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?
a) To find the probability that the total weight is less than 16 ounces, we can convert this problem into a z-score problem and use the z-table or a calculator to find the probability.
First, let's calculate the z-score using the formula:
z = (x - μ) / σ
Where x is the desired value (16 ounces in this case), μ is the mean (16.15 ounces), and σ is the standard deviation (0.12 ounces).
z = (16 - 16.15) / 0.12
z = -0.15 / 0.12
z = -1.25
Next, we can use the z-table or a calculator to find the probability associated with the z-score of -1.25.
Looking up -1.25 in the z-table, we find that the probability is approximately 0.1056, or 10.56%.
Therefore, the probability of randomly selecting a bag with a total weight less than 16 ounces is approximately 10.56%.
b) To calculate the mean and standard deviation of the total weight of a box containing 6 bags of chips, we need to consider that the weight of each bag follows an approximately normal distribution.
The total weight of the box will be the sum of the weights of the 6 bags. Since the bags are independent, the mean of the total weight will be the sum of the individual bag means, and the standard deviation of the total weight will be the square root of the sum of the variances of the individual bags.
The mean of the total weight of the box can be calculated as follows:
Mean of total weight = 6 * mean of individual bag weight
Mean of total weight = 6 * 16.15 ounces
Mean of total weight = 96.9 ounces
The standard deviation of the total weight of the box can be calculated as follows:
Standard deviation of total weight = square root (6 * variance of individual bag weight)
Standard deviation of total weight = square root (6 * (0.12 ounces)^2)
Standard deviation of total weight = square root (6 * 0.0144 ounces^2)
Standard deviation of total weight = square root (0.0864 ounces^2)
Standard deviation of total weight = 0.2946 ounces
Therefore, the mean of the total weight of a box containing 6 bags of chips is 96.9 ounces, and the standard deviation is 0.2946 ounces.
c) To find the mean weight that Buckley Farms should use in order for only 5% of the bags to have weights less than 16 ounces, we need to find the z-score associated with this desired probability and use it to calculate the mean.
First, we need to find the z-score using the z-table or a calculator. We want the probability to be 0.05.
Looking up a z-score that corresponds to a probability of 0.05, we find that it is approximately -1.645.
Now we can use the z-score formula to solve for the mean:
z = (x - μ) / σ
-1.645 = (16 - μ) / 0.12
-1.645 * 0.12 = 16 - μ
-0.1974 = 16 - μ
μ = 16 - (-0.1974)
μ = 16.1974
Therefore, Buckley Farms should use a mean weight of approximately 16.1974 ounces in order for only 5% of the bags to have weights less than 16 ounces.
b) To find the mean and standard deviation of the total weight of a box containing 6 bags of chips, we need to use the properties of means and standard deviations.
The total weight of a box is simply the sum of the weights of the 6 bags. Since the bags are independent, we know that the mean of the total weight is the sum of the means of the individual bags.
Mean of the total weight = Mean of 1 bag × Number of bags = 16.15 ounces × 6 bags = 96.9 ounces.
Next, to find the standard deviation of the total weight, we need to use the fact that the bags are independent. The standard deviation of the sum of independent variables is equal to the square root of the sum of the variances of the individual variables.
Standard deviation of the total weight = sqrt(Variance of 1 bag × Number of bags) = sqrt((0.12 ounces)^2 × 6) = sqrt(0.0144 ounces^2 × 6) = sqrt(0.0864 ounces^2) = 0.294 ounces.
Therefore, the mean weight of a box containing 6 bags of chips is 96.9 ounces and the standard deviation is 0.294 ounces.
c) To find the mean weight that Buckley Farms should use to ensure that only 5% of the bags have weights less than 16 ounces, we need to find the z-score corresponding to this 5% probability.
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to a 5% probability is approximately -1.645.
The z-score formula is given by:
z = (x - μ) / σ
Rearranging the formula to solve for x, the mean weight we want to find:
x = z * σ + μ
Plugging in the known values:
x = -1.645 * 0.12 ounces + 16.15 ounces.
Calculating:
x ≈ -0.1974 ounces + 16.15 ounces
x ≈ 15.9526 ounces
Therefore, Buckley Farms should increase the mean weight of each bag of chips to approximately 15.9526 ounces to ensure that only 5% of the bags have weights less than 16 ounces, assuming the standard deviation remains 0.12 ounces.