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 180-lbs and a standard deviation of 29-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 6825-lbs.

Assume that there are 35 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 35 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) 3 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

Answer 1

Solution:-

a) The average weight beyond which the elevator would be considered overloaded is 195 lbs

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

x = 195, μ = 180, σ = 29

By applying normal distribution:-

z = 0.52

P(z > 0.52) = 0.3015

c) If we assume that 35 male occupants in the elevator are the result of a random selection, then the probability that the evelator will be overloaded is 0.0011.

x = 195, μ = 180, σ = 29

By applying normal distribution:-

z = 3.06

P(z > 3.06) = 0.0011

d)

p = 0.0011

Number of times elevator is full in one year = 3 × 365 = 1095

Number of times overloaded = 0.0011 × 1095 = 1.2045 or 2 times

e) No, the current overload limit is adequate to insure the safety of the passengers.