Solving Forearm Length Stats Problem

[pinksapphire168]

Hi Everyone,

I am new to the forum and really need so help before I have a test tomorrow. I have been going over my book and working through the problems but I am stuck on this problem. Would someone please help me? thank you very much for your time

From a study
average height of men =68in sd=2.7inch
average forearm length=18in sd 1inch r=0.80

a) what percentage of men have forearms which are 18 inches long to the nearest inch? for this problem I figured out 38%

b) of the men who are 68inches tall, what percentage have forearms which are 18 inches long to the nearest inch?

I know that the nearest inch refers to 17.5 and 18.5 and the answer in the back of the book is 60% however, I do not know how they solve the problem to get 60%

 

[HallsofIvy]

Since I don't believe that "forearm length" and "height" are independent, I don't see any way to do this with knowing the correlation.

 

[EnumaElish]

I wonder whether "r=0.80" in the OP stands for corr.

For indep. normal distributions, since sd = 1 for the forearm length, the question is, what is the probability around the mean plus/minus half a sd? You can look it up from a standard normal table (e.g. http://www.math.unb.ca/~knight/utility/NormTbl0.htm ). I believe 38.3%.

For correlated distributions you need to think about multivariate normal.

 

[pinksapphire168]

found the answer

Thank you for the replies, I already found a way to solve the problem.

r stands for coefficient correlation
and the answer as I included is 60%