Question

NOTE: Answers using z-scores rounded to 3 (or more) decimal places will work for this problem....

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 167-lbs and a standard deviation of 27-lbs. An elevator in the health club is limited to 32 occupants, but it will be overloaded if the total weight is in excess of 5824-lbs.
Assume that there are 32 men in the elevator. What is the average weight beyond which the elevator would be considered overloaded?
average weight =  
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 32 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) 6 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.)

Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with a mean of 105.2 ft and a standard deviation of 4.5 ft.
A tree of this type grows in my backyard, and it stands 94 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.
P(X<94) =
My neighbor also has a tree of this type growing in her backyard, but hers stands 104.3 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.
P(X>104.3) =

Homework Answers

Answer #1

1)a ) average weight = 5824/32=182

b) P(one man exceeds) =P(X> 182)=P(Z>0.556)=0.2893

c) P(elevator overloaded) = P(Xbar>182)=P(Z>(182-167)*sqrt(32)/32)=P(Z>3.143)=0.0008

d) number of times overloaded = 6*365*0.0008=1.752~2

2)

a)probability that the height of a randomly selected tree is as tall as mine or shorter.
P(X<94) = P(Z<-2.49)=0.0064

b)

probability that the full height of a randomly selected tree is at least as tall as hers.

P(X>104.3) = 0.5793

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