document.write( "Question 849900: Hi, \r
\n" ); document.write( "\n" ); document.write( "I needed help with the following:\r
\n" ); document.write( "\n" ); document.write( "The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is
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\n" ); document.write( "1) more than 500 minutes to 4 dp
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\n" ); document.write( "2) less than 440 minutes to 4 dp
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\n" ); document.write( "3) less than 555 minutes to 4 dp
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\n" ); document.write( "4) between 500 and 555 minutes 0.5342 to 4 dp
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\n" ); document.write( "5) between 440 minutes and 555 minutes. to 4 dp
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\n" ); document.write( "6) Given that 67% of students spend less than M minutes,
\n" ); document.write( " calculate the value of M, correct to 1 decimal place. to 1 dp
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\n" ); document.write( "\n" ); document.write( "Thank You.
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Algebra.Com's Answer #511830 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is \r
\n" ); document.write( "\n" ); document.write( "1) more than 500 minutes to 4 dp
\n" ); document.write( "z(500) = (500-490)/42 = 10/42
\n" ); document.write( "P(x > 500) = P(z > 10/42) = normalcdf(10/42,100) = 0.4059
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\n" ); document.write( "2) less than 440 minutes to 4 dp
\n" ); document.write( "Same procedure using z(440) = (440-490)/42 = -50/42
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\n" ); document.write( "3) less than 555 minutes to 4 dp
\n" ); document.write( "same \r
\n" ); document.write( "\n" ); document.write( "4) between 500 and 555 minutes to 4 dp
\n" ); document.write( "z(555)=(555-490)/42 = 65/42
\n" ); document.write( "Etc.
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\n" ); document.write( "\n" ); document.write( "5) between 440 minutes and 555 minutes. to 4 dp
\n" ); document.write( "Find P(-50/42< z < 65/42) = normalcdf(-50/42,65/42) = 0.7863
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\n" ); document.write( "The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is \r
\n" ); document.write( "\n" ); document.write( "1) more than 500 minutes to 4 dp \r
\n" ); document.write( "\n" ); document.write( "2) less than 440 minutes to 4 dp \r
\n" ); document.write( "\n" ); document.write( "3) less than 555 minutes to 4 dp \r
\n" ); document.write( "\n" ); document.write( "4) between 500 and 555 minutes 0.5342 to 4 dp \r
\n" ); document.write( "\n" ); document.write( "5) between 440 minutes and 555 minutes. to 4 dp \r
\n" ); document.write( "\n" ); document.write( "6) Given that 67% of students spend less than M minutes,
\n" ); document.write( "calculate the value of M, correct to 1 decimal place. to 1 dp
\n" ); document.write( "Find the z-value with a left-tail of 0.67::
\n" ); document.write( "invNorm(0.67) = 0.4399
\n" ); document.write( "Find M = 0.4399*42+500 = 518.48
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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