Question

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 0.4 ounces. A new 16-ounce cup that is being considered for use actually holds 16.8 ounces of drink. Question 1. What is the probability that a new cup will overflow when filled by the automatic dispenser? (Use 4 decimal places) .4207 Incorrect: Your answer is incorrect. Question 2. 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? (Use 2 decimal places in your answer and use 0.4 ounces for the standard deviation).

Answer 1

Answer:

a) 0 .0228  

b) 14.69 ounces                    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 16 ounces

Standard Deviation, σ = 0.4 ounces

We 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 ounces

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$  

a) new cup will overflow when filled by the automatic dispenser

P( x > 16.8)

$P( x > 16.8) = P( z > \displaystyle\frac{16.8 - 16}{0.4}) = P(z > 2)$

$= 1 - P(z \leq 2)$

Calculation the value from standard normal z table, we have,  

$P(x > 16.8) = 1 - 0.9772 = 0.0228 = 2.28\%$

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 wish

We have to find the value of x such that the probability is 0.006

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{16.8 - \mu}{0.4})=0.006$  

$= 1 -P( z \leq \displaystyle\frac{16.8 - \mu}{0.4})=0.006$  

$=P( z \leq \displaystyle\frac{16.8 - \mu}{0.4})=0.994$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{16.8 - \mu}{0.4} = 2.512\\\\\mu = 14.688\\\mu \approx 14.69$

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