SOLUTION: 67.5% of the us population were born in their state of residence. in a random sample of 200 americans find the probability that: a) at least 175 were born in their state of reside

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Question 1010631: 67.5% of the us population were born in their state of residence. in a random sample of 200 americans find the probability that:
a) at least 175 were born in their state of residence
b) exactly 150 were born in their state of residence
Answer by mathmate(429) About Me  (Show Source):
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Question:
67.5% of the us population were born in their state of residence. in a random sample of 200 americans find the probability that:
a) at least 175 were born in their state of residence
b) exactly 150 were born in their state of residence

Solution:
This question qualifies for the binomial distribution, assuming all samples are independent, and population size is large compared to sample.
given:
n=200
p=0.675

Using Binomial Distribution
A. P(X>=175)
P%28X%3E=175%29=sum%28C%28n%2Cx%29%2Ap%5Ex%2A%281-p%29%5E%28n-x%29%29+for+x=175+to+200
where C(n,x)=n!/(x!(n-x)!) is the number of combination of n choose x.
so after summing the terms (laborious job), we get
P(X>=175)=5.36*10^(-11)

B. P(X=150)
P%28X=150%29=C%28n%2C150%29%2Ap%5E150%2A%281-p%29%5E%2850%29=0.004432

Using normal approximation:
mean=mu=np=135
variance=np(1-p)=43.875
standard deviation=sigma=sqrt(variance)=6.62382
Use continuity correction
A.
z=(174.5-135)/6.62382=5.9423
upper tail = 1.355*10^(-9)

B. it is more accurate and simpler to work with the binomial distribution.