Four boxes of cereal are to be selected at random from a continuous process that
fills boxes labeled "Contents: 340 g." The contents of the first three boxes are
found to weigh 339.0g, 337.7 g, and 338.9 g.
(a) What minimum weight of cereal must the fourth box contain so that the sample mean equals the amount specified in the label?
(b) Assume the fourth box will contain more cereal than any of the three boxes already tested. "What weight of cereal should the fourth box contain if the sample mean and median are to be identical?
(a) An individual used a correct method to calculate a population standard deviation, reporting a value of 1.100 when n = 12. Unfortunately, the sample
standard deviation was required. Determine the correct sample standard
deviation.
(b) Another individual used a correct method to calculate a sample standard
deviation, reporting a value of 1.732 when n = 4. Determine the population standard deviation, assuming N = 4.
(a) To find the minimum weight of cereal the fourth box must contain for the sample mean to equal the amount specified in the label, we need to consider the total weight of the four boxes. Let's denote the weight of the fourth box as x.
The sum of the weights of the four boxes can be calculated as follows:
339.0g + 337.7g + 338.9g + x = 340g * 4
Simplifying the equation, we get:
1015.6g + x = 1360g
Subtracting 1015.6g from both sides, we have:
x = 1360g - 1015.6g
x = 344.4g
Therefore, the minimum weight the fourth box must contain for the sample mean to equal the amount specified in the label is 344.4g.
(b) To find the weight of cereal the fourth box should contain for the sample mean and median to be identical, we need to consider the position of the median in the set of four boxes' weights. Since the first three boxes' weights are already given, the median will be the weight of the third box.
Given that the weights of the first three boxes are 339.0g, 337.7g, and 338.9g, the median will be 338.9g. This means the fourth box should also have a weight of 338.9g for the sample mean and median to be identical.
(a) The population standard deviation can be estimated from the sample standard deviation using the formula:
Population standard deviation (σ) = Sample standard deviation (s) * √(n/(n-1))
Given that the sample standard deviation (s) is 1.100 and the sample size (n) is 12, we can calculate the population standard deviation as follows:
Population standard deviation (σ) = 1.100 * √(12/(12-1))
Population standard deviation (σ) ≈ 1.100 * √(12/11)
Population standard deviation (σ) ≈ 1.100 * √(1.0909)
Population standard deviation (σ) ≈ 1.100 * 1.043
Population standard deviation (σ) ≈ 1.1463
Therefore, the correct sample standard deviation is approximately 1.1463.
(b) To find the population standard deviation (σ) given the sample standard deviation (s) and a sample size (n), we can use the formula:
Population standard deviation (σ) = Sample standard deviation (s) * √(n/(n-1))
Given that the sample standard deviation (s) is 1.732 and the sample size (n) is 4, we can calculate the population standard deviation as follows:
Population standard deviation (σ) = 1.732 * √(4/(4-1))
Population standard deviation (σ) ≈ 1.732 * √(4/3)
Population standard deviation (σ) ≈ 1.732 * √(1.3333)
Population standard deviation (σ) ≈ 1.732 * 1.1547
Population standard deviation (σ) ≈ 1.9991
Therefore, assuming N = 4, the population standard deviation is approximately 1.9991.
(a) To find the minimum weight of cereal the fourth box must contain so that the sample mean equals the amount specified in the label, we can start by calculating the mean weight of the three boxes that have already been weighed.
Step 1: Add the weights of the three boxes: 339.0g + 337.7g + 338.9g = 1015.6g
Step 2: Divide the sum by the number of boxes: 1015.6g / 3 = 338.53g (approx.)
Step 3: To find the minimum weight the fourth box must have to match the label, we subtract the calculated mean from the target amount specified on the label: 340g - 338.53g = 1.47g (approx.)
Therefore, the minimum weight the fourth box must contain is approximately 1.47 grams.
(b) To find the weight of cereal the fourth box should contain if the sample mean and median are to be identical, we need to first calculate the median weight of the three boxes that have already been weighed.
Step 1: Arrange the weights in ascending order: 337.7 g, 338.9 g, 339.0 g
Step 2: Since we have an odd number of values, the median is the middle value, which is 338.9 g.
Step 3: To have the sample mean and median be identical, the fourth box must also weigh 338.9 g.
Therefore, the weight of the cereal the fourth box should contain is 338.9 grams.
(a) To determine the correct sample standard deviation when the population standard deviation was mistakenly calculated, we need to use the formula to convert between the population and sample standard deviations, given by:
Sample Standard Deviation (s) = Population Standard Deviation (σ) / √(n-1)
Given that the population standard deviation (σ) is reported as 1.100 and the sample size (n) is 12, we can substitute these values into the formula:
s = 1.100 / √(12-1) = 1.100 / √11 = 0.330 (approx.)
Therefore, the correct sample standard deviation is approximately 0.330.
(b) To determine the population standard deviation when the sample standard deviation is reported, we can use the same formula as before in its inverse form:
Population Standard Deviation (σ) = Sample Standard Deviation (s) * √(n-1)
Given that the sample standard deviation (s) is reported as 1.732 and the sample size (n) is 4 (which coincides with the population size N in this case), we can substitute these values into the formula:
σ = 1.732 * √(4-1) = 1.732 * √3 = 1.732 * 1.732 = 3.000 (approx.)
Therefore, the population standard deviation is approximately 3.000.