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

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 176-lbs and a standard deviation of 31-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 6426-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-bs 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) 7 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? 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 Points possible: 2 This is attempt 1 of 3 orum? Submit

Explanation / Answer

1)average weight =6426/34=189

)

P(one exceed)=P(X>189)=P(Z>(189-176)/31)=P(Z>0.42)=0.3372

3)

P(elevator overloaded)=P(Xbar>189)=P(Z>(189-176)/(31/sqrt(34))=P(Z>2.45)=0.0071

4)

number of times overloaded =7*365*0.0071=18

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