Question

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.

Answer 1

Answer:

The value of mean is  $\mu = 15.64$

Step-by-step explanation:

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

                   $X$ ~ $N (\mu ,0.191^2)$

The standard form representation is

           $P(X$

     Now we need to obtain $\mu$ in such a way that

                   $P(X>16) = 0.01$

   In standard form

       Since

                 $P(Z>16 -\frac{\mu}{0.191} ) = 0.03$

   This means

              $P(Z$

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

    i.e

            $16 - \frac{\mu}{0.191} = 1.88$

Making $\mu\ the \ subject$

        $16 - \mu = 1.88 *0.191$

       $\mu = 16 -(1.88*0.191)$

          $=15.64$