document.write( "Question 1078255: The time spent watching tv on a weekend by USC students is normally distributed with a
\n" ); document.write( " mean of 180 minutes and a standard deviation of 25 minutes. Find the probability that
\n" ); document.write( " such a student selected at random will spend\r
\n" ); document.write( "\n" ); document.write( " (a) More than 230 minutes watching tv
\n" ); document.write( " (b) Less than 205 minutes watching tv
\n" ); document.write( " (c) Between 130 minutes to 155 minutes of watching tv
\n" ); document.write( " (d) More than 155 minutes of watching tv
\n" ); document.write( "
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Algebra.Com's Answer #692758 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The time spent watching tv on a weekend by USC students is normally distributed with amean of 180 minutes and a standard deviation of 25 minutes. Find the
\n" ); document.write( " probability that such a student selected at random will spend
\n" ); document.write( "(a) More than 230 minutes watching tv
\n" ); document.write( "Find the z-value::
\n" ); document.write( "z(230) = (230-180)/35 = 50/35 = 1.4286
\n" ); document.write( "P(x > 230) = P(z > 1.4286) = normalcdf(1.4286,100) = 0.0066
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\n" ); document.write( "Use the same procedure::
\n" ); document.write( "Find the z-values ; then find the probability.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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\n" ); document.write( "(b) Less than 205 minutes watching tv
\n" ); document.write( "(c) Between 130 minutes to 155 minutes of watching tv
\n" ); document.write( "(d) More than 155 minutes of watching tv
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