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put this solution on YOUR website!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.
a) What is the probability that a call lasts less than180 seconds?
b) What is the probability that a particular call lasts between 180 and 300 seconds?
c) What is the length of a call if only 1% of all calls are shorter?
WE DO IT BY USING NORMAL DISTRIBUTION FORMULAE
MEAN=M=240......STANDARD DEVIATION =S=40... X IS THE VALUE UNDER CONSIDERATION..
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
0.5-0.4332=0.0668.
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2.180
X=300.........T=(300-240)/40=1.5.............F(1.5)=0.4332
SO WE WANT AREA UNDER F(-1.5) TO F(1.5)=2*0.4332=0.8664
HENCE PROBABILITY OF THE CALL LASTING BETWEEN 180 AND 300 SECS IS 0.8664
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3.1% OF ALL CALLS ARE SHORTER...PROBABILITY =0.01...AREA UNDER STD.CURVE ....
0.5-0.01=0.49.....CORRESPONDING VALUE OF T IS -2.33
HENCE -2.33=(X-240)/40
X-240=-93.2...X=146.8 SECS.
HENCE LENGTH OF CALL IS LESS THAN 146.8 SECS FOR 1 % OF CALLS
