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习题题目
答案
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=14xiPi
E ( x 2 ) = ∑ i = 1 4 x i 2 P i E(x^2)=\sum_{i=1}^4x_i^2P_i E(x2)=∑i=14xi2Pi
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,
而 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(Y≥0)
= P ( 7 − 4 X ≥ 0 ) =P(7-4X≥0) =P(7−4X≥0)
= P ( X ≤ 7 4 ) =P(X≤\frac{7}{4}) =P(X≤47)
= 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
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