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 174-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 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 elevator will be overloaded? P(elevator overloaded) = (Report answer accurate to 4 decimal places.) If the elevator is full (on average) 2 times a day, how many times will the elevator 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

Answer #1

a)

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

b)

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

probability = | P(X>190) | = | P(Z>0.52)= | 1-P(Z<0.52)= | 1-0.6971= | 0.3029 |

c)

sample size =n= | 34 |

std error=σ_{x̅}=σ/√n= |
5.3165 |

probability that the elevator will be overloaded

probability = | P(X>190) | = | P(Z>3.01)= | 1-P(Z<3.01)= | 1-0.9987= | 0.0013 |

d)

number of times overloaded =2*365*0.0013=0.9490

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

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