document.write( "Question 1174515: Urgent help needed. please help!!\r
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document.write( "NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem.\r
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document.write( "The population of weights for men attending a local health club is normally distributed with a mean of 169-lbs and a standard deviation of 30-lbs. An elevator in the health club is limited to 35 occupants, but it will be overloaded if the total weight is in excess of 6370-lbs.\r
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document.write( "Assume that there are 35 men in the elevator. What is the average weight beyond which the elevator would be considered overloaded?
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document.write( "average weight =
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document.write( "What is the probability that one randomly selected male health club member will exceed this weight?
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document.write( "P(one man exceeds) = \r
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document.write( "(Report answer accurate to 4 decimal places.)\r
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document.write( "If we assume that 35 male occupants in the elevator are the result of a random selection, find the probability that the evelator will be overloaded?
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document.write( "P(elevator overloaded) = \r
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document.write( "(Report answer accurate to 4 decimal places.)\r
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document.write( "If the evelator is full (on average) 6 times a day, how many times will the evelator be overloaded in one (non-leap) year?
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document.write( "number of times overloaded = \r
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document.write( "(Report answer rounded to the nearest whole number.)\r
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document.write( "Is there reason for concern?
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document.write( "no, the current overload limit is adequate to insure the safety of the passengers
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document.write( "yes, the current overload limit is not adequate to insure the safey of the passengers \n" );
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Algebra.Com's Answer #799942 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! average weight >182 lbs is the point for overloading. \n" ); document.write( "- \n" ); document.write( "probability of 1 man being more than that is z >(182-169)/30 or z > 13/30 probability of 0.3324 \n" ); document.write( "- \n" ); document.write( "with 35 z =(182-169)/30/(sqrt(35), or z > 13*sqrt(*35)/30 or 2.56. That probability is 0.0052 \n" ); document.write( "2190 uses in a year \n" ); document.write( "0.0052 probability of being overloaded \n" ); document.write( "expected value of number of times is 11 times a year or almost once a month. \n" ); document.write( "- \n" ); document.write( "Last question is not statistical. Given the issues of elevator safety, that appears to be more frequent than should be allowed, but one can argue from both sides. \n" ); document.write( " |