Question 1142977: A survey of 1020 workers in a certain year found that 54% of the respondents spend a total of $40 or less on lunch each week. If 10 of the workers who participated in the survey were chosen at random, what is the probability that at most 3 of them spend a total of $40 or less on lunch each week?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a binomial where we are looking at probability 0,1,2,3 out of 10 with p=0.54
so prob. 0 is .46^10 or 0.0004
prob. 1 is (10C1)(0.46^9)*.54=0.0050
prob. 2 is 10C2*0.46^8*.54^2=0.0263
prob.3 is 10C3*0.46^7*.54^3=0.0824
That sum is 0.1141
rough check with normal approximation
continuity correction factor use 3.5
mean is 5.4, variance is 5.4*.46 or 2.48, so sd is 1.58
z< (3.5-5.4)/1.58
z< -1.20
that probability is 0.1151
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