Question 1193213: a) Past records indicate that the probability of online retail orders that turn out to be fraudulent is
0.06. Suppose that, on a given day, 10 online retail orders are placed. Assume that the number of
online retail orders that turn out to be fraudulent is distributed as a binomial random variable.
I. What are the mean and standard deviation of the number of online retail orders that turn out to
be fraudulent?
II. What is the probability that zero online retail orders will turn out to be fraudulent?
III. What is the probability that greater than or equal two online retail order will turn out to be
fraudulent?
b) According to the recent research 52% of American adults own tablets. If you survey 7 American
adults. Find the probability of
I. Four will own a tablet
II. No-one own tablet
III. At least two will own a tablet
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean is np=10*0.06=0.6
variance is np(1-p)=0.6*0.94=0.504
sd is sqrt (V)=0.701
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0 are fraudulent mean that 10 are not=0.94^10=0.5386
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1 is fraudulent is 10C1*0.06*0.94^9=0.3438
The sum of p(0) and p(1) is 0.8824
>=2 is 1- the above sum or 0.1176, the answer
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4 will own is 7C4*0.52^4*0.48^3=0.2830
no one will is 0.48^7=0.0059
1 will is 7C1*0.48^1*0.52^6=0.0664
the last two sum to 0.0723
the complement is probability 0.9277, and that is 2 or more, the answer.
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