## bhargavi Group Title 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? one year ago one year ago

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.