document.write( "Question 1143157: A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.5 years, and standard deviation of 1.2 years.\r
\n" ); document.write( "\n" ); document.write( " If you randomly purchase 16 items, what is the probability that their mean life will be longer than 8 years? \n" ); document.write( "

Algebra.Com's Answer #763935 by rothauserc(4718) $\"\"$ $\"About$

You can put this solution on YOUR website!
The probability(P) that their mean life will be longer than 8 years = 1 - P(X < 8)
\n" ); document.write( ":
\n" ); document.write( "Since the population standard deviation is known, we can use the normal distribution for the sample with the standard error
\n" ); document.write( ":
\n" ); document.write( "standard error(SE) = population standard deviation/square root of sample size
\n" ); document.write( ":
\n" ); document.write( "SE = 1.2/square root(16) = 1.2/4 = 0.3
\n" ); document.write( ":
\n" ); document.write( "z-score(8) = (8 - 8.5)/0.3 = −1.6667 is approximately -1.67
\n" ); document.write( ":
\n" ); document.write( "P associated with z-score of −1.67 is 0.0475 (use z-score tables)
\n" ); document.write( ":
\n" ); document.write( "************************************************
\n" ); document.write( "P(X > 8) = 1 - P(X < 8) = 1 - 0.0475 = 0.9525
\n" ); document.write( "***********************************************
\n" ); document.write( ": \n" ); document.write( "

\n" );