An automatic machine is used to dispense a particular soft drink in bottles with a nominal quantity (volume) of 1.5l. If the machine puts more than 1.6l of the soft drink in a bottle, the excess quantity is wasted. The actual volume that the machine dispenses in the bottles is a random variable

Question

An automatic machine is used to dispense a particular soft drink in bottles with a nominal quantity (volume) of 1.5l. If the machine puts more than 1.6l of the soft drink in a bottle, the excess quantity is wasted.
The actual volume that the machine dispenses in the bottles is a random variable which has the normal distribution with mean 𝜇 (in liters) and a standard deviation 0.02l. What is the mean quantity 𝜇 that should be set for the machine so that the probability that it dispenses more than 1.6l in a bottle is 0.2%?

Accepted Answer

Let X denote the random variable corresponding to the amount dispensed by the machine in a bottle. We are given that X ∼ N(𝜇,(0.02)²), and we then seek the value of 𝜇 such that

[latex]P(X \gt 1.6) = 0.002.[/latex]

Since, by Proposition 7.6,

[latex]P(X\gt 1.6)=1-\Phi\left({\frac{1.6-\mu}{0.02}}\right),[/latex]

we want to find 𝜇 such that

[latex]1-\Phi\left({\frac{1.6-\mu}{0.02}}\right)=0.002[/latex]

which gives

[latex]\Phi\left({\frac{1.6-\mu}{0.02}}\right)=1-0.002=0.998.[/latex]

From the tables of the standard normal distribution, we see that the value z such that Φ(z) = 0.998 is z = 2.05. Since Φ is a one-to-one function (as it is strictly increasing), we deduce that we must have

[latex]{\frac{1.6-\mu}{0.02}}=2.05.[/latex]

Upon solving this, we obtain 𝜇 = 1.6 − (0.02)(2.05) = 1.559, which is the desired result.

Question 7.7: An automatic machine is used to dispense a particular soft d......

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