According to a recent study, 38% of all women will suffer a hipfracture because

Question

According to a recent study, 38% of all women will suffer a hipfracture because of osteporosis by the age of 85. If six women aged85 are randomly selected, what is the probability that.

a) None of them will suffer(or has suffered) a hip fracture due toosteo...
b) At least four of them will suffer( or have suffered) a hipfracture due to osteop...
c) Fewer than two of them will suffer(or have suffered) a hipfracture due to osteoporosis?

For A) is this correct?

P(x=0)= 6C0= 6!/ 0!(6-0)!= 6!/0!(6)!=
6*5*4*3*2*1/1*6*5*4*3*2*1= 720/720= 1

B) Not sure

C) Not sure

Explanation / Answer

a)      Ans to part A is

P(a particular woman aged 85 has suffered fracture) = 0.38(given )

We want no one to suffer

P(X=0) = this follow a binomial distribution with n = 6 and p =0.38 and (1-p) = 1-0.38 = 0.62

Therefore P(X=k) = 6Ck 0.38k0.62(6-k)

P(x=0) = 6c0 0.626 = 1* 5.68% =5.68%

B) At least 4 of them have suffered or will suffer.

P(X4) =P(x=4)+P(x=5)+P(x=6)

= 6C4 0.384 0.62(6-4)+6C5 0.385 0.62(6-5)+6Ck 0.386 0.62(6-6) =15*2.085%*38.44% +6*0.7924%*62%+1*0.3011%*1

=15.27%

C) Fewer than 2 of them will suffer or have suffered.

P(X<2) =P(x=1)+P(x=0)

=6C1 0.381 0.62(6-1) +6c0 0.626   = 20.89% + 5.68% =26.57%

Hope it helps you. Feel free to ask for any clarifications

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