document.write( "Question 1197941: In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.\r
\n" ); document.write( "\n" ); document.write( "It is estimated that 3.5% of the general population will live past their 90th birthday. In a graduating class of 781 high school seniors, find the following probabilities. (Round your answers to four decimal places.)
\n" ); document.write( "(a) 15 or more will live beyond their 90th birthday\r
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\n" ); document.write( "\n" ); document.write( "(b) 30 or more will live beyond their 90th birthday\r
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\n" ); document.write( "\n" ); document.write( "(c) between 25 and 35 will live beyond their 90th birthday\r
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\n" ); document.write( "\n" ); document.write( "(d) more than 40 will live beyond their 90th birthday
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Algebra.Com's Answer #831414 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
Binomial Distribution:
\n" ); document.write( "n = 781 , p = .035
\n" ); document.write( " 781*.035 and 781*.965 are > 5
\n" ); document.write( "Using the normal approximation and continuity correction factor.
\n" ); document.write( "(the continuity correction factor used as a Binomial Distribution is not continuous)
\n" ); document.write( " mean =781*.035= 27.335
\n" ); document.write( " sd = \"sqrt%28781%2A.035%2A.965%29\" = 5.136
\n" ); document.write( "Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
\n" ); document.write( "P(x ≥ 15 ) = normpdf(-9999 14.5, 27.335,5.136)= .0062
\n" ); document.write( "P(x ≥ 30 ) = normpdf(-9999 29.5, 27.335,5.136)= .6633
\n" ); document.write( "P( 25 ≤ x ≤ 35) = normcdf( 24.5,35.5, 27.335,5.136) = .6536
\n" ); document.write( "P(x > 40) = normpdf(40.5,9999, 27.335,5.136) = .0052
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