The volume of apple juice in cans filled by a certain machine is normally distri

Question

The volume of apple juice in cans filled by a certain machine is normally distributed with mean 12.05 oz and standard deviation 0.03 oz. (a) What proportion of cans contain less than 12 oz of apple juice? (b) The mean volume of apple juice can be adjusted by calibrating the machine. If the standard deviation is fixed at 0.03 oz, to what value should the mean be set so that 99% of the cans will contain 12 oz or more of apple juice? (c) If instead we modify the standard deviation but fix the mean at 12.05 oz, to what value must we set the standard deviation so that 99% of the cans will contain 12 oz or more of apple juice?

Explanation / Answer

A) mean m = 12.05 oz

Standard deviation sd= 0.03 oz

Z= (X-m)/sd

P(X<12) = P(Z<(12-12.05)/0.03) = P(z<-1.67)

= 0.0475

So,4.75% will have less than 12 oz

B) For 99% to have more than12 oz, P(z<z1) = 0.01

Z1 = -2.327

So, (12- M)/0.03 = -2.327

M = 12.07 oz

The mean shall be set as 12.07 oz

C) Z1 = -2.327

(12-12.05)/sd = -2.327

sd = 0.021

So, the standard deviation must be set as 0.021 oz

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