The Nelson Company makes the machines that automatically dispense soft drinks into cups. Many national fast food chains such as McDonald's and Burger King use these machines. A study by the company shows that the actual volume of soft drink that goes into a 16-ounce cup per fill can be approximated by a normal model with mean 16 ounces and standard deviation 0.31 ounces. A new 16-ounce cup that is being considered for use actually holds 16.62 ounces of drink.a. What is the probability that a new cup will overflow when filled by the automatic dispenser? .0228b. The company wishes to adjust the dispenser so that the probability that a new cup will overflow is .006. At what value should the mean amount dispensed by the machine be set to satisfy this wish? [ looking for help with this answer ] ounces. (Use 2 decimal places in your answer and use 0.31 ounces for the standard deviation).

Answer:a) 0 .0228  b) 14.69 ounces                     Step-by-step explanation:We are given the following information in the question:Mean, μ = 16 ouncesStandard Deviation, σ = 0.4 ouncesWe are given that the distribution of volume of soft drink is a bell shaped distribution that is a normal distribution.Volume of new cup = 16.8 ouncesFormula:[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]   a) new cup will overflow when filled by the automatic dispenserP( x > 16.8)[tex]P( x > 16.8) = P( z > \displaystyle\frac{16.8 - 16}{0.4}) = P(z > 2)[/tex][tex]= 1 - P(z \leq 2)[/tex]Calculation the value from standard normal z table, we have,  [tex]P(x > 16.8) = 1 - 0.9772 = 0.0228 = 2.28\%[/tex]0.0228 is the probability that a new cup will overflow when filled by the automatic dispenser.b) mean amount dispensed by the machine be set to satisfy this wishWe have to find the value of x such that the probability is 0.006P(X > x)  [tex]P( X > x) = P( z > \displaystyle\frac{16.8 - \mu}{0.4})=0.006[/tex]  [tex]= 1 -P( z \leq \displaystyle\frac{16.8 - \mu}{0.4})=0.006 [/tex]  [tex]=P( z \leq \displaystyle\frac{16.8 - \mu}{0.4})=0.994 [/tex]  Calculation the value from standard normal z table, we have,  [tex]\displaystyle\frac{16.8 - \mu}{0.4} = 2.512\\\\\mu = 14.688\\\mu \approx 14.69[/tex] Thus, the mean amount dispensed should be set to approximately 14.69 ounces so that the probability that a new cup will overflow is 0.006