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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?

Respuesta :

MissPhiladelphia
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
fichoh

Using the normal probability distribution principle given the mean and standard deviation of Buckley's farm ; the solutions to the exercise are :

  • P(Z < z) = 10.6%
  • Mean = 106.9 ounces ; Standard deviation = 0.298 ounces
  • Mean weight that should be used = 15.95 ounces

Given the Parameters :

  • Mean, μ = 16.15 ounces
  • Standard deviation, σ = 0.12 ounces

Recall :

  • Zscore = [tex] \frac{\bar{x} - \mu}{σ} [/tex]

Zscore = [tex] \frac{16 - 16.15}{0.12} = -1.25 [/tex]

P(Z < - 1.25) = 0.1056 = 10.6% (normal distribution table)

2.)

Empty box :

  • Weight = 10 ounces ; Standard deviation = 0.05 ounces

Mean weight of 6 bags :

10 + (6 × 16.15)

10 + 96.90 = 106.90 ounces

Standard deviation of 6 bags :

√(0.05² + 0.12² + 0.12² + 0.12² + 0.12² + 0.12² + 0.12²) = 0.298 ounces

3.)

Using the normal distribution table ;

Zscore for 5% = 0.05 = - 1.645

Using the Zscore formula :

[tex] -1.645 = \frac{x - 16.15}{0.12} [/tex]

Cross multiply :

x - 16.15 = -1.645 × 0.12

x = - 19.74 + 16.15

x = 15.95

Therefore, the mean weight that should be used is 15.95 ounces

Learn more:https://brainly.com/question/8165716

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