39 of postmenopausal women have low bone density osteopenia

39% of postmenopausal women have low bone density (osteopenia), placing them at risk for osteoporosis with ensuing spontaneous fractures. Osteoporosis is estimated to cost $14 billion per year in medical expenses alone, and yet it can be prevented if treated early enough. A physicians group has 245 postmenopausal primary patients and has diagnosed 80 with low bone density.

a) Find the probability that only 80 or fewer of the 245 patients have low bone density [hint: use the Normal approximation to the binomial distribution].
(Use 4 decimal places.)

(b) Based on your answer, what would you suggest at the next physicians meeting?

No conclusions can be drawn. A larger sample is needed. Because of the low cumulative probability for 80 patients or fewer, it is likely that cases of osteopenia have been missed (not diagnosed).     Because of the high cumulative probability for 80 patients or fewer, it is likely that cases of osteopenia have been misdiagnosed (too many diagnosed).The diagnosis of the 80 patients is consistent with expectations, and no change is needed.

Solution

binomial probability model
n = 249 , p = 0.39 , q = 0.61

P[k] = nCk*p^k*q^(n-k)

P[<=81] = P[0] + P[1] + ... + P[81]

using a binomial calculator, P[<=81] = 0.0204 or 2.04 %

qb
for the normal approximation to the binomial distribution,
mean = np = 97.11, SD = sqrt(npq) = 7.697

applying the continuity correction
for using a continuous distribution
for a discrete distribution,
<= 81 becomes <= 81.5
z value = [81.5 - 97.11] / 7.697 = - 2.03

reqd. probability = 0.0212 or 2.12 %