document.write( "Question 1203213: ​Full-time college students report spending a mean of 25 hours per week on academic​ activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 4 hours. If you select a random sample of 25 ​full-time college​ students, what is the probability that the mean time spent on academic activities is at least 24 hours per​ week? \n" ); document.write( "
Algebra.Com's Answer #838610 by Theo(13342)\"\" \"About 
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population mean is 25
\n" ); document.write( "population standard deviation is 4
\n" ); document.write( "sample size is 25
\n" ); document.write( "sample mean is 24
\n" ); document.write( "you want to know the probability that the mean time spent on academic activities is less than 24 hous per week.\r
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\n" ); document.write( "\n" ); document.write( "z = (x - m) / s
\n" ); document.write( "z is the z-score
\n" ); document.write( "x is the smaple mean
\n" ); document.write( "m is the population mean
\n" ); document.write( "s is the standad error.\r
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\n" ); document.write( "\n" ); document.write( "standard error = standard deviation / sqrt(sample size) = 4 / sqrt(25) = 4/5 = .8
\n" ); document.write( "z-score formula becomes:
\n" ); document.write( "z = (24 - 25) / .8 = -1/25.\r
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\n" ); document.write( "\n" ); document.write( "area to the right of z-score of -1.25 = .89435
\n" ); document.write( "that's the probability that the mean time spent on academic activitie is at least 24 hours per week.\r
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\n" ); document.write( "\n" ); document.write( "here's what it looks like on a normal distribution graph.\r
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