Question 1081199
t=(81.8-80)/12/sqrt (36)
1.8*6/12=0.9
I want the probability that a t df=35 is >0.9 which is 0.1871.  If I use a z, I get 0.184, so 18.4%
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look at 0 of them which is .6^18=0.0001
1 of them which is 18*0.6^17*0.4=0.0012
2 of them which is 18C2=153*0.6^16*0.4^2=0.007
3 of them which is 18C3=816*0.6^15*0.4^3=0.0246.  They add to 0.033 or 3.3%, the second choice
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The interval is 1.96 SE =1.96*sqrt (p*(1-p)/n)=1.96* sqrt (.426*.574/305)=0.056 on either side of 0.426
(0.370, 0.481), the last choice.
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The last one is a CI for a z-test.  The % is not stated--will assume 95%
(73.32, 75.08), the first choice.
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The last one can't be done without more information.