SOLUTION: 3. Cans of regular Coke are labeled as containing 12.00 oz. Assume that the actual contents are normally distributed with a mean of 12.19 oz and a standard deviation of 0.11 oz.

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Question 357946: 3. Cans of regular Coke are labeled as containing 12.00 oz. Assume that the actual contents are normally distributed with a mean of 12.19 oz and a standard deviation of 0.11 oz.
a. What percentage of cans contain less than 12.10 oz ?




b. What amount of Coke separates the top 10% of the cans from the others?




c. To save some of the Coke in cans that are filled too much, the quality-control manager is instituting a plan to remove some of the Coke from cans that are in
the top 2%. What is the minimum amount of Coke in such cans that will be “reworked”?




d. What is the maximum amount of Coke in the cans that are in the bottom 15% ?

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Cans of regular Coke are labeled as containing 12.00 oz. Assume that the actual contents are normally distributed with a mean of 12.19 oz and a standard deviation of 0.11 oz.
a. What percentage of cans contain less than 12.10 oz ?
---
z(12.10) = (12.10-12.19)/0.11 = -0.8182
---
P(x<12.10) = P(z< -0.8182) = 0.2066
=============================================
b. What amount of Coke separates the top 10% of the cans from the others?
Find the z with a right tail of 10%: z = 1.2816
----
x = 1.2816*0.11+12.19 = 12.33 ounces
=============================================
c. To save some of the Coke in cans that are filled too much, the quality-control manager is instituting a plan to remove some of the Coke from cans that are in the top 2%.
What is the minimum amount of Coke in such cans that will be “reworked”?
Find the z-value that has a 2% right tail.
invNorm(0.98) = 2.0537
---
x = 2.0537*0.11+12.19 = 12.42 ounces
------------------------------------------
d. What is the maximum amount of Coke in the cans that are in the bottom 15% ?
Find the z with a left tail of 15%
invNorm(0.15) = -1.0364
x = -1.0364*0.11+12.19 = 12.0760
===================================
Cheers,
Stan H.

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