NOTE: Answers using z -scores rounded to 2 (or more) decimal places will work fo
ID: 3358390 • Letter: N
Question
NOTE: Answers using z-scores rounded to 2 (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 168-lbs and a standard deviation of 27-lbs. An elevator in the health club is limited to 35 occupants, but it will be overloaded if the total weight is in excess of 6230-lbs.
Assume that there are 35 men in the elevator. What is the average weight per man beyond which the elevator would be considered overloaded?
average weight = lbs Round to the nearest pound.
What is the probability that one randomly selected male health club member will exceed this weight?
P(one man exceeds) = Round to 4 decimal places.
If we assume that 35 male occupants in the elevator are a random sample of all male club members, find the probability that the evelator will be overloaded?
P(elevator overloaded) = Round to 4 decimal places.
If the evelator is full (on average) 8 times a day, how many times do we expect the elevator will be overloaded in one (non-leap) year?
number of times overloaded = Round to the nearest whole number.
Is there reason for concern?
no, the current overload limit is adequate to ensure the safety of the passengers
yes, the current overload limit is not adequate to ensure the safey of the passengers
Explanation / Answer
Ans:
mean = 168-lbs
standard deviation of 27-lbs.
a)
The elevator is overloaded if the total weight is over 6240, so it's overloaded is the average is over 6230/35 = 178 lb.
b)
z=(178-168)/27=10/27=0.37
P(z>0.37)=0.3557
c)
standard error of mean=27/sqrt(35)=4.564
z=(178-168)/4.564=2.19
P(z>2.19)=0.0143
d)
The expected number of overloads is (number of trials)*P(overoad).
The number of trials is 8 times the number of days in a year.
8*365=2920
The expected number of overloads=2920*0.0143=41.756
Approximately 42 times will be overloaded.
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