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?
To calculate the mean and standard deviation of the total weight of a box containing 6 bags of chips, we can use the properties of the sum of independent random variables.
First, let's calculate the mean. Since each bag has the same mean weight, we can simply multiply the mean weight of a single bag by the number of bags in a box. The mean weight of a bag is 10 ounces, so the mean weight of a box containing 6 bags is:
Mean = 10 ounces x 6 bags = 60 ounces.
Next, let's calculate the standard deviation. Since the bags are independent, the variance of the total weight is equal to the sum of the variances of the individual bags. Since the standard deviation is the square root of the variance, we need to calculate the standard deviation of a single bag and then multiply by the square root of the number of bags.
The standard deviation of a bag is given as 0.05 ounces. So, the variance of a single bag is:
Variance = (0.05 ounces)^2 = 0.0025 ounces^2.
The standard deviation of the total weight of a box containing 6 bags is:
Standard deviation = square root of (variance x number of bags) = square root of (0.0025 ounces^2 x 6 bags) = square root of 0.015 ounces^2 = 0.1225 ounces.
Therefore, the mean weight of a box containing 6 bags of chips is 60 ounces, and the standard deviation of the total weight is 0.1225 ounces.
Now, let's move on to the second part of the question.
Buckley Farms wants to increase the mean weight of each bag of chips such that only 5% of the bags have weights less than 16 ounces. We assume that the standard deviation remains at 0.12 ounces.
To find the mean weight that satisfies this condition, we need to calculate the z-score associated with the 5th percentile of the standard normal distribution. We can then use the z-score to find the corresponding weight value.
The z-score is calculated using the formula:
z = (x - μ) / σ,
where x is the given value (16 ounces in this case), μ is the mean, and σ is the standard deviation.
We need to find the z-score that corresponds to the 5th percentile, which is -1.645.
Solving the equation for z, we have:
-1.645 = (16 - μ) / 0.12.
Rearranging the equation, we get:
16 - μ = -1.645 x 0.12,
16 - μ = -0.1974.
Solving for μ, we have:
μ = 16 + 0.1974 = 16.1974.
Therefore, Buckley Farms should use a mean weight of approximately 16.1974 ounces to ensure that only 5% of the bags have weights less than 16 ounces, assuming the standard deviation remains 0.12 ounces.
To calculate the mean and standard deviation of the total weight of a box containing 6 bags of chips, we need to consider the weights of individual bags and the fact that there are 6 bags in each box.
Let's denote the mean weight of an individual bag as μ and the standard deviation as σ.
Given that the empty boxes have a mean weight of 10 ounces with a standard deviation of 0.05 ounces, it implies that the weight of each bag will add an additional 10/6 ounces to the total weight of the box.
Therefore, the mean weight of the total box would be:
Mean weight of box = 10 + (10/6) ounces = 10.67 ounces
To calculate the standard deviation of the total weight of the box, we need to utilize the fact that the weights of individual bags are independent of each other. We can apply the formula for the sum of independent random variables to calculate the standard deviation.
The standard deviation of the total weight of the box would be:
Standard deviation of box = σ/sqrt(n)
where n is the number of bags in the box, which is 6.
Given that σ (standard deviation) of the individual bags is 0.05 ounces, we can substitute the values into the equation:
Standard deviation of box = 0.05/sqrt(6) = 0.0204 ounces (rounded to four decimal places)
For part c), Buckley Farms wants only 5% of the bags to have weights less than 16 ounces, assuming the standard deviation remains the same at 0.12 ounces.
To find the corresponding mean weight μ, we can use a standard normal distribution table or a statistical calculator.
As 5% corresponds to a Z-score of -1.645 (approximately), we can rearrange the equation for the Z-score:
(Z - μ)/σ = -1.645
Substituting the known values, we have:
(-1.645 - μ)/0.12 = -1.645
Simplifying the equation, we find:
-1.645 - μ = -0.1974
Solving for μ, we find:
μ = 1.645 - 0.1974 ≈ 1.4486
Therefore, Buckley Farms should use a mean weight of approximately 1.4486 ounces per bag in order for only 5% of the bags to have weights less than 16 ounces.