## bhargavi 2 years ago If spine bone density is normally distributed in young women with a mean of 1.0 g/cm2 and a standard deviation of .10 g/cm2, then how many young women out of 100 would you expect to have bone densities less than .85 g/cm2?

1. Jamierox4ev3r

probably not the majority because it is not close to the mean amount of 1.0, it is less than average

2. kropot72

The z-score for 0.85 g/cm^2 is found as follows: $z=\frac{X-\mu}{\sigma}=\frac{0.85-1}{0.1}=-1.5$ From a normal distribution table the cumulative probability when z = -1.5 is 0.0668. Therefore the number of young women out of 100 expected to have bone densities less than 0.85 g/cm^2 is $0.0668\times 100=you\ can\ calculate$

3. mww

You must convert the problem into a standard N(0,1) form. $x < 0.85g/cm^2$ $Z = \frac{ x - \mu }{ \sigma }$ is the transformation required with mean of 1 and sd of 0.1 $Z = \frac{ x - \mu }{ \sigma } < \frac{ 0.85 - 1 }{ 0.1 } = -1.5$ So find Z<-1.5 on the tables which is 6.7% Multiply this by 100 to find the number of people.