SOLUTION: 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

Question 423697: 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
A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
ACCORDING TO A RECENT STUDY 38% OF ALL WOMEN WILL SUFFER A HIP FRACTURE BECAUSE OF OSTEOPOROSIS BY THE AGE OF 85.
---
P(will suffer) = 0.38
P(will not suffer) = 0.62
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IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT
A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(x = 0) = (0.62)^6
------------------------------
b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(4<= x <= 6) = 1 - P(x<= 0 <=3) = 1-binomcdf(6,0.38,3) = 0.1527
--------
c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(0<= x <=2) = binomcdf(6,0.38,2) = 0.5857
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Cheers,
Stan H.

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