S1试讲讲稿

2023-12-15 23:30:19

习题题目

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答案

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用到的概念:
概率之和等于1
E ( x ) = ∑ i = 1 4 x i P i E(x)=\sum_{i=1}^4x_iP_i E(x)=i=14?xi?Pi?
E ( x 2 ) = ∑ i = 1 4 x i 2 P i E(x^2)=\sum_{i=1}^4x_i^2P_i E(x2)=i=14?xi2?Pi?
V a r ( X ) Var(X) Var(X) = ∑ i = 1 4 ( x i ? x  ̄ ) 2 ? P i =\sum_{i=1}^4(x_i-\overline{x})^2·P_i =i=14?(xi??x)2?Pi?
V a r ( A X + B ) = A 2 ? V a r ( X ) Var(AX+B)=A^2·Var(X) Var(AX+B)=A2?Var(X)

习题(a)
E ( x ) = ? 2 a + 0 ? b + 2 a + 4 c = 0.6 E(x)=-2a+0*b+2a+4c=0.6 E(x)=?2a+0?b+2a+4c=0.6
c = 0.15 c=0.15 c=0.15
E ( x 2 ) = 4 a + 0 b + 4 a + 16 c = 4 E(x^2)=4a+0b+4a+16c=4 E(x2)=4a+0b+4a+16c=4
a = 0.2 a=0.2 a=0.2
∵ a + b + a + c = 1 ∵a+b+a+c=1 a+b+a+c=1
∴ b = 0.45 ∴b=0.45 b=0.45
其中 E ( x ) E(x) E(x)是总体的平均,
x  ̄ 是从总体抽取到的样本的平均 \overline{x}是从总体抽取到的样本的平均 x是从总体抽取到的样本的平均
这里 d E ( x ) = x  ̄ = 0.6 dE(x)=\overline{x}=0.6 dE(x)=x=0.6

习题 ( b ) 习题(b) 习题(b)
V a r ( X ) Var(X) Var(X)
= ∑ i = 1 4 ( x i ? x  ̄ ) 2 ? P i =\sum_{i=1}^4(x_i-\overline{x})^2·P_i =i=14?(xi??x)2?Pi?
= ∑ i = 1 4 ( x i ? E ( x ) ) 2 ? P i =\sum_{i=1}^4(x_i-E(x))^2·P_i =i=14?(xi??E(x))2?Pi?
= ( ? 2 ? 0.6 ) 2 ? 0.2 + ( 0 ? 0.6 ) 2 ? 0.45 + ( 2 ? 0.6 ) 2 ? 0.2 + ( 4 ? 0.6 ) 2 ? 0.15 =(-2-0.6)^2*0.2+(0-0.6)^2*0.45+(2-0.6)^2*0.2+(4-0.6)^2*0.15 =(?2?0.6)2?0.2+(0?0.6)2?0.45+(2?0.6)2?0.2+(4?0.6)2?0.15
= 3.64 =3.64 =3.64

习题 ( c ) 习题(c) 习题(c)

E ( Y ) = E ( 7 ? 4 X ) = 7 ? 4 E ( x ) = 7 ? 4 ? 0.6 = 4.6 E(Y)=E(7-4X)=7-4E(x)=7-4*0.6=4.6 E(Y)=E(7?4X)=7?4E(x)=7?4?0.6=4.6

习题 ( d ) 习题(d) 习题(d)
V a r ( Y ) Var(Y) Var(Y)
= V a r ( 7 ? 4 x ) =Var(7-4x) =Var(7?4x)
= 16 V a r ( x ) =16Var(x) =16Var(x)
= 16 ? 3.64 =16*3.64 =16?3.64
= 58.24 =58.24 =58.24

习题 ( e ) 习题(e) 习题(e)
P ( Y ≥ 0 ) P(Y≥0) P(Y0)
= P ( 7 ? 4 X ≥ 0 ) =P(7-4X≥0) =P(7?4X0)
= P ( X ≤ 7 4 ) =P(X≤\frac{7}{4}) =P(X47?)
= P ( x = ? 2 ) + P ( x = 0 ) =P(x=-2)+P(x=0) =P(x=?2)+P(x=0)
= 0.2 + 0.45 = 0.65 =0.2+0.45=0.65 =0.2+0.45=0.65

文章来源:https://blog.csdn.net/appleyuchi/article/details/135026069
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