Health and Beauty

Osteoporosis Standard Deviation

Making Sense of Today's Best Bone-building Strategies

Making Sense of Today's Best Bone-building Strategies The results are reported as a sum up, which tells us how far off our BMD is from a healthy adult without osteoporosis. A result of –2.5 SD (standard deviation) or greater indicates osteoporosis. A consequence between –1 SD to –2.5 SD means there is some

Hip Arthroplasty Risk in Men Higher With Increased Dairy Product Consumption

Consuming more dairy products may raise the risk for total hip arthroplasty for osteoarthritis (OA) in men, according to a study published in the Journal of Rheumatology .

Increased consumption of dairy products, which are high in calcium, is associated with increased bone mineral density (BMD). Although dairy product consumption is beneficial for bone health, higher BMD may be linked to a greater risk for hip OA because of lower bone resorption and turnover, which changes the shape of hip bones. Data on the effect of dairy product consumption on the risk for hip OA were lacking until recently.

Researchers led by Sultana M. Hussain, PhD, and Yuanyuan Wang, PhD, from Monash University in Australia, examined the relationship between dairy product consumption and the risk for total hip arthroplasty for OA, a surrogate marker for severe hip OA. Men and women were evaluated separately because BMD, bone shape, and incidence and severity of OA differ between the sexes.

Statistics Question?

The WHO criterion for osteoporosis is a BMD 2.5 standard deviations below the mean for young adults.
a) What percentage of healthy young adults have osteoporosis by the WHO criterion?

b) Women aged 70-79 are of course not young.


a) The quick and dirty way is to use Ti-83 here:
2nd,vars, choice 2
normalcdf(-9999,-2.5,0,1)
where 0 is your mean 1 is your standard deviation which are the properties of normal distribution
normalcdf(-9999,-2.5,0,1)= 0.0062096799


a) The quick and dirty way is to use Ti-83 here:
2nd,vars, choice 2
normalcdf(-9999,-2.5,0,1)
where 0 is your mean 1 is your standard deviation which are the properties of normal distribution
normalcdf(-9999,-2.5,0,1)= 0.0062096799

Please help, this is urgent, and I really have tried many times, without any success. Please!?

Eleanor scores 641 on the mathematics part of the SAT. The distribution of SAT math scores in recent years has been Normal with mean 534 and standard deviation 110. Gerald takes the ACT mathematics test and scores 25.8. ACT math scores are Normally distributed


Formula: X-mean/SD Convert your answers to percents using Table A.
Part 1: (641-534)/110 = .9727 about 83.5% Part 2: (25.8-16.9)/6.4 = 1.3906 about 92%
Gerald Wins!

Use table A as well.
SD=Z
So the percent of healthy


Formula: X-mean/SD Convert your answers to percents using Table A.
Part 1: (641-534)/110 = .9727 about 83.5% Part 2: (25.8-16.9)/6.4 = 1.3906 about 92%
Gerald Wins!

Use table A as well.
SD=Z
So the percent of healthy