问一道求数学期望和方差的题流水生产线上每个产品不合格的概率为P,0

问一道求数学期望和方差的题流水生产线上每个产品不合格的概率为P,0
问一道求数学期望和方差的题
流水生产线上每个产品不合格的概率为P,0

问一道求数学期望和方差的题流水生产线上每个产品不合格的概率为P,0
设X=n+k,即n个“合格品”和k个“不合格品”.那么,n服从“负二项分布”,即
P(n=i) = C(i+k-1, k-1) x p^k x (1-p)^i.
这个分布的均值和方差分别是
E(n) = k(1-p)/p;
D(n) = k(1-p)/p^2.
所以, X的均值和方差分别是
E(X) = E(n)+k = k(1-p)/p + k;
D(X) = D(n) = k(1-p)/p^2.
负二项分布
当r是整数时,负二项分布又称帕斯卡分布,其概率质量函数为 它表示,已知一个事件在伯努利试验中每次的出现概率是p,在一连串伯努利试验中,一件事件刚好在第r + k次试验出现第r次的概率.
p{X=k} = f(k;r,p) = (k+r-1)!/[k!(r-1)!]p^r(1-p)^k, k=0,1,2,..., 0正无穷）kf(k;r,p) = sum(k=1->正无穷）k(k+r-1)!/[k!(r-1)!]p^r(1-p)^k = sum(k=1->正无穷）(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
= r(1-p)/p*sum(k=1->正无穷）(k-1 + r+1 -1)!/[(k-1)!(r+1 -1)!]p^(r+1)(1-p)^(k-1)【把k-1看做1个整体,r+1看做1个整体,p和(1-p)的指数凑成(k-1)和(r+1)的形式】
= r(1-p)/p*sum(n=k-1=0->正无穷）(n+s-1)!/[n!(s-1)!]p^s(1-p)^n【n=k-1,s=r+1】
= r(1-p)/p*sum(n=0->正无穷）f(n;s,p)
= r(1-p)/p*1【由归一性,sum(n=0->正无穷）f(n;s,p)=1】
= r(1-p)/p
EX^2 = sum(k=0->正无穷）k^2f(k;r,p) = sum(k=1->正无穷）k^2(k+r-1)!/[k!(r-1)!]p^r(1-p)^k = sum(k=1->正无穷）k(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
= sum(k=1->正无穷）(k-1+1)(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
= sum(k=1->正无穷）(k-1)(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
+ sum(k=1->正无穷）(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
sum(k=1->正无穷）(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
=EX= r(1-p)/p
sum(k=1->正无穷）(k-1)(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
=sum(k=2->正无穷）(k-1)(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
=sum(k=2->正无穷）(k+r-1)!/[(k-2)!(r-1)!]p^r(1-p)^k
=r(r+1)(1-p)^2/p^2sum(k=2->正无穷）(k-2 + r+2 -1)!/[(k-2)!(r+2 -1)!]p^(r+2)(1-p)^(k-2)
=r(r+1)(1-p)^2/p^2sum(n=k-2=0->正无穷）(n+s-1)!/[n!(s-1)!]p^s(1-p)^n 【n=k-2,s=r+2】
=r(r+1)(1-p)^2/p^2sum(n=0->正无穷）f(n;s,p)
=r(r+1)(1-p)^2/p^2,
EX^2 = sum(k=1->正无穷）(k-1)(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
+ sum(k=1->正无穷）(k+r-1)!/[(k-1)!(r-1)!]p^r(1-p)^k
= r(r+1)(1-p)^2/p^2 + r(1-p)/p
DX = EX^2 - (EX)^2
= r(r+1)(1-p)^2/p^2 + r(1-p)/p - [r(1-p)/p]^2
= [r(r+1)(1-p)^2 + rp(1-p) - r^2(1-p)^2]/p^2
= r(1-p)[(r+1)(1-p) + p - r(1-p)]/p^2
= r(1-p)[1-p + p]/p^2
= r(1-p)/p^2
几何分布
p{X=k} = p(1-p)^(k-1), k=1,2,...,0正无穷）x^k = 1/(1-x), 0 < x < 1.
g'(x) = sum(k=1->正无穷）kx^(k-1) = [1/(1-x)]' = 1/(1-x)^2,
EX = sum(k=1->正无穷）kp(1-p)^(k-1) = psum(k=1->正无穷）k(1-p)^(k-1)
= pg'(1-p) = p/[1-(1-p)]^2 = p/p^2 = 1/p,
EX^2 = sum(k=1->正无穷）k^2p(1-p)^(k-1) = sum(k=1->正无穷）k(k-1)p(1-p)^(k-1) + sum(k=1->正无穷）kp(1-p)^(k-1)
= sum(k=1->正无穷）k(k-1)p(1-p)^(k-1) + EX
g''(x) = sum(k=1->正无穷）k(k-1)x^(k-2) = [1/(1-x)^2]' = 2/(1-x)^3
EX^2 = sum(k=1->正无穷）k(k-1)p(1-p)^(k-1) + EX
= p(1-p)sum(k=1->正无穷）k(k-1)(1-p)^(k-2) + EX
= p(1-p)g''(1-p) + 1/p
= p(1-p)*2/[1-(1-p)]^3 + 1/p
= 2(1-p)/p^2 + 1/p
DX = EX^2 - [EX]^2 = 2(1-p)/p^2 + 1/p - (1/p)^2 = 1/p^2 - 1/p
= (1-p)/p^2