A company wishes to make 1 ounce candy bars. According to government regulations, in order to label the candy bars as 1 ounce, at least 95 % of the bars must weigh 1 ounce or more. Our process standard deviation is .08 ounces. What do we need to make the mean weight be so that 95% weigh 1 ounce or more?
To determine the mean weight needed for 95% of the candy bars to weigh 1 ounce or more, we can use the concept of the normal distribution.
Step 1: Find the Z-score corresponding to the desired proportion.
The Z-score can be calculated using the formula:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the desired outcome (in this case, 1 ounce)
μ is the mean weight
σ is the standard deviation
Since we want at least 95% of the bars to weigh 1 ounce or more, we need to find the Z-score that corresponds to the 5th percentile (100% - 95% = 5%).
Using a standard normal distribution table or calculator, we can find the Z-score for the 5th percentile. The Z-score that corresponds to the 5th percentile is approximately -1.645.
Step 2: Rearrange the formula to solve for the mean (μ).
Rearranging the formula gives us:
μ = X - Z * σ
Plugging in the values:
μ = 1 - (-1.645) * 0.08
Calculating:
μ ≈ 1 + 0.1316
Final Answer:
To ensure that at least 95% of the candy bars weigh 1 ounce or more, the mean weight should be approximately 1.1316 ounces.