Question

Buckley Farms produces homemade potato chips that it sells in bags labeled 16 ounces. The total weight of each bag follows an approximately normal distribution with a mean of 16.15 ounces and a standard deviation of 0.12 ounces. a)If you randomly selected 1 bag of these chips what is the probability that the total weight is less than 16 ounces? (I know this one. I got 11%. I need help with the other two. I have no clue how to continue on...) b) Buckley Farms ships its chips in boxes that contain 6 bags. The empty boxes have a mean weight of 10 ounces and a standard deviation of 0.05 ounces. Calculate the mean and standard deviation of the total weight of a box containing 6 bags of chips. c) Buckley Farms decides to increase the mean weight of each bag of chips so that only 5% of the bags have weights that are less than 16 ounces. Assuming that the standard deviation remains 0.12 ounces what mean weight should Buckley Farms use?

Answer 1

Answer:

Step-by-step explanation:

b) Here sample size = n = no of bags = 6

x bar = sample mean = 10 oz

std dev = 0.05 oz.

By central limit theorem, we can say that population mean for all samples containing 6 bags would be

10 oz.

Std deviation for 6 bags = 0.05/sq rt 6=0.0204

c) Mean weight to be increased such that only 5% of bags weights that are <16 oz

Let new mean weight after increase by a

Then z = (a-16)/0.12/sqrt 6

=(a-16)/0.04899

|z|<0.0987

-0.00484<a-16<0.00484

a<16.484

Mean should be increased to 16.484 oz.

Answer 2

We transform the mean to
X* = X1 + X2
E(X*) = E(X1 + X2)
Assuming the bags are chosen randomly
X* =  E(X1) + E(X2)
X* = 16 + 10
X* = 26 ounces
SD = 0.12 * (26/16) = 0.20