document.write( "Question 853158: I need help solving this problem. Can you please help me?\r
\n" ); document.write( "\n" ); document.write( "A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20 adult women are randomly selected, find the probability that fewer than 15 of them have had a physical examination in the past two years.\r
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Algebra.Com's Answer #513919 by stanbon(75887)\"\" \"About 
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A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20 adult women are randomly selected, find the probability that fewer than 15 of them have had a physical examination in the past two years.
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\n" ); document.write( "Binomial Problem with n = 20 ; p(visit) = 0.7 ; p(not visit) = 0.3
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\n" ); document.write( "P(0<= x <=14) = binomcdf(20,0.7,14) = 0.5836
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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