The population of weights for men attending a local health club is normally dist

Question

The population of weights for men attending a local health club is normally distributed with a mean of 185-lbs and a standard deviation of 26-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 6766-lbs.

What is the probability that one randomly selected male health club member will exceed this weight? Answer rounded to four decimal places

If we assume that 34 male occupants in the elevator are the result of a random selection, find the probability that the elevator will be overloaded? Answer rounded to four decimal places

If the elevator is full (on average) 5 times a day, how many times will the evelator be overloaded in one (non-leap) year? Answer rounded to nearest whole number

Explanation / Answer

average weight to be overloaded =6766/34=199

1)probability that one randomly selected male health club member will exceed this weight=P(X>199)

=P(Z>(199-185)/26)=P(Z>0.5385)=0.2951

2)average mean weight =185

and std error of mean for 34 passengers =26/(34)1/2 =4.459

therefore probability that the elevator will be overloaded =P(X>199)=

=P(Z>(199-185)/4.459)=P(Z>3.1397)=0.000845

3) total number of times elevator is full in a year =365*5=1825

hence number of  times will the evelator be overloaded in one (non-leap) year =np=1825*0.000845=1.5 ~2 times

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