Asample of size 200 will be taken at random from an infinite population. given that the population proportion is 0.60, the probability that the sample proportion will be greater than 0.58 is

Let p be the population proportion. 
We have p=0.60, n=200 and we are asked to find P(^p<0.58). 
The thumb of the rule is since n*p = 200*0.60 and n*(1-p)= 200*(1-0.60) = 80 are both at least greater than 5, then n is considered to be large and hence the sampling distribution of sample proportion-^p will follow the z standard normal distribution. Hence this sampling distribution will have the mean of all sample proportions- U^p = p = 0.60 and the standard deviation of all sample proportions- δ^p = √[p*(1-p)/n] = √[0.60*(1-0.60)/200] = √0.0012. 
So, the probability that the sample proportion is less than 0.58 
= P(^p<0.58) 
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0... 
= P(z<-0.58) 
= P(z<0) - P(-0.58<z<0) 
= 0.5 - 0.2190 
= 0.281 
So, there is 0.281 or 28.1% probability that the sample proportion is less than 0.58. 

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