SOLUTION: The true proportion of residents at small barangay that favored the genuine opposition senatorial candidates is 85%. What is the probability that at least 75% of the 60 residents f

Question 1161983: The true proportion of residents at small barangay that favored the genuine opposition senatorial candidates is 85%. What is the probability that at least 75% of the 60 residents favored the genuine opposition?
Found 2 solutions by Boreal, MOO04182020:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
can do with binomial, since very little remaining probability. This is pdf
p(44)=0.0077
p(43)=0.0035
p(42)=0.0015
p(41)=0.0006
P(40)=0.0002
cdf is1- binomcdf (60,.85,44) since we want 44 and everything less.
That is 0.9864 probability, the exact value
Normal approximation
The expected value is 51 (60*0.85)
the variance is np (1-p)=51(0.15)=7.65
sd is sqrt (V)=2.766
z>(44.5-51)/2.766=-2.35
probability is 0.9906 of z>=-2.36, a decent estimate

Answer by MOO04182020(3) (Show Source): You can put this solution on YOUR website!
assume the population has a mean of $29321 with sd 2120
z=(x bar-mean)/sd
z<(29000-29321)/2120/sqrt(100)
<-321*10/2120
<-1.51
that probability is 0.0655

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