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?
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To determine 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 and the standard normal distribution.
Step 1: Convert the given quantile (1% or 0.01) to a z-score:
The z-score represents the number of standard deviations an element is from the mean in a standard normal distribution.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.01 is approximately -2.33.
Step 2: Use the z-score formula to find the required mean (average) weight:
The z-score formula is:
z = (x - μ) / σ
where z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, the observed value (x) is 7.3 ounces, the standard deviation (σ) is 0.8 ounces, and the z-score (z) is -2.33.
Rearranging the formula to solve for the mean (μ), we have:
μ = x - (z * σ)
Plugging in the values:
μ = 7.3 - (-2.33 * 0.8)
μ = 7.3 + 1.864
μ ≈ 9.164
Therefore, the average weight should be approximately 9.164 ounces if the production manager wants no more than 1% of the products to weigh more than 7.3 ounces.