E e e px 2 px 0 px 1 px 2 00183 00733 02381 02381 4

This preview shows page 25 - 36 out of 51 pages.

e e e P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0183 + 0.0733 + 0.2381 = 0.2381 4-25
Image of page 25

Subscribe to view the full document.

(a) Tính xác su t có 2 tai n n xu t hi n/tháng? (b) Xác su t có 1 tai n n/tháng? 12 18 = 1.5 tai n n/tháng P( X = 2) = ! x e x λ λ - ! 2 5 . 1 2 5 . 1 - = e = 0.2510 (4sf) answer Có 12 tháng/n ă m nên λ = P( X 1) = P(X=1) + P(X=2) + P(X=3) + …. Infinite. Tính xác su t P(X=0) ! x e x λ λ - = ! 0 5 . 1 0 5 . 1 - = e 4-26 Ví d 4 : Trung bình có 18 tai n n xu t hi n m t giao l m i n ă m. Xác su t
Image of page 26
! 0 5 . 1 0 5 . 1 - = e 5 . 1 - = e 1 1 5 . 1 × = - e = 0.223130… Nên P(X>1) = 1 – P(X=0) = 1 – 0.223130… = 0.7769 (4sf) answer (c) Xác su t có nhi u h ơ n 2 tai n n/tháng? P( X > 2) = There are an infinite number of cases so instead consider X < 2 1 – P( X < 2) λ = 1.5 4-27 Xác su t
Image of page 27

Subscribe to view the full document.

B ng phân b Poisson Xác su t ghi b ng là giá tr tích l ũ y P( X x ) . Xác su t λ 0.5 1.0 1.5 2.0 2.5 x = 0 0.6065 0.3679 0.2231 0.1353 0.0821 x = 1 0.9098 0.7358 0.5578 0.4060 0.2873 x = 2 0.9856 0.9197 0.8088 0.6767 0.5438 x = 3 0.9982 0.9810 0.9344 0.8571 0.7576 x = 4 0.9998 0.9963 0.9814 0.9473 0.8912 x = 5 1.0000 0.9994 0.9955 0.9834 0.9580 x = 6 1.0000 0.9999 0.9991 0.9955 0.9858 X ~ Po(1.5), P(X 4) = 0.9814 4-28
Image of page 28
V i X ~ Po(1.5), P( X = 2) = P( X 2) – P( X 1) = 0.8088 – 0.5578 = 0.251 λ 0.5 1.0 1.5 2.0 2.5 x = 0 0.6065 0.3679 0.2231 0.1353 0.0821 x = 1 0.9098 0.7358 0.5578 0.4060 0.2873 x = 2 0.9856 0.9197 0.8088 0.6767 0.5438 x = 3 0.9982 0.9810 0.9344 0.8571 0.7576 x = 4 0.9998 0.9963 0.9814 0.9473 0.8912 x = 5 1.0000 0.9994 0.9955 0.9834 0.9580 x = 6 1.0000 0.9999 0.9991 0.9955 0.9858 Xác su t 4-29 B ng phân b Poisson
Image of page 29

Subscribe to view the full document.

N ế u Y ~ Po(2), P( Y > 1) = P( Y = 2, 3, 4, …) = 1 – P( Y 1) = 1 – 0.4060 = 0.594 λ 0.5 1.0 1.5 2.0 2.5 x = 0 0.6065 0.3679 0.2231 0.1353 0.0821 x = 1 0.9098 0.7358 0.5578 0.4060 0.2873 x = 2 0.9856 0.9197 0.8088 0.6767 0.5438 x = 3 0.9982 0.9810 0.9344 0.8571 0.7576 x = 4 0.9998 0.9963 0.9814 0.9473 0.8912 x = 5 1.0000 0.9994 0.9955 0.9834 0.9580 x = 6 1.0000 0.9999 0.9991 0.9955 0.9858 Xác su t 4-30 B ng phân b Poisson
Image of page 30
Trung bình và độ l ch chu n Th i gian trung bình gi a hai s ki n: 1/ λ Dãy th i gian t, s l ượ ng s ki n k v ng: λ t = = µ ( ) λ E x x Trung bình Ví d 5 : T c độ xu t hi n virus λ = 2. S l ượ ng virus xu t hi n laptop trong 1 tu n? μ = E(X) = 2 4-31
Image of page 31

Subscribe to view the full document.

Trung bình và độ l ch chu n Độ l ch chu n = = 2 σ σ λ t Ví d 6 : T c độ xu t hi n virus λ = 2. Độ l ch chu n trong 1 tu n? = = = = 2 σ σ λ 2 1.41 t 4-32
Image of page 32
Ví d 7 : B ng bên d ướ i th ng kê s l ượ ng bàn th ng c a Premiership t August 21 st đế n September 12 th 2010. Trung bình và độ l ch chu n X 0 1 2 3 4 5 T n su t f 21 19 10 3 3 0 V i gi s s bàn th ng phân ph i Poisson, tính tr trung bình và ph ươ ng sai d li u? Nh n xét k ế t qu ? Solution Trung bình (0×21)+(1×19)+(2×10)+(3×3)+(4×3) 60 = = =1.071 56 56 x 4-33
Image of page 33

Subscribe to view the full document.

Ph ươ ng sai Trung bình và độ l ch chu n 2 2 2 2 = (0 ×21)+(1 ×19)+...+(4 ×3) = 134 x f x f x σ - = - 2 2 2 2 1 134 60 = n 56 56 = 1.245 (4 s.f.) Nh n xét tr trung bình và ph ươ ng sai t ươ ng đồ ng, t c ch n phân ph i Poisson cho d li u này là h p lý. 4-34
Image of page 34
Dùng phân ph i Poisson, v i d li u trên, tính xác su t xu t hi n s bàn th ng là 0, 1, 2, 3, 4, hay ít nh t 5 bàn/tr n?
Image of page 35

Subscribe to view the full document.

Image of page 36
  • Spring '12
  • NguyenXuanLong,JohnLafferty

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern