The weight of a product is normally distributed with a standard deviation of .8 ounces. What should the average weight be if the production manager wants no more than 1% of the products to weigh more than 7.3 ounces?
If the mean is 5.45 oz, and the standard deviation is 0.8, 99% will weigh less than 7.3 oz.
That was obtained with a Java normal distribution web site by trial and error, varying the mean value.
You could also use a rule that the upper bound should be 2.3 sigma from the the mean for 99% inclusion.
To find the average weight that will ensure no more than 1% of the products weigh more than 7.3 ounces, we need to use the concept of z-scores.
First, let's find the z-score that corresponds to the 1% cutoff. We can do this by looking up the cumulative probability in the standard normal distribution table.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
z is the z-score
x is the value we are interested in (7.3 ounces)
μ is the mean (average weight)
σ is the standard deviation (0.8 ounces)
Rearranging the formula, we can solve for the mean:
μ = x - (z * σ)
Now, let's substitute in the values we know:
x = 7.3 ounces
σ = 0.8 ounces
Next, find the z-score using the standard normal distribution table. Since we want no more than 1% of the products to weigh more than 7.3 ounces, we need to find the z-score that corresponds to a cumulative probability of 0.99, since the table provides values for the area to the left of the z-score.
By looking up 0.99 in the table, we find that the z-score is approximately 2.33.
Plugging in the values:
μ = 7.3 - (2.33 * 0.8)
Calculating:
μ = 7.3 - 1.864
Therefore, the average weight should be approximately 5.436 ounces to ensure no more than 1% of the products weigh more than 7.3 ounces.