Question

The optical density of bone is often imaged to give an indication of bone density, and hence the risk of fracture. One method of reporting the results of the imaging is using a z-score, which compares the patient to a population of people of the same age, gender and ethnicity. These results are normally distributed with a mean of 0, and a standard deviation of 1, so they follow the standard normal curve.

a) What percentage of the population has a z-score between .24 and 1.37?

b) What z-score would be given to a patient who is in the top 11.9% of the population?

Answer 1

Solution :

Given that ,

mean = = 0

standard deviation = = 1

P(0.24< x < 1.37) = P[(0.24 -0) /1 < (x - ) / < (1.37 -0) /1 )]

= P( 0.24< Z <1.37 )

= P(Z < 1.37) - P(Z <0.24 )

Using z table   

= 0.9147 -0.5948

= 0.3199

answer=31.99%

(B)

Using standard normal table,

P(Z > z) = 11.9%

= 1 - P(Z < z) = 0.119

= P(Z < z ) = 1 - 0.119

= P(Z < z ) = 0.881

= P(Z < 1.18 ) = 0.881

z = 1.18 (using standard normal (Z) table )