document.write( "Question 841806: A statistical analysis of 1,000 long distance telephone calls made from the headquarters of a corporation indicates that the length of these calls is normally distributed with a mean of 240 seconds and standard deviation of 40 seconds. \r
\n" ); document.write( "\n" ); document.write( "What is the probability that a particular call lasted between 110 and 180 seconds? \r
\n" ); document.write( "\n" ); document.write( "How many of the calls lasted less than 180 seconds or more than 300 seconds? \n" ); document.write( "
Algebra.Com's Answer #507218 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" );
document.write( "Hi,
\n" );
document.write( "Using TI:
\n" );
document.write( "*Note: \"z+=blue+%28x+-+240%29%2Fblue%2840%2Fsqrt%281000%29%29=+%28x-240%29%2F1.27\"
\n" );
document.write( "probability that a particular call lasted between 110 and 180 seconds? 
\n" );
document.write( "P(110>x<180) 
\n" );
document.write( "Find z:
\n" );
document.write( "  z = (110-240)/1.27 = -130/1.27
\n" );
document.write( "  z = (180-240)/1.27 = -60/1.27
\n" );
document.write( "P(between smaller z and larger z): The syntax is normalcdf(smaller z, larger z).
\n" );
document.write( "normalcdf(-130/1.27, -60/1.27)\r
\n" );
document.write( "\n" );
document.write( "How many of the calls lasted less than 180 seconds or more than 300 seconds
\n" );
document.write( " 1000(1 - normalcdf(-60/1.27, 60/1.27))
\n" );
document.write( " \n" );
document.write( "
\n" );