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The amount dispensed by a soft-drink dispensing machine has a normal distribution with mean μ and standard deviation 0.191 ounce. The mean amount dispensed, μ , can be controlled. At what value should μ be set so that a 16 ounce cup will overflow with probability 0.03? Use RStudio and round your answer to two decimal places.

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okpalawalter8

Answer:

The value of mean is  [tex]\mu = 15.64[/tex]

Step-by-step explanation:

The normal distribution of the soft-drink dispensing machine can be stated statistically like

                   [tex]X[/tex] ~ [tex]N (\mu ,0.191^2)[/tex]

The standard form representation is

           [tex]P(X<x) = P(Z<x- \frac{\mu}{\sigma} )[/tex]

     Now we need to obtain [tex]\mu[/tex] in such a way that

                   [tex]P(X>16) = 0.01[/tex]

   In standard form

       Since

                 [tex]P(Z>16 -\frac{\mu}{0.191} ) = 0.03[/tex]

   This means

              [tex]P(Z<16 - \frac{\mu}{0.191} ) = 0.97[/tex]

Now looking at the z-table for probability of 0.99 we obtain 1.88

    i.e

            [tex]16 - \frac{\mu}{0.191} = 1.88[/tex]

Making [tex]\mu\ the \ subject[/tex]

        [tex]16 - \mu = 1.88 *0.191[/tex]

       [tex]\mu = 16 -(1.88*0.191)[/tex]

          [tex]=15.64[/tex]

   

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