7) A wizard has a dragon which produces delicious eggs, which he sells at a market once a year. The mass of the eggs are normally distributed with a mean of 185g and standard deviation 10g. Eggs with a mass below 100g are not sold. Eggs with a mass between 160g and 200g are sold for 40 schmeckles. Eggs with a mass above 200g are sold for 75 schmeckles. The dragon produces 52 eggs a year. What is the wizard's expected income at the market? Solution? We use normalcdf with a lower bound of 160, upper bound 200, mean 185, and standard deviation 10 to get approximately 0.92698. We use normalcdf with a lower bound of 200, upper bound 999999, mean 185, and standard deviation 10 to get approximately 0.066807. The expected number of eggs between 160g and 200g will be 52 × 0.92698 × 48.20 eggs. The expected number of eggs greater than 200g will be 52 × 0.066807 × 3.474 eggs. The expected income is therefore approximately 40 x 48.20 + 75 × 3.474 × 2188.55 schmeckles. 7b) What is the probability that at least 4 of the dragon's 52 eggs will be more than 200g? Solution? (Note: some of the work from 7a is used. Assume that it is correct in your analysis of this problem.) Let X be the number of eggs which are more than 200g. X~B(0.066807,52) Using binomialcdf, we find the probability is approximately 0.461
7) A wizard has a dragon which produces delicious eggs, which he sells at a market once a year. The mass of the eggs are normally distributed with a mean of 185g and standard deviation 10g. Eggs with a mass below 100g are not sold. Eggs with a mass between 160g and 200g are sold for 40 schmeckles. Eggs with a mass above 200g are sold for 75 schmeckles. The dragon produces 52 eggs a year. What is the wizard's expected income at the market? Solution? We use normalcdf with a lower bound of 160, upper bound 200, mean 185, and standard deviation 10 to get approximately 0.92698. We use normalcdf with a lower bound of 200, upper bound 999999, mean 185, and standard deviation 10 to get approximately 0.066807. The expected number of eggs between 160g and 200g will be 52 × 0.92698 × 48.20 eggs. The expected number of eggs greater than 200g will be 52 × 0.066807 × 3.474 eggs. The expected income is therefore approximately 40 x 48.20 + 75 × 3.474 × 2188.55 schmeckles. 7b) What is the probability that at least 4 of the dragon's 52 eggs will be more than 200g? Solution? (Note: some of the work from 7a is used. Assume that it is correct in your analysis of this problem.) Let X be the number of eggs which are more than 200g. X~B(0.066807,52) Using binomialcdf, we find the probability is approximately 0.461
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Solve 7B using information from 7a.
Transcribed Image Text:7) A wizard has a dragon which produces delicious eggs, which he sells at a market once a year. The mass of the
eggs are normally distributed with a mean of 185g and standard deviation 10g. Eggs with a mass below 100g are
not sold. Eggs with a mass between 160g and 200g are sold for 40 schmeckles. Eggs with a mass above 200g are
sold for 75 schmeckles. The dragon produces 52 eggs a year. What is the wizard's expected income at the market?
Solution?
We use normalcdf with a lower bound of 160, upper bound 200, mean 185, and standard deviation 10 to get
approximately 0.92698.
We use normalcdf with a lower bound of 200, upper bound 999999, mean 185, and standard deviation 10 to get
approximately 0.066807.
The expected number of eggs between 160g and 200g will be 52 × 0.92698 × 48.20 eggs.
The expected number of eggs greater than 200g will be 52 x 0.066807 = 3.474 eggs.
The expected income is therefore approximately 40 x 48.20 + 75 x 3.474 z 2188.55 schmeckles.
7b) What is the probability that at least 4 of the dragon's 52 eggs will be more than 200g?
Solution? (Note: some of the work from 7a is used. Assume that it is correct in your analysis of this problem.)
Let X be the number of eggs which are more than 200g.
X~B(0.066807,52)
Using binomialcdf, we find the probability is approximately 0.461
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