The fill volume of an automated filling machine used for filling cans of carbona

Question

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces and a standarod deviation of 0.1 fluid ounce. Round the answers to 3 significant digits. a) What is the probability a fill volume is less than 12 fluid ounces? b) If all cans less than 12.1 or greater than 12.6 ounces are scrapped, what proportion of cans is scrapped? Round your answer to 4 decimal places.) % (Round your answer to 2 decimal places.) c) Determine specifications (in ounces) that are symmetric about the mean that include 99% of all cans. ) (Round your answers to 3 decimal places.)

Explanation / Answer

1. The fill volume...

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    12      
u = mean =    12.4      
          
s = standard deviation =    0.1      
          
Thus,          
          
z = (x - u) / s =    -4      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -4   ) =    3.16712*10^-5 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    12.1      
x2 = upper bound =    12.6      
u = mean =    12.4      
          
s = standard deviation =    0.1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.97589997      

Thus, those outside this interval is the complement =    0.02410003 = 2.41% [ANSWER]

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c)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.99      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.005      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -2.575829304      
By symmetry,          
z2 =    2.575829304      
          
As          
          
u = mean =    12.4      
s = standard deviation =    0.1      
          
Then          
          
x1 = u + z1*s =    12.14241707      
x2 = u + z2*s =    12.65758293      

Hence, between 12.142 and 12.658. [ANSWER]

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