Solutions to Coin Questions 2

If the Master of the Mint sets things up so that the guineas weigh only 127.7 gr., then the population of interest remains the same as in the last question but with a mean of 127.7 gr. and SD still 1 gr.

The expected value of the sum S of the weights of the 100 coins in the sample is now E(S) = n* E(S) = 12770 gr, with SE(S)=10 gr. As before the long run histogram of S is normal, centered at 12770 gr and with variance equal to 100 (10 squared).

Then

P(survive) = P(12768 < S < 12832) 
           = P( (12768-12770)/10 < Z < (12832-12770)/10 )
           = P( -.2 < Z < 6.2 )
which is approximately 58 %. Therefore, his chances of surviving are better than 1 in 2.

However, if he manages to survive he will keep a lot of gold. It will take him n * E(X) = 12,770,000 gr to manufacture the 100,000 coins, give or take SD * (square root of n) = 1 * (square root of 1000,000) = 316 gr. He will receive 100000 * 128 = 12,800,000 gr to manufacture them, so he will end up with 12,800,000 - 12,770,000 = 30,000 gr, give or take about 316 gr.

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