SOLUTION: Hi,I am trying to do a practice sheet to study for the final and I am really struggling, your help is really appreciated. The amount of soda a dispensing machine pours into a 12

Question 311295: Hi,I am trying to do a practice sheet to study for the final and I am really struggling, your help is really appreciated.
The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with mean of 12.03 ounces and a standard deviation of .02 ounces. The cans can hold only 12.05 ounces of soda. Every can that has more than 12.05 ounces of soda poured into it causes aa spill and the can needs to go through a special cleaning process before it can be sold.
A. What proportion of the cans need to go through the cleaning process?
B. How much soda would the cans need t hold if the company wanted to limit the percentage that needed to be cleaned because of spillage to 3%?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Hi,I am trying to do a practice sheet to study for the final and I am really struggling, your help is really appreciated.
----------------------------
The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with mean of 12.03 ounces and a standard deviation of .02 ounces.
----------------------------
The cans can hold only 12.05 ounces of soda. Every can that has more than 12.05 ounces of soda poured into it causes aa spill and the can needs to go through a special cleaning process before it can be sold.
----------------------------

A. What proportion of the cans need to go through the cleaning process?
Find the z-value of 12.05
z(12.05) = (12.05-12.03)/0.02 = 1
---
P(x > 12.05) = P(z > 1) = 0.1587
====================================
B. How much soda would the cans need to hold if the company wanted to limit the percentage that needed to be cleaned because of spillage to 3%?
---
Find the z-value with a 3% right tail:
invNorm(0.97) = 1.8808
-------------------------
Find the corresponding x-value:
x = zs + u
x = 1.8808*0.02 + 12.03
x = 12.07
=============
Cheers,
Stan H.
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