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. Suppose, that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 fluid ounce. (a) At what value should the mean be set so that 99.9% of all cans exceed 10 fluid ounces? Round your answer to three decimal places (e.g. 98.765) fluid ounces (b) At what value should the mean be set so that 99.9% of all cans exceed 10 fluid ounces if the standard deviation can be reduced to 0.07 fluid ounce? Round your answer to three decimal places (e.g. 98.765) fluid ounces

Explanation / Answer

Solution : Given that standard deviation ? = 0.1

(a) P(x > 10) = 0.999

=> P((x - ?)/? > (10 - ?)/0.1) = 0.999

=> P(Z > (10 - ?)/0.1) = 0.999

=> 1 - P(Z <= (10 - ?)/0.1) = 0.999

=> P(Z <= (10 - ?)/0.1) = 1 - 0.999 = 0.001

=> (10 - ?)/0.1 = -3.09

=> 10 - ? = -0.309

=> ? = 10 + 0.309

=> ? = 10.309

(b) standard deviation ? = 0.07

P(x > 10) = 0.999

=> P((x - ?)/? > (10 - ?)/0.07) = 0.999

=> P(Z > (10 - ?)/0.07) = 0.999

=> 1 - P(Z <= (10 - ?)/0.07) = 0.999

=> P(Z <= (10 - ?)/0.07) = 1 - 0.999 = 0.001

=> (10 - ?)/0.07 = -3.09

=> 10 - ? = -0.216

=> ? = 10 + 0.216

=> ? = 10.216

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