Question

The population of weights for men attending a local health club is normally distributed with a mean of 173-lbs and a standard deviation of 30-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 6615-lbs. Assume that there are 35 men in the elevator. What is the average weight of the 35 men 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 elevator will be overloaded? P(elevator overloaded) = (Report answer accurate to 4 decimal places.) If the elevator is full (on average) 5 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.)

Answer 1

Solution :

Given that ,

mean = = 173 lbs

standard deviation = = 30 lbs

n = 35

= = 173 lbs

= / n = 30 / 35 = 5.07

a) = 6615 / 35 = 189 lbs

b) P(x > 189) = 1 - p( x< 189)

=1- p P[(x - ) / < (189 - 173) / 30]

=1- P(z < 0.53)

Using z table,

= 1 - 0.7019

= 0.2981

c) P( > 189) = 1 - P( < 189)

= 1 - P[( - ) / < (189 - 173) / 5.07 ]

= 1 - P(z <3.16 )

Using z table,    

= 1 - 0.9992

= 0.0008

d) 5 * 365 * 0.0008 = 1.46

1 times will the elevator be overloaded in one (non-leap) year