in a large restaurant an average of 3 out of 5 customers will ask for water...
in a large restaurant an average of 3 out of 5 customers will ask for water with their waiter. a random sample of 50 customers is randomly selected. what is the probability that less than 34 ask for water with their meals?
Homework Answers
P(Ask for water) = 3/5 = 0.60
Mean = np = 50*0.60 = 30
Standard deviation = 
= 3.464
Hence by using normal approximation to binomial:
P(Less than 34 asks for water)
= P(X < 33.5) [Continuity Correction]
= P(z < (33.5 - 30)/3.464)
= P(z < 1.01)
= 0.8438
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