Question

NOTE: Answers using z-scores rounded to 2 (or more) decimal places will work for this problem. The population of weights for men attending a local health club is normally distributed with a mean of 171-lbs and a standard deviation of 31-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 6510-lbs. Assume that there are 35 men in the elevator. What is the average weight per man beyond which the elevator would be considered overloaded? average weight = lbs Round to the nearest pound. What is the probability that one randomly selected male health club member will exceed this weight? P(one man exceeds) = Round to 4 decimal places. If we assume that 35 male occupants in the elevator are a random sample of all male club members, find the probability that the evelator will be overloaded? P(elevator overloaded) = Round to 4 decimal places. If the evelator is full (on average) 3 times a day, how many times do we expect the elevator will be overloaded in one (non-leap) year? number of times overloaded = Round to the nearest whole number. Is there reason for concern? no, the current overload limit is adequate to ensure the safety of the passengers yes, the current overload limit is not adequate to ensure the safey of the passengers

Answer 1

1)average weight per man beyond which the elevator would be considered overloaded =6510/35=

=186

2)probability that one randomly selected male health club member will exceed this weight

=P(X>186)=P(Z>(186-171)/31)=P(Z>0.48)=0.3156

3)

probability that the evelator will be overloaded =P(Xbar>186)=P(Z>(186-171)*sqrt(35)/31)

=P(Z>2.86)=0.0021

4)

total number of times =365*3=1095

hence number of time we expect the elevator will be overloaded in one year =1095*0.0021~ 2

no, the current overload limit is adequate to ensure the safety of the passengers