40% of postmenopausal women have low bone density (osteopenia), placing them at

Question

40% 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 248 postmenopausal primary patients and has diagnosed 83 with low bone density.

(a) Find the probability that only 83 or fewer of the 248 patients have low bone density [hint: use the Normal approximation to the binomial distribution].

If possible, please explain how to do it with TI-84

Explanation / Answer

We have the params , lets solve by taking the params first:

pcap = 83/248 = .3347

p= .4

n=248

Calc. test statistic:
Z=pcap-p / sqrt(p*p'/n)

= (.3347-.4)/sqrt(.4*.6/248)

=-2.1

P(Z<-2.1) = 1-.9821 = .0179 ( from Z tables)

or 1.79% chance that only 83 or fewer patients out of the 248 patients have low bone density

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