document.write( "Question 1128016: The Federal Reserve reports that the mean lifespan of a five dollar bill is 4.9 years. Let’s suppose that the standard deviation is 1.9 years and that the distribution of lifespans is normal
\n" ); document.write( "Find:
\n" ); document.write( "(a) the probability that a $5 bill will last more than 4 years.\r
\n" ); document.write( "\n" ); document.write( "(b) the probability that a $5 bill will last between 3 and 5 years.\r
\n" ); document.write( "\n" ); document.write( "(c) the 97th percentile for the lifespan of these bills (a time such that 97% of bills last less than that time).\r
\n" ); document.write( "\n" ); document.write( "(d ) the probability that a random sample of 37 bills has a mean lifespan of more than 4.5 years.
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Algebra.Com's Answer #744479 by Boreal(15235)\"\" \"About 
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z=(x-mean)/sd
\n" ); document.write( "a. z=(4-4.9)/1.9=-0.47
\n" ); document.write( "probability z>-0.47 is 0.6808
\n" ); document.write( "b. this is a z between -1.9/1.9 or -1 and .1/1.9 or 0.05
\n" ); document.write( "probability is z between -1 and 0.05, or 0.3613
\n" ); document.write( "c. 97th percentile is z>1.88
\n" ); document.write( "1.88=(x-4.9)/1.9
\n" ); document.write( "=8.47 years
\n" ); document.write( "d. This is a z> (4.5-4.9)/1.9/sqrt(35), sampling distribution of sample means.
\n" ); document.write( "z> -0.4*sqrt(35)/1.9=-1.25
\n" ); document.write( "probability is 0.8944
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