The population of weights for men attending a local health club is normally dist
ID: 3172809 • Letter: T
Question
The population of weights for men attending a local health club is normally distributed with a mean of 167-lbs and a standard deviation of 30-lbs. An elevator in the health club is limited to 33 occupants, but it will be overloaded if the total weight is in excess of 6039-lbs.
Assume that there are 33 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 33 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) 4 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?
yes, the current overload limit is not adequate to insure the safey of the passengers
no, the current overload limit is adequate to insure the safety of the passengers
Explanation / Answer
ans=
If this is a problem, you have no chance with the rest. The elevator is overloaded if the total weight is over 6240, so it's overloaded is the average is over 6039/33 = 183 lb.
2. That's 183-167 = 16 lb above the mean. 16/30 = 0.5333 S.D. above the mean. Now look that up or use your calculator to find the probability that a random normal variable is 0.53333 S.D. or more above the mean. It's not on my calculator. Call this probability p.
3. 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 33 randomly-selected men is
30 / sqrt(33) = 30/5.2223 =5.744
That makes the 16 pounds above mean equal to 16/5.744 = 2,7855 S.D. above the mean on this distribution. Go to the same table or calculator to find the probablity that a normally-distributed random variable is more than 2.7855 S.D. above the mean.
4. The expected number of overloads is (number of trials)*P(overoad). You computed P(overload) in step 3. The number of trials is 4 times the number of days in a year.
yes, the current overload limit is not adequate to insure the safey of the passengers
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