document.write( "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
\n" ); document.write( "A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
\n" ); document.write( "b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
\n" ); document.write( "c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
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Algebra.Com's Answer #295442 by stanbon(75887)\"\" \"About 
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.
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\n" ); document.write( "P(will suffer) = 0.38
\n" ); document.write( "P(will not suffer) = 0.62
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\n" ); document.write( "IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT
\n" ); document.write( "A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
\n" ); document.write( "P(x = 0) = (0.62)^6
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\n" ); document.write( "\n" ); document.write( "b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
\n" ); document.write( "P(4<= x <= 6) = 1 - P(x<= 0 <=3) = 1-binomcdf(6,0.38,3) = 0.1527
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\n" ); document.write( "c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
\n" ); document.write( "P(0<= x <=2) = binomcdf(6,0.38,2) = 0.5857
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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