The weight of a product is normally distributed with a mean of 4 ounces and a variance of .25 "sq ounces." The company wants to classify the unit as a scrap in a maximum of 1% of the units if the weight is below a desired value. Determine the desired weight such that no more than 1% of the units are below it.
4-(2.33)(√.25)=2.835 ounces
Where did you get this equation from?
Use excel to find the inverse of .01 to get the z value of -2.32635
and plug that into the equation for the z=(x-mu)/SD
To determine the desired weight, we need to find the value that corresponds to the 1st percentile of the normal distribution.
First, we need to find the standard deviation (σ) from the given variance:
Variance = σ^2
0.25 = σ^2
Taking the square root of both sides, we get:
σ = √0.25
σ = 0.5
Now, we can calculate the z-score corresponding to the 1st percentile (1%):
z = InvNorm(α)
z = InvNorm(0.01)
Using a standard normal distribution table or a statistical calculator, we can find that the z-score is approximately -2.33.
The z-score formula is:
z = (x - μ) / σ
Since we want to find the desired weight (x), we rearrange the formula:
x = z * σ + μ
x = -2.33 * 0.5 + 4
Calculating the expression on the right side of the equation, we get:
x = -1.165 + 4
x ≈ 2.835
Therefore, the desired weight should be approximately 2.835 ounces so that no more than 1% of the units are below it.