In the food industry, quality assurance is an important business practice. One m

Question

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) 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 n = 40?

what is the mean?

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.

ii. Report the final probability to 4 decimal places using Table A (z-table).

(d) Find the probability that the sample mean weight for the 40 packages is between 13.2 ounces and 13.9 ounces.

i. Report the z-scores corresponding to 13.2 ounces and 13.9 ounces.

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 40 package 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 40 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 1%? round to nearest 2 decimal places

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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Thanks and have a good day!

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