Question 623540: ACCORDING TO A RECENT STUDY 38% OF ALL WOMEN WILL SUFFER A HIP FRACTURE BECAUSE OF OSTEOPOROSIS BY THE AGE OF 85. IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
please show how you obtained answer without The binomcdf( Command as that will not help me Thanx
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Your post is too difficult to read when typed in all caps (aside from the fact that all caps is shouting and is therefore both annoying and rude). Repost so that it is easier to read.
John
My calculator said it, I believe it, that settles it
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! if the woman is 85, then the probability of having a hip fracture is .38
this means that the probability of not having a hip fracture is 1 - .38 = .62
p(having) + p(not having) = .38 + .62 = 1.0 as it should.
since this is a binomial distribution, you have to use the binomial distribution formula to get the answers.
that formula is:
p(x) = C(n,x) * p^x * q^(n-x)
n is equal 6 because you're dealing with 6 women.
x is equal to the number of hip injuries among the 6 women.
p is equal to the probability of having a hip injury.
q is equal to the probability of not having a hip injury.
C(n,x) is the combination formula of n! / (x! * (n-x)!)
as an example:
when x is equal to 0, the formula tells you that the probability is:
p(0) = C(6,0) * .38^0 * .62^6
when x is equal to 1, the formula tells you that the probability is:
p(1) = C(6,1) * .38^1 * .62^5
when x is equal to 2, the formula tells you that the probability is:
p(2) = C(6,2) * .38^2 * .62^4
etc down to x is equal to 6, the formula tells you that the probability is:
p(6) = C(6,6) * .38^6 * .62^0
you can't solve the problem without using the formula.
when you apply the formula, you get the following:
x (n-x) C(6,x) p^x q^(n-x) p^x*q^(n-x) C(6,x)*p^x*q^(n-x)
0 6 1 1 0.056800236 0.056800236 0.056800236
1 5 6 0.38 0.091613283 0.034813048 0.208878286
2 4 15 0.1444 0.14776336 0.021337029 0.320055438
3 3 20 0.054872 0.238328 0.013077534 0.26155068
4 2 15 0.02085136 0.3844 0.008015263 0.120228942
5 1 6 0.007923517 0.62 0.00491258 0.029475482
6 0 1 0.003010936 1 0.003010936 0.003010936
sum of all probabilities equals >>>>>>>>> 1
you now have all the information you need to solve the problem.
the problem states:
IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
if fewer than 2 of them suffer, then that would be equal to the probability of p(0) + p(1).
from the chart, p(0) = 0.056800236 and p(1) = 0.208878286
add these together and you get the probability that fewer than 2 will suffer is equal to .265678522.
you only need to calculate the probability of 0 or 1 sufferers because the problem stated that you wanted the probability that fewer than 2 suffered.
i shows you the whole table to show you how the formula works and to demonstrate that the total probability had to be equal to 1 as it was based on the use of those formulas.
in the table, that last column on the right is equal to p(x).
example:
when x = 0, p(x) = 0.056800236
when x = 6, p(6) = 0.003010936
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