The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a normal distribution with a standard deviation of 0.12 ounces. Every can that has more than 12.30 ounces of soda poured into it causes a spill and the can needs to go through a special cleaning process before it can be sold. What is the mean amount of soda the machine should dispense if the company wants to limit the percentage that needs to be cleaned because of spillage to 3%?
Let the mean amount of soda dispensed be μ. We want to find μ such that P(X > 12.30) = 0.03, where X follows a normal distribution with mean μ and standard deviation 0.12 ounces. This is equivalent to solving for μ with the equation:
P((X - μ) / 0.12 > (12.30 - μ) / 0.12) = 0.03.
To solve for μ, we first need to find the z-score corresponding to a 3% probability in the upper tail of a standard normal distribution. Using a z-table or calculator, we find that the z-score is approximately 1.88. Therefore,
(12.30 - μ) / 0.12 = 1.88
Now, we can solve for μ:
12.30 - μ = 1.88 * 0.12
12.30 - μ = 0.2256
μ = 12.30 - 0.2256 = 12.0744
So, the machine should dispense a mean amount of approximately 12.07 ounces of soda to limit the percentage of cans that need to be cleaned because of spillage to 3%.
To find the mean amount of soda the machine should dispense, we need to determine the cutoff point that separates the top 3% of the distribution from the rest.
Step 1: Convert the percentage to a standard score (z-score).
Since the distribution follows a normal distribution, we can use z-scores to find the cutoff point. The z-score represents the number of standard deviations away from the mean a particular value is.
We want to find the z-score that corresponds to the top 3% of the distribution. The remaining 97% will be below this value.
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to the top 3% is approximately 1.88.
Step 2: Convert the z-score back to the raw data value.
The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the raw data value, μ is the mean, and σ is the standard deviation.
We can rearrange the formula to solve for x:
x = (z * σ) + μ
Substituting the known values, we have:
12.30 = (1.88 * 0.12) + μ
Simplifying the equation, we can solve for μ:
12.30 = 0.2256 + μ
μ = 12.30 - 0.2256
μ ≈ 12.0744
Therefore, the mean amount of soda the machine should dispense to limit the percentage that needs cleaning because of spillage to 3% is approximately 12.0744 ounces.
To find the mean amount of soda the machine should dispense, we need to determine the threshold value that corresponds to the 3% spillage rate in the normal distribution.
Step 1: Convert the desired spillage percentage to a z-score.
The z-score represents the number of standard deviations away from the mean a particular value falls.
The spillage rate of 3% can be written as a decimal, 0.03.
Since we want to find the threshold value in the right tail of the distribution, we subtract the decimal from 1:
1 - 0.03 = 0.97
Looking up the z-score for 0.97 in the standard normal distribution table, we find it to be approximately 1.88.
Step 2: Calculate the mean value using the z-score formula.
The formula to convert a z-score to the corresponding raw value is:
raw value = (z-score * standard deviation) + mean
Since we're given the standard deviation of 0.12 ounces, we substitute the values into the formula:
12.30 = (1.88 * 0.12) + mean
Simplifying the equation:
12.30 = 0.2256 + mean
Subtracting 0.2256 from both sides:
mean = 12.30 - 0.2256
mean ≈ 12.0744
Therefore, the mean amount of soda the machine should dispense is approximately 12.0744 ounces to limit the spillage rate to 3%.