A sample of size 200 will be taken at random from an infinite population. given that the population proportion is 0.60, the prob

Let p represent the proportion of the population. <span>
Given p=0.60, n=200 we need to calculate P(^p<0.58).
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According to the rule of thumb, since n*p = 200*0.60 and n*(1-p)= 200*(1-0.60) = 80, both values exceed 5, confirming that n is large enough for the sampling distribution of sample proportion-^p to comply with the z standard normal distribution. The mean for the sampling distribution will be U^p = p = 0.60, and the standard deviation, δ^p = √[p*(1-p)/n] = √[0.60*(1-0.60)/200] = √0.0012.
Thus, the probability that the sample proportion is below 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>Thus, the likelihood that the sample proportion is smaller than 0.58 stands at 0.281 or 28.1%.</span>