NOTE: Answers using z -scores rounded to 2 (or more) decimal places will work fo
ID: 3252330 • 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 183-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 6630-lbs.
Assume that there are 34 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 34 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) 2 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?
yes, the current overload limit is not adequate to ensure the safey of the passengers
no, the current overload limit is adequate to ensure the safety of the passengers
Explanation / Answer
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 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 6630-lbs.
1.Assume that there are 34 men in the elevator. What is the average weight per man beyond which the elevator would be considered overloaded?
average weight =6630/34
average weight =195 lbs
2.What is the probability that one randomly selected male health club member will exceed this weight?
That's 183-195 = -12 lb above the mean.
-12/26 = - 0.4615 S.D. above the mean.
P(one man exceeds) =0.3228 (from standard normal table)
3.If we assume that 34 male occupants in the elevator are a random sample of all male club members, find the probability that the evelator will be overloaded?
The mean of n samples from a normal distribution with mean m and standard deviation s has the same mean m, but a standard deviation of s/sqrt(n).
The S.D. for 34 randomly-selected men is 26/ sqrt(34) = 4.4589 lb.
That makes the 12 pounds above mean equal to -12/4.4589 = -2.6912 S.D. above the mean on this distribution.
P(elevator overloaded) = 0.0036 (from standard normal table)
4.If the evelator is full (on average) 2 times a day, how many times do we expect the elevator will be overloaded in one (non-leap) year?
It's impossible to know how many times the elevator will be overloaded. The question probably wants you to find the expected number of times the elevator will be overloaded
The expected number of overloads is (number of trials)*number of days in year *P(overoad).
So the P(overload) is computed in step 3. The number of trials is 2 times the number of days in a year.
number of times overloaded = 2*365*0.0036 =3 times
5.Is there reason for concern?
no, the current overload limit is adequate to ensure the safety of the passengers
Unlikely, so you should only be concerned if life are at risk.
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