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 179-lbs and a standard deviation of 29-lbs. An elevator in the health club is limited to 34 occupants, but it will be overloaded if the total weight is in excess of 6460-lbs.

Assume that there are 34 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 34 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) 2 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

average weight beyond which the elevator would be considered overloaded=6460/34= 190

---------------

P ( X ≥   190   ) = P( (X-µ)/σ ≥ (190-179) / 29)      
= P(Z ≥   0.38   ) = P(Z<-0.38 ) =    0.3522   (answer)

=================

n=   34          
              
X =   190          
Z =(X - µ )/(σ/√n) =    (190-179)/(29/√34)=       2.21  
P(X>190) = P(Z >2.21) = P(Z < -2.21 ) =           0.0135   (answer)
================

number of times overloaded =2*365*0.0135=9.8 ≈ 10

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