document.write( "Question 1204642: U.S. internet users spend an average of
\n" ); document.write( "18.3 hours a week online. If 95% of users spend between 13.1 and 23.5 hours a week, what is the probability that a randomly selected user is online less than 15 hours a week?
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Algebra.Com's Answer #841072 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "To solve this problem, we need to use the properties of the normal distribution.\r
\n" ); document.write( "\n" ); document.write( "First, we need to calculate the standard deviation (\"sigma\") of the distribution. We know that \"95\"% of the data lies between \"13.1\" and \"23.5\" hours.
\n" ); document.write( "This range corresponds to the interval from \"-1.96\" to \"1.96\" standard deviations from the mean in a normal distribution (since \"95\"% of the data in a normal distribution is within \"1.96+\"standard deviations of the mean).\r
\n" ); document.write( "\n" ); document.write( "We can set up the following equations to solve for \"sigma\":\r
\n" ); document.write( "\n" ); document.write( "\"mu+-+1.96sigma=+13.1\"
\n" ); document.write( "\"mu+%2B+1.96sigma+=+23.5\"
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\n" ); document.write( "\"2mu=13.1%2B23.5\"
\n" ); document.write( "\"2mu=36.6\"
\n" ); document.write( "\"mu=18.3\"\r
\n" ); document.write( "\n" ); document.write( "\"sigma=%2823.5-18.3%29%2F1.96\"\r
\n" ); document.write( "\n" ); document.write( "\"sigma=2.65\"\r
\n" ); document.write( "\n" ); document.write( "Next, we need to calculate the z-score for \"X=15\" hours:\r
\n" ); document.write( "\n" ); document.write( "\"z=%28X-mu%29%2Fsigma\"\r
\n" ); document.write( "\n" ); document.write( "\"z=%2815-18.3%29%2F2.65\"\r
\n" ); document.write( "\n" ); document.write( "\"z=-1.245\"\r
\n" ); document.write( "\n" ); document.write( "The z-score tells us how many standard deviations an element is from the mean. A z-score of
\n" ); document.write( "\"-1.245\" means that 15 hours is \"1.245\" standard deviations below the mean.\r
\n" ); document.write( "\n" ); document.write( "Finally, we need to find the probability that a randomly selected user is online less than \"15+\"hours a week. This is the same as finding the area to the left of \"z+=+-1.245\" in the standard normal distribution.\r
\n" ); document.write( "\n" ); document.write( "You can find this value in a standard normal distribution table or use a calculator with a normal distribution function. \r
\n" ); document.write( "\n" ); document.write( "probabilities for the normal distribution:
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\n" ); document.write( "endpoint: \"15\"
\n" ); document.write( "mean: \"18.3\"
\n" ); document.write( "standard deviation: \"2.65\"\r
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\n" ); document.write( "\n" ); document.write( "\"P%28X%3C15%29+=+0.1065\"\r
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\n" ); document.write( "\n" ); document.write( "So, the probability that a randomly selected user is online less than \"15+\"hours a week is approximately \"0.1065\", or \"10.65\"%.\r
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