document.write( "Question 1161992: Suppose in fact 60% of all Filipino adults between the ages of 30 and 39 exercises at least 3 days per week. If we take a random sample of 100 Filipino adults in their 30s, what is the probability that at least 70% of them exercise at least 3 days per week? \n" ); document.write( "
Algebra.Com's Answer #785650 by Theo(13342)![]() ![]() You can put this solution on YOUR website! p = .6 \n" ); document.write( "q = 1 - .6 = .4 \n" ); document.write( "n = 100 \n" ); document.write( "mean of population = .6 \n" ); document.write( "standard deviation of distribution of sample means = sqrt(.6 * .4 / 100) = .04899 rounded to 5 decimal places. \n" ); document.write( "z = (.7 - .6) / .04899 = 2.04 rounded to 2 decimal places. \n" ); document.write( "area to the left of that z-score is equal to .97932, based on the table that can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf \n" ); document.write( "area to the right of that z-score is equal to 1 - .97932 = .02068\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( ".02068 should be your answer.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "p is the probability that the population ratio is as stated. \n" ); document.write( "q is the probability that the population ratio is not as stated. \n" ); document.write( "that's why you get p = .6 and q = .4 \n" ); document.write( "the standard distribution of sample means is equal to sqrt (p*q/n) which makes it equal to sqrt(.6 * .4 / 100) which gets you .04899 rounded to 5 decimal places. \n" ); document.write( "the z-score formula is z = (x - m) / s \n" ); document.write( "z is the z-score. \n" ); document.write( "x is the raw score \n" ); document.write( "m is the raw mean \n" ); document.write( "s is the standard deviation of the distribution of sample means. \n" ); document.write( "in this problem, x is equal to .7 and m is equal to .6 and s is equal to .04899. \n" ); document.write( "that gets you the z-score of 2.04 rounded to 2 decimal places. \n" ); document.write( "it's rounded to 2 decimal places because that's the accuracy of the z-score table used. \n" ); document.write( "the table gives you the area to the left of that z-score which is the probability of getting that z-score or less. \n" ); document.write( "the area to the right of that z-score is equal to 1 minus the area to the left of that z-score. \n" ); document.write( "that tells you the probability of getting that z-score or more. \n" ); document.write( "since the problem states at least, then you want the area to the right of that z-score, which is what was calculated above.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if you had used a calculator, you would have used unrounded numbers. \n" ); document.write( "in that case: \n" ); document.write( "s = sqrt(.6 * .4 / 100) = .0489897949 \n" ); document.write( "z = (.7 - .6) / that = 2.041241452. \n" ); document.write( "p(z > 2.041241452) = .0206133484. \n" ); document.write( "that's pretty close to the result from the table of .02068. \n" ); document.write( ".02068 is only off from .0206133484 by 0.32% rounded to 2 decimal places. \n" ); document.write( "that's considerably less than a 1% error.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |