SOLUTION: About 12.5% of restaurant bills are incorrect. If 200 bills are selected at random, find the probability that at least 22 will contain an error.

Question 1128023: About 12.5% of restaurant bills are incorrect. If 200 bills are selected at random, find the probability that at least 22 will contain an error.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The mean of this is 200*.125=25
the variance is np(1-p)=25*.875=21.875
sd is sqrt (variance)=4.68
use normal approximation with mean 25 and sd 4.68
continuity correction, use 21.5 to cover at least 22.
z>=(21.5-25)/4.68
z>-0.75 for probability 0.7734

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