Question 744773
the probability that a random individual has blue eyes is 0.35.
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You have 100 random selected students . What is the probability half of the group has blue eyes ?
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z(0.5) = (0.5-0.35)/sqrt(0.35*0.65/100) = 3.1450
P(x > = 0.5) = P(z > 3.1450) = normalcdf(3.1450,100) = 0.0008309
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What's the probability at least 25 percent of the group has blue eyes?
z(0.25) = (0.25-0.35)/sqrt(0.35*0.65/100) = -2.0966 = P(z > -2.0966)
= normalcdf(-2.0966,100) = 0.9820
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Draw probability distribution for this exercise .
Draw a normal curve with mean = 0.35 and std = sqrt[0.35*0.65/100) = 0.0477
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How many more people need to have blue eyes to be considered an outlier ??
If you take z >=2 as outlier region, determine the probability of the
distribution that is >= 2. 
normalcdf(2,100) = 0.023
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Then the total # of the 200 with blue eyes would have to be 
be >= 200(1-0.023) = 196,
Cheers,
Stan H.
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