Question

# NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem....

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 183-lbs and a standard deviation of 27-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 6630-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?
(Report answer accurate to 4 decimal places.)

If the evelator is full (on average) 6 times a day, how many times will the evelator be overloaded in one (non-leap) year?
(Report answer rounded to the nearest whole number.)

Is there reason for concern?

• no, the current overload limit is adequate to insure the safety of the passengers
• yes, the current overload limit is not adequate to insure the safey of the passengers

1) average weight =6630/34= 195

2)

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

 probability =P(X>195)=P(Z>(195-183)/27)=P(Z>0.44)=1-P(Z<0.44)=1-0.67=0.3300

3)

 sample size       =n= 34 std error=σx̅=σ/√n= 4.6305
 probability =P(X>195)=P(Z>(195-183)/4.63)=P(Z>2.59)=1-P(Z<2.59)=1-0.9952=0.0048

4)

expected number of times =6*365*0.0048 =10.51 ~11

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

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