Question 1160227: 1. A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects, and defects occur independently of each other.
(i). Find the probability of exactly 3 defective computers in a shipment of twenty. Ans: 0.0596 (ii). Find the probability of at most 2 defective computers in a shipment of twenty. Ans: 0.9245
(iii). Find the probability that least 2 defective computers in a shipment of twenty. Ans: 0.2642
(iv). Find mean number of defective computers and its standard deviation in a shipment of twenty. Ans: 𝜇 = 1, 𝜎 = 0.975
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. thesis 20C3*0.95^17*0.05^3=0.0596
b. at most 2 means 0,1,2 defective
for 0, it is .95^20 or 0.3585
for 1, it is 20*0.95^19*0.05=0.3774
for 2, it is 190*0.95^18*0.05^2=0.1887
this is 0.9246 and the 0.0001 difference was rounding each of those rather than rounding at end.
c. At least 2 defective computers means 1- p(0)-p(1)=1-0.3585-0.3774=0.2641 (rounding error again)
d. mean is np=20*0.05=1 computers
variance is np(1-p)=0.95
sd is sqrt (V)=0.9747 or 0.975 computers
If probability were 60% a computer would not be attacked, for at least 1 would mean 1-p(0 computers attacked), since that is the only other possibility. p(0)=0.00004, so probability is 0.9999 that at least 1 was attacked.
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