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4) According to the NIH, 32% of all women will fracture their hip by age 90. If 8 women aged 90 are selected at random, what is the probability that exactly 5 of them will have suffered a hip fracture?
p=0.32 ; q=0.68; n=8; r=5
(nCr)(p^r)[q^(1-r)]
(8C5)(0.32^5)(0.68^3)=0.059
How did you get q=0.68?
It took me a while to figure out where we learnt how do do this example.
To get q you do 1-p so in this case 1-0.32 = 0.68 also the equation is written wrong. It is (nCr) (p^r) (q^n-r) so (8nCr5)(0.32^5)(0.68^3)=0.059 or 5.9%
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