document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=21657>Question 21657</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Here's another that i think i know how to do, but am not really sure. please help me out!!!\r<BR>\n" );
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document.write( "A statistical analysis of 1,000 long distance telephone calls made from the headquarters of the Bricks and Clicks Computer Corporation indicates that the length of these calls is normally distributed with m = 240 seconds and s = 40 seconds.\r<BR>\n" );
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document.write( "a)	What is the probability that a call lasts less than180 seconds?<BR>\n" );
document.write( "b)	What is the probability that a particular call lasts between 180 and 300 seconds?<BR>\n" );
document.write( "c)	What is the length of a call if only 1% of all calls are shorter? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #11758 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=11758\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A statistical analysis of 1,000 long distance telephone calls made from the headquarters of the Bricks and Clicks Computer Corporation indicates that the length of these calls is normally distributed with m = 240 seconds and s = 40 seconds. <BR>\n" );
document.write( "a) What is the probability that a call lasts less than180 seconds?<BR>\n" );
document.write( "b) What is the probability that a particular call lasts between 180 and 300 seconds?<BR>\n" );
document.write( "c) What is the length of a call if only 1% of all calls are shorter?<BR>\n" );
document.write( "WE DO IT BY USING NORMAL DISTRIBUTION FORMULAE<BR>\n" );
document.write( "MEAN=M=240......STANDARD DEVIATION =S=40... X IS THE VALUE UNDER CONSIDERATION..<BR>\n" );
document.write( "1.X<180..SAY X=180...T=(X-M)/S=(180-240)/40=-1.5..NOW WE FIND PROBABILITY DENSITY FUNCTION FOR THIS VALUE OF T.......F(-1.5)=0.5-F(1.5)..WE FIND FROM TABLES THAT F(1.5)=0.4332..HENCE PROBABILITY THAT CALL LASTS LESS THAN 180 SECS IS<BR>\n" );
document.write( "0.5-0.4332=0.0668.<BR>\n" );
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document.write( "2.180 X=300.........T=(300-240)/40=1.5.............F(1.5)=0.4332<BR>\n" );
document.write( "SO WE WANT AREA UNDER F(-1.5) TO F(1.5)=2*0.4332=0.8664<BR>\n" );
document.write( "HENCE PROBABILITY OF THE CALL LASTING BETWEEN 180 AND 300 SECS IS 0.8664<BR>\n" );
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document.write( "3.1% OF ALL CALLS ARE SHORTER...PROBABILITY =0.01...AREA UNDER STD.CURVE ....<BR>\n" );
document.write( "0.5-0.01=0.49.....CORRESPONDING VALUE OF T IS -2.33<BR>\n" );
document.write( "HENCE -2.33=(X-240)/40<BR>\n" );
document.write( "X-240=-93.2...X=146.8 SECS.<BR>\n" );
document.write( "HENCE LENGTH OF CALL IS LESS THAN 146.8 SECS FOR 1 % OF CALLS </SPAN>\n" );
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