Question 1171094: Let the random variable X denote the number of Sunday papers sold by a paperboy in a city. Assume that X is b(500,1/5). If 50 such paperboys are randomly selected approximate the probability that the sample mean is less than 99.
I found a formula for using the central limit theorem on binomial distributions.
it is find the value in a normal distribution table of (b-np)/(sqrt(npq).
I know b is 99, p = 1/5, and q = 1/5. I think n is 500 but I am not sure. If n = 500 I get the z-score value as -0.11 and using the table I get the
P(x < 99) = 0.4562
I am not sure if I did this right or need to change my n value.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Almost
p=1/5 and q=4/5 or (1-p)
mean is np=100
variance is np(1-p)=80
sd is sqrt (V)=8.944
You need to use the continuity continuation factor, so <99 means a high of 98.5
for 50, it is
sd is 8.944/sqrt(50)
z=(98.5-100)/8.944/sqrt(50)
=-1.5*sqrt(50)/8.944
=-1.18
probability is 0.1190.
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