Question 671442: 10% of Americans are left handed.
a) If 6 people are selected at random, what is the probability that more than 3 of them are left-handed?
b) Suppose a group of 600 mathematicians get together for a conference. What is the probability that more than 80 of them are left-handed? (Use the normal approximation to the binomial)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 10% of Americans are left handed.
a) If 6 people are selected at random, what is the probability that more than 3 of them are left-handed?
Binomial Problem with n = 6 ; p(left) = 0.10
P(4<= x <=6) = 1 - binomcdf(6,0.10,3) = 0.0013
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b) Suppose a group of 600 mathematicians get together for a conference. What is the probability that more than 80 of them are left-handed? (Use the normal approximation to the binomial)
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mean = np = 600*0.10 = 60
std = sqrt(npq) = 7.35
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P(81<= x <= 600) = 1 - P(x < 80.5)
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z(80.5) = (80.5-60)/7.35 = 2.79
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1 - P(z < 2.79) = 0.0026
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Cheers,
Stan H.
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