SOLUTION: In a large hospital an average of 3 out of every 5 patients ask for water with their meal. A random sample of 10 customers is selected. Assuming a binomial distribution

Question 1166247: In a large hospital an average of 3 out of every 5 patients ask for water with their meal. A random sample of 10 customers is selected. Assuming a binomial distribution, find the probability that i. Exactly 6 patients ask for water with their meal ii. At most 4 patients ask for water with their meal iii. At least 3 patients ask for water with their meal iv. Find the mean of this distribution v. Find the standard deviation of this distribution.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
binomial with n=10 p=0.6
mean is np=6
variance is np(1-p)=6*0.4=2.4
sd is sqrt(V)=1.55

exactly 6 is 10C6*0.6^6*0.4^4=0.2508
At most 4 is 0,1,2,3,4 and that probability is binomcdf(10,0.6,4) or 0.1662
At least 3 patients ask for water
That is everyone but 0,1,2. That probability (for 0,1,2) is 0.0123, so the answer is 1-0.0123 or 0.9877

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