2. Potato Chips. In the food industry, quality assurance is an important busines

Question

2. Potato Chips. In the food industry, quality assurance is an important business practice. One manufacturer produces packages of potato chips, which are advertised to contain 13 ounces of product. The process of filling packages with potato chips is automated, but not perfect, so some natural variability exists in the amount of product in each package. When the filling process is working correctly the actual weight of contents in a package is random and follows a skewed distribution with mean 13.5 ounces and standard deviation 1.2 ounces (a) (Free Response) Given the tools we have learned in class, can you find the probability that a randomly selected package contains less than 13 ounces? Explain why or why not. (b) Now suppose you have a random sample of n 50 packages i. What is the mean of the sampling distribution of the sample mean when n = 50? Mean: (Round up to 2 decimal places) ii. What is the standard error of the sampling distribution of the sample mean when n = 50? Standard Error: (Round up to 2 decimal places) iii. What is the shape of the sampling distribution of the sample mean when n = 50 Choose one)! » Normal Approximately Normal » Not Normal (c) Find the probability that the sample mean weight for the 50 packages is less than 13 ounces i. Report the z-score corresponding to 13 ounces. (Round up to 2 decimal places) ii. Report the final probability to 4 decimal places using Table A (z-table) d) uppose most consumers want between 13.25 and 14 ounces in each package of chips. Find he probability that the sample mean weight for the 50 packages is between 13.25 ounces and 14 ounces i. Report the z-scores corresponding to 13.25 ounces and 14 ounces. (Round up to 2 decimal place ii.Report the final probability to 4 decimal places using Table A (e) The company will not ship the day's production if the sample mean weight of the 50 pack- ages is too low. Managers would like to have a "cut off" value whereby they will decide not to ship the day's production if the sample mean weight of the 50 packages falls below the cut-off. What should the cut-off value be so that the chance of not shipping the day's production is 2%? Compute this value and round the answer to 2 decimal places

Explanation / Answer

Ans:

d)i)

standard error of mean=1.2/sqrt(50)

z(13.25)=(13.25-13.5)/(1.2/sqrt(50))=-1.47

z(14)=(14-13.5)/(1.2/sqrt(50))=2.95

ii)P(-1.47<z<2.95)=P(z<2.95)-P(z<-1.47)

=0.9984-0.0704=0.9280

e)P(Z<=z)=0.02

z=-2.054

cut-off=13.5-2.054*1.2/sqrt(50)=13.15

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