Solution: Given that n = 6, r = 0,1,2,3,4,5,6, p = 0.05
Binomial distribution :P(X=r) = nCr*p^r*q^(n-r)
=> option C.
x P(X)
0 0.7351
1 0.2321
2 0.0305
3 0.0021
4 0.0001
5 0.0000
6 0.0000
(d) μ = 0.30
=> μ = ∑x*p(x)
=
(0*0.7351)+(1*0.2321)+(2*0.0305)+(3*0.0021)+(4*0.0001)+(5*0)+(6*0)
= 0.2998
= 0.30 (rounded)
Expected number = n*p = 6*0.05 = 0.3
(c) P(X >= 2) = 0.8906 = 0.891 (rounded)
(d) standard deviation = sqrt(npq) = sqrt(6*0.05*0.95) = 1.225 (rounded)
L M NULES ASR FUUF Tedurer The quality-control inspector of a production plant will reject a...
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