(P=.30) Random Sample (n=100 families) What is Probability that sample portion is between (.20) & (.40)?

A) .0146


B) .0292


C) .4854


D) .9708


E) None of above

(P=.30) Random Sample (n=100 families) What is Probability that sample portion is between (.20) %26amp; (.40)?
First, convert each p^ (sample proportion) value to a z value using





z = (p^ - p)/s,





where s is the standard deviation for the sampling distribution of p^, which can be found using





s = sqrt(pq/n),





where q = 1-p = 1-.30 = .70.





You get s = 0.045825757 (approximately)





Your first p^ value to convert is .20:





z = (.20 - .30)/0.045825757 = -2.18





Your second p^ value to convert is .40:


z = (.40 - .30)/0.045825757 = 2.18.





I can't do the rest right now, because I don't have a chart to look up these z values. Unless you are doing this exclusively with a calculator, you shoud have a chart where you can look up the values for these z's. Look them up and subtract them to get your final answer.





EDIT: If it's not E, then it has to be D. The Empirical rule says that about 95% of the values are within two standard deviations of the mean in a normal distribution (your p^ distribution is normal because the sample size is large enough: both np = 100(.30) = 30 and nq = 100(.70) = 70 are greater than 5). Your p^ values are 2.18 standard deviations away from the mean (because that's what z says: the number of standard devations you are away from the mean). So that answer has to be more than 95%.