Question

NOTE: Answers using z-scores rounded to 3 (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 170-lbs and a standard deviation of 28-lbs. An elevator in the health club is limited to 33 occupants, but it will be overloaded if the total weight is in excess of 6171-lbs.

Assume that there are 33 men in the elevator. What is the average weight beyond which the elevator would be considered overloaded?
average weight =  lbs

What is the probability that one randomly selected male health club member will exceed this weight?
P(one man exceeds) =  
(Report answer accurate to 4 decimal places.)

If we assume that 33 male occupants in the elevator are the result of a random selection, find the probability that the evelator will be overloaded?
P(elevator overloaded) =  
(Report answer accurate to 4 decimal places.)

If the evelator is full (on average) 4 times a day, how many times will the evelator be overloaded in one (non-leap) year?
number of times overloaded =  
(Report answer rounded to the nearest whole number.)

Is there reason for concern?

yes, the current overload limit is not adequate to insure the safey of the passengers

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

Answer 1

Solution:- Given that mean = 170-lbs, standard deviation = 28 lbs , n = 33

=> average weight = 6171/33 = 187

=> P(one man exceeds) = P(X > 187)

  P(X > 187) = P((x-mu)/sd > (187-170)/28)

= P(Z > 0.6071)

= 0.2709

=> P(elevator overloaded) =

P(X > 187) = P((x-mu)/(sd/sqrt(n)) > (187-170)/(28/sqrt(33))

= P(Z > 3.4878)

= 0.0002

=> EXpected number of times = 365*4*0.0002 = 0.292

=> option B> no, the current overload limit is adequate to insure the safety of the passengers