Question 816586: 8% of the employees are left handed. If 20 employees are randomly selected, what is the probability that none of them are left handed? what is the probability that at most 2 are left-handed? what is the standard deviation and mean for the number of left handed employees?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 8% of the employees are left handed.
If 20 employees are randomly selected, what is the probability that none of them are left handed?
Ans: 0.92^20 = 0.1887
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what is the probability that at most 2 are left-handed?
P(0<= x <=2) = binomcdf(20,0.08,2) = 0.7879
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what is the standard deviation and mean for the number of left handed employees?
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mean = np = 20*0.08 = 1.6
std = sqrt(npq) = sqrt(1.6*0.2) = 0.5657
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Cheers,
Stan H.
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