document.write( "Question 827862: 4. Tennis-Warehouse recently conducted a study of long distance phone calls made by its employees. The study showed that the length of the calls has a mean of 3.2 minutes, a standard deviation of 0.50 minutes and it follows the normal probability distribution.
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\n" ); document.write( "(a) What portion of the calls was between 3.2 and 4 minutes?
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\n" ); document.write( "\n" ); document.write( "(b) What portion of the calls was more than 4 minutes?\r
\n" ); document.write( "\n" ); document.write( "(c) What portion of the calls was between 4 and 4.5 minutes?\r
\n" ); document.write( "\n" ); document.write( "(d) What portion of the calls was between 3 and 4.5 minutes?
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Algebra.Com's Answer #498909 by stanbon(75887)\"\" \"About 
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The study showed that the length of the calls has a mean of 3.2 minutes, a standard deviation of 0.50 minutes and it follows the normal probability distribution.
\n" ); document.write( "For each of these problems:
\n" ); document.write( "1st: Find the z-value of the figures
\n" ); document.write( "2nd: Find the probability that z satisfies the condition of the problem:
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\n" ); document.write( "\n" ); document.write( "(a) What portion of the calls was between 3.2 and 4 minutes?
\n" ); document.write( "z(3.2) = (3.2-3.2)/0.5 = 0
\n" ); document.write( "z(4) = (4-3.2)/0.5 = 0.8/0.5 = 1.6
\n" ); document.write( "P(3.2< x <4) = P(0< z < 1.6) = normalcdf(0,1.6) = 0.4452
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.======================
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\n" ); document.write( "(b) What portion of the calls was more than 4 minutes?
\n" ); document.write( "(c) What portion of the calls was between 4 and 4.5 minutes?
\n" ); document.write( "(d) What portion of the calls was between 3 and 4.5 minutes?
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