document.write( "Question 805605: A very reliable baseball player is known to get a hit when at bat 43% of the time.
\n" ); document.write( "He is expected to have 102 more times at bat before the end of the season, and he is 48 hits away from breaking a league hitting record.\r
\n" ); document.write( "\n" ); document.write( "What is the probability of him breaking that record before the end of the season?\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #485357 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
A very reliable baseball player is known to get a hit when at bat 43% of the time.
\n" ); document.write( "He is expected to have 102 more times at bat before the end of the season, and he is 48 hits away from breaking a league hitting record.\r
\n" ); document.write( "\n" ); document.write( "What is the probability of him breaking that record before the end of the season?
\n" ); document.write( "
\r\n" );
document.write( "This is a binomial probability with n=102 trials\r\n" );
document.write( "with probability of 1 hit in 1 trial is p=0.43\r\n" );
document.write( "we want to find P(x>=48) \r\n" );
document.write( "\r\n" );
document.write( "Using a TI-84,\r\n" );
document.write( "\r\n" );
document.write( "1-binomcdf(102,.43,47)\r\n" );
document.write( "\r\n" );
document.write( "Answer:  .2327171125\r\n" );
document.write( "\r\n" );
document.write( "Or if you have to use tables and the normal approximation, find the \r\n" );
document.write( "z-score of 47.5, with m=np=(102)(.43) = 43.86 and s=√np(1-p)=√(102)(.43)(1-.43) = 5.00002, use 5\r\n" );
document.write( "\r\n" );
document.write( "z = (x - m)/s\r\n" );
document.write( "\r\n" );
document.write( "z = (47.5-43.86)/5 = 3.64/5 = .728\r\n" );
document.write( "\r\n" );
document.write( "There are two kinds of tables\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Look in the table for 0.73 and \r\n" );
document.write( "\r\n" );
document.write( "if you find in your table that the area to the left of \r\n" );
document.write( "0.73 is 0.7673, then you subtract from 1 and get .2327.\r\n" );
document.write( "\r\n" );
document.write( "or\r\n" );
document.write( "\r\n" );
document.write( "if you find that the area between 0 and 0.73 is 0.2673, \r\n" );
document.write( "then you subtract from .5000 and get .2327.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );