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 required for 95% of the candy bars to weigh 1 ounce or more, we can use the concept of z-scores and the standard normal distribution.
First, let's calculate the z-score corresponding to the desired proportion of 95%. The z-score is a measure of how many standard deviations a particular value is from the mean. In this case, we want to find the z-score that corresponds to having 95% of the data above it.
We can find this value using a standard normal distribution table or a statistical software. From the standard normal distribution table, a z-score of approximately 1.645 corresponds to a cumulative proportion of 0.95.
Next, let's use the formula for z-score to find the mean weight required:
z-score = (x - mean) / standard deviation
In this case, we want the z-score to be 1.645, the standard deviation is given as 0.08 ounces, and we want to solve for the mean weight (x).
Substituting the values into the equation, we have:
1.645 = (1 - x) / 0.08
Now let's solve for x:
1.645 * 0.08 = 1 - x
0.1316 = 1 - x
x = 1 - 0.1316
x = 0.8684
Therefore, to ensure that 95% of the candy bars weigh 1 ounce or more, the mean weight should be approximately 0.8684 ounces.