Let p be
the population proportion. <span>
We have p=0.60, n=200 and we are asked to find
P(^p<0.58). </span>
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 <span>So, there is 0.281 or 28.1% probability that the
sample proportion is less than 0.58. </span>
