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Question 618:



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For this questions we are given both the population mean and population standard deviation. In real life this is very rare. We almost never know these values and instead have to estimate them from samples, usually small samples. The common times you do know the population parameters are in cases like IQ test or standardized admission tests like the SAT, GRE, GMAT or LSAT.


So, the concept here is to change the z-score formula just a bit to take into account the sample size. The z-score for a point (instead of an average) is (x-mean)/standard deviation. To take into account the sample size we divide by the standard error of the mean (SEM) which is the sample standard deviation divided by the square root of the sample size. In this example, or SEM is 12.8/SQRT(20) = 2.862.  We now substitute it into the z-score formula for the mean of 90. (90-84.2)/2.862 = 2.0264. To find the probability of this value we use a normal table or the z-score to percentile calculator and enter 2.0264 for 1 sided area. We get 97.864%. So the probability is .978 that the mean of 20 students will be less than 90.

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