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 = 40 packages i. What is the mean of the sampling distribution of the sample mean when n40? (Round up to 2 decimal places) ean: ii. What is the standard error of the sampling distribution of the sample mean when n=40? Standard Error (Round up to 2 decimal places) iii. What is the shape of the sampling distribution of the sample mean when n = 40 Choose one)? Normal . Approximately Normal Not Normal (c) Find the probability that the sample mean weight for the 40 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)

Explanation / Answer

a) mean = 13.5

sd= 1.2

but since this is skewed , we can not find the probability

P(X <13)

= P (Z<0.42)

=0.3372

b) i) n = 40

mean = 13.5

sd =s/sqrt(n) = 1.2/sqrt(40) = 0.18973665

iii) approximately normal

due to central limit theorem (n>30)

c)

Z =(X - mean)/sd

= (13 - 13.5)/0.18973665

= -2.635231517

P(X < 13)

=P(Z< -2.64)

= 0.004203996

d)

P(13.2 <X< 13.9)

= P ( 1.58<Z<2.11 )=0.9255

e) P(Z<z*) = 0.01

z* = -2.32634

X* = 13.5 - 2.23634 * 0.18973665

= 13.0586

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