SOLUTION: 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. Assum

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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.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
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.
Let S be the random variable for the total mass. The z score is found by the equation
z+=+%28S-650%29%2F9.5
The z-score for 598 is z+=+%28598-650%29%2F9.5+=+-5.47, while the z-score for
606 is z+=+%28606-650%29%2F9.5+=+-4.63.
Now use the standard normal table to find the area between the two z-scores.