二項分配示例
Author: 藍邦偉  老師

$ 二項分配示例 $

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$ 若X∼B(2,\frac{1}{2}) ，\;則：\;E(X)=2\cdot \frac{1}{2},\;V(X)=2\cdot \frac{1}{2}{\cdot (1-\frac{1}{2})}$

$ ∵X的機率質量函數為 $

$ X\;\;\;0\;\;\;1\;\;\;2 $
$p\;\;\;\;\frac{1}{4}\;\;\;\frac{1}{2}\;\;\frac{1}{4}$

$ E(X)=1\cdot \frac{1}{2}+2\cdot \frac{1}{4}=1=2\cdot  \frac{1}{2} $

$ V(X)=(1^2\cdot \frac{1}{2}+2²\cdot \frac{1}{4})-1^2=\frac{3}{2}-1=\frac{1}{2} $

$ 若X∼B(3,\frac{1}{2}) ，\;則：\;E(X)=3\cdot \frac{1}{2},\;V(X)=3\cdot \frac{1}{2}{\cdot (1-\frac{1}{2})}$ $

$ X\;\;\;0\;\;\;1\;\;\;2\;\;3 $
$p\;\;\;\;\frac{1}{8}\;\;\;\frac{3}{8}\;\;\frac{3}{8}\;\;\frac{1}{8}$

$ E(X)=1\cdot \frac{3}{8}+2\cdot \frac{3}{8}+3\cdot \frac{1}{8}=\frac{3}{2} $

$ V(X)=(1^2\cdot \frac{3}{8}+2²\cdot \frac{3}{8}+3^2\cdot \frac{1}{8})-(\frac{3}{2})^2=3-\frac{9}{4}=\frac{3}{4} $

$ Pf $

$ X∼B(2,\frac{1}{2}) $

$ X∼B(3,\frac{1}{2}) $

a = true
b = false
A = (-3.4, 5.28)
n = 0 (slider from 0 to 4, step 1)
B: A + (0, -2)
C: B - (0, 1.5)
D: B + (2.1, 0)
E: C + (2.1, 0)
c_0 = Segment(B, C)
c_3 = Segment(B, D)
c_2 = Segment(E, D)
c_1 = Segment(E, C)
Q: B - (0, 0.6)
Q_2: D - (0, 0.6)
c_5 = Segment(Q, Q_2)
c_6 = Segment(B + (0.6, 0), C + (0.6, 0))
c_7 = Segment(B + (1.15, 0), C + (1.15, 0))
c_8 = Segment(B + (1.65, 0), C + (1.65, 0))
B_1: B + (2.6, 0)
B_2: Q + (2.6, 0)
B_3: C + (2.6, 0)
d_1 = Segment(B, B_1)
d_2 = Segment(B_3, B_1)
d_3 = Segment(B_3, C)
d_4 = Segment(B, C)
d_5 = Segment(B_2, Q)
B_4: B + (0.6, 0)
B_5: C + (0.6, 0)
d_6 = Segment(B_4, B_5)
d_7 = Segment(B_4 + (0.55, 0), B_5 + (0.55, 0))
d_8 = Segment(B_4 + (1.05, 0), B_5 + (1.05, 0))
d_9 = Segment(B_4 + (1.5, 0), B_5 + (1.5, 0))