document.write( "Question 820453: What is the mean number of rolls of a die before a 1 is observed? Roll a die until a 1 is observed. Repeat this process 30 times and answer the following questions.\r
\n" ); document.write( "\n" ); document.write( " Obtain a point estimate of the mean number of rolls of a die before a 1 is observed.\r
\n" ); document.write( "\n" ); document.write( " The population standard deviation for the number of rolls before a 1 is observed. Use this result to construct a 90% Z-interval for the mean number of rolls required before a 1 is observed.\r
\n" ); document.write( "\n" ); document.write( " The population mean number of rolls before a 1 is observed is 6. Does your interval include 6? What proportion of the Z-intervals in the class included 6? How many did you expect to include 6?\r
\n" ); document.write( "\n" ); document.write( " Construct a 90% t-interval for the mean number of rolls required before a 1 is observed.\r
\n" ); document.write( "\n" ); document.write( " The population mean number of rolls before a 1 is observed is 6. Does your interval include 6? What proportion of the t-intervals in the class includes 6? How many did you expect to include 6?\r
\n" ); document.write( "\n" ); document.write( " Compare the Z-interval with the t-interval. Which has the smaller margin of error?
\n" ); document.write( "i tried this and so far I came up with this:
\n" ); document.write( "6+1+3+1+4+4+5+2+6+3
\n" ); document.write( "5+4+2+1+2+5+4+2+6+5
\n" ); document.write( "5+3+4+5+5+6+6+6+3+3=117
\n" ); document.write( "add and divide by 30=3510 \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #532398 by loubennett(1) $\"\"$ $\"About$

You can put this solution on YOUR website!
1-p/^p^2=30=sqrt.0)6+-1.645 sqrt.(30)/sqrt.(30) =6+-1.645 \n" ); document.write( "

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