 V = 1 k=1 x k (x k µ) 2 k=1 σ = V. V = σ 2 = 1 x 2 k µ 2 k=1 1 µ, V σ. (1) 4, 7, 3, 1, 9, 6 (2) 14, 17, 13, 11, 19, 16 (3) 12, 21, 9, 3, 27, 18 (4) 27.2, 29.3, 29.1, 26.0,
01_.g.r..
I II III IV V VI VII VIII IX X XI I II III IV V I I I II II II I I YS-1 I YS-2 I YS-3 I YS-4 I YS-5 I YS-6 I YS-7 II II YS-1 II YS-2 II YS-3 II YS-4 II YS-5 II YS-6 II YS-7 III III YS-1 III YS-2
,, 2. Matlab Simulink 2018 PC Matlab Scilab 2
(2018 ) ( -1) TA Email : ohki@i.kyoto-u.ac.jp, ske.ta@bode.amp.i.kyoto-u.ac.jp : 411 : 10 308 1 1 2 2 2.1............................................ 2 2.2..................................................
i ii iii iv v vi vii ( ー ー ) ( ) ( ) ( ) ( ) ー ( ) ( ) ー ー ( ) ( ) ( ) ( ) ( ) 13 202 24122783 3622316 (1) (2) (3) (4) 2483 (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) 11 11 2483 13
X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2
(I) (II) 2 (I) 2 (II) 2 (III) (I) (II) (II) : 2 Typeset by Akio Namba usig Powerdot. 2 / 47
4 Typeset by Akio Namba usig Powerdot. / 47 (I) (II) 2 (I) 2 (II) 2 (III) (I) (II) (II) : 2 Typeset by Akio Namba usig Powerdot. 2 / 47 (I) (II) 2 (I) 2 (II) 2 (III) (I) (II) (II) : 2 (radom variable):
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
3 5 18 3 5000 1 2 7 8 120 1 9 1954 29 18 12 30 700 4km 1.5 100 50 6 13 5 99 93 34 17 2 2002 04 14 16 6000 12 57 60 1986 55 3 3 3 500 350 4 5 250 18 19 1590 1591 250 100 500 20 800 20 55 3 3 3 18 19 1590
困ったときのQ&A
ii iii iv NEC Corporation 1997 v P A R T 1 vi vii P A R T 2 viii P A R T 3 ix x xi 1P A R T 2 1 3 4 1 5 6 1 7 8 1 9 1 2 3 4 10 1 11 12 1 13 14 1 1 2 15 16 1 2 1 1 2 3 4 5 17 18 1 2 3 1 19 20 1 21 22 1
178 5 I 1 ( ) ( ) 10 3 13 3 1 8891 8 3023 6317 ( 10 1914 7152 ) 16 5 1 ( ) 6 13 3 13 3 8575 3896 8 1715 779 6 (1) 2 7 4 ( 2 ) 13 11 26 12 21 14 11 21
I 178 II 180 III ( ) 181 IV 183 V 185 VI 186 178 5 I 1 ( ) ( ) 10 3 13 3 1 8891 8 3023 6317 ( 10 1914 7152 ) 16 5 1 ( ) 6 13 3 13 3 8575 3896 8 1715 779 6 (1) 2 7 4 ( 2 ) 13 11 26 12 21 14 11 21 4 10 (
9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x
2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin
Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 = ( p µ γ µ + m)(p ν γ ν + m) (5.1) γ = p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 = 1 2 p µp ν {γ µ, γ ν } + m
Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 p µ γ µ + mp ν γ ν + m 5.1 γ p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 1 2 p µp ν {γ µ, γ ν } + m 2 5.2 p m p p µ γ µ {, } 10 γ {γ µ, γ ν } 2η µν 5.3 p µ γ µ + mp
ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
10:30 12:00 P.G. vs vs vs 2
1 10:30 12:00 P.G. vs vs vs 2 LOGIT PROBIT TOBIT mean median mode CV 3 4 5 0.5 1000 6 45 7 P(A B) = P(A) + P(B) - P(A B) P(B A)=P(A B)/P(A) P(A B)=P(B A) P(A) P(A B) P(A) P(B A) P(B) P(A B) P(A) P(B) P(B