document.write( "Question 1020224: A baker makes digestive biscuit whose masses are normally distributed with a mean of 26g and a standard deviation of 1.9g. The biscuits are packed by hand into packets of 25. Assuming the biscuits are a random sample from the population, what is the distribution of the total mass of biscuits in a packet and what is the probability that it lies between 598 and 606. \n" ); document.write( "

Algebra.Com's Answer #636128 by robertb(5830) $\"\"$ $\"About$

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The distribution of the total mass would still be normally distributed with mean 25*26 =650, and variance $\"25%2A%281.9%29%5E2\"$ , or standard deviation $\"sqrt%2825%2A%281.9%29%5E2%29+=+5%2A%281.9%29+=+9.5\"$ , assuming i.i.d. (identical and independently distributed) for each biscuit.
\n" ); document.write( "Let S be the random variable for the total mass. The z score is found by the equation\r
\n" ); document.write( "\n" ); document.write( " $\"z+=+%28S-650%29%2F9.5\"$ \r
\n" ); document.write( "\n" ); document.write( "The z-score for 598 is $\"z+=+%28598-650%29%2F9.5+=+-5.47\"$ , while the z-score for
\n" ); document.write( "606 is $\"z+=+%28606-650%29%2F9.5+=+-4.63\"$ .
\n" ); document.write( "Now use the standard normal table to find the area between the two z-scores. \n" ); document.write( "

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