Hi,

b. 10*.53 = 5.3  Expect 5

c. sqrt(.53*.47/10)

d. p = .53 ,  n = 10  P(x=6)= binompdf(10, .53, 6)= .2271

e. P( x < 6) =  binomcdf(10, .53, 5) = .5474

f.  P( x < 120) =  binomcdf(250, .53, 119) = .4987

recommend stattrek.com as a  reference 

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
 according to a may 2013 gallup poll, 53% of americans support the legalization of same sex marriage. 

b. if a random sample of 10 americans is found, how many people would we expect to support the legalization of same sex marriage

E(x) = np = 10*0.53 = 5.3

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c. what would the standard deviation of this 10 person distribution be

std = sqrt(npq) = sqrt(5.3*0.47) = 1.58

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d. if we would randomly sample 10 people, what is the probability that exactly 6 would support the legalization of same sex marriage.

P(x = 6) = 10C6(0.53)^6(0.47)^4 = binompdf(10,0.53,6) = 0.2271

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e if we would randomly sample 10 people, what is the probability less than 6 would support the legalization of same sex marriage. 

---

P(0<= x <=5) = binomcdf(10,0.53,5) = 0.5474

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f. if we secure a simple randome sample of 250 americans, what is the probability that less than 120 would support the legalization of same sex marriage



P(x < 120) = binomcdf(250,0.53,119) = 0.0499

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Cheers,

Stan H.

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