document.write( "Question 846156: 22. Smokers, According to Information Please Almanac, 80% of adult smokers started smoking before they were 18 years old. Suppose 100 smokers 18 years old or older are randomly selected. Use the normal approximation to the binomial to:
\n" ); document.write( "(a) Approximate the probability that exactly 80 of them started smoking before they were 18 years old.
\n" ); document.write( "(b) Approximate the probability that at least 80 of them started smoking before they were 18 years old.
\n" ); document.write( "(c) Approximate the probability that fewer than 70 of them started smoking before they were 18 years old.
\n" ); document.write( "(d) Approximate the probability that between 70 and 90 of them, inclusive, started smoking before they were 18 years old.
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #509638 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
\n" );
document.write( "Hi,
\n" );
document.write( ".0993 vs .1034, .5595 vs .5517, .0061 vs .0043, .9916 vs .9914 ???
\n" );
document.write( "book may find few who would agree with those answers given
\n" );
document.write( "for ex. long hand: P(x=80) = 100C80*  = .0993
\n" );
document.write( " p = .80, n = 100
\n" );
document.write( "Using TI Calculator
\n" );
document.write( "P(x = 80)= binompdf(n, p, x-value) = binompdf(100, .80, 80)= .0993  0r  9.93%
\n" );
document.write( "P(x ≥ 80)= 1 – binomcdf(n, p, largest x-value)= 1 - binomcdf(100, .80,79)= .5595  0r 55.95%
\n" );
document.write( "P( x < 70)=  binomcdf(n, p, largest x-value)= binomcdf(100, .80, 69) = .0061 0r  .66%
\n" );
document.write( "P( 70≤ x ≤90) = binomcdf(100, .80, 90) - binomcdf(100, .80, 70)
\n" );
document.write( "   .9977- .0061 = .9916   0r 99.16% \n" );
document.write( "

\n" );