A bone mineral density test can be helpful in identifying the presence of osteop

Question

A bone mineral density test can be helpful in identifying the presence of osteoporosis. The result of the test is commonly measured as a z score. A randomly selected adult undergoes a bone density test. Find the probability that the result is a reading greater than 1.50 points. Using the same bone density test, find the probability that a randomly selected person has a result below -1.25. A bone density reading between -1.05 and -1.75 indicates the subject is high risk for osteoporosis. Find this probability. Using the same bone density test, find the bone density scores that separates that bottom 5% and find the bone density score that separates the top 5%

Explanation / Answer

as from property of z distribution: its mean =0; and std deviation =1

from z value table:

probability that the result is a reading greater than 1.50 points =P(z>1.50) =1-P(Z<1.50) =1-0.9332=0.0668

probability that a randomly selected person has a result below -1.25 =P(Z<-1.25) =0.10565

probability of bone density reading between -1.05 and -1.75 indicates the subject is high risk for osteoporosis

=P(-1.75<X<1.05) =0.1469 -0.0401 =0.1068

as for lower 5% , z =-1.64485

hence bone density scores that separates that bottom 5% =-1.64485

bone density score that separates the top 5% =1.64485

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