The actual amount of potato chips that a filling machine puts into an “8-ounce” bag varies from bag to bag. If we consider the amount put into each bag as a random variable having a normal distribution with σ = 0.05 ounce, find the mean such that only 1% of the bags contain less than 8 ounces of potato chips.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.0100) and its corresponding Z score. Use in equation above.
To find the mean such that only 1% of the bags contain less than 8 ounces of potato chips, we need to use the concept of z-scores and the standard normal distribution.
Step 1: Understanding the Problem
In this problem, the amount of potato chips put into each bag is a random variable with a normal distribution. The standard deviation (σ) is given as 0.05 ounces.
Step 2: Convert the Probability to a Z-score
We want to find the mean such that only 1% of the bags contain less than 8 ounces of potato chips. Since the normal distribution is symmetric, we can calculate the z-score that corresponds to the 1% probability in the tail of the distribution.
Using a standard normal distribution table or calculator, we find that the z-score for a 1% probability in the left tail is approximately -2.33.
Step 3: Apply the Z-score Formula
The z-score formula is given by:
z = (x - μ) / σ
Here, z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
We know the z-score is -2.33, the observed value is 8 ounces, and the standard deviation is 0.05 ounces. We want to find the mean, so we rearrange the formula and solve for μ.
-2.33 = (8 - μ) / 0.05
Step 4: Solve for the Mean
Now we can solve for the mean (μ) by rearranging the equation and isolating the variable on one side.
-2.33 * 0.05 = 8 - μ
-0.1165 = 8 - μ
μ = 8 + 0.1165
μ ≈ 8.12 ounces
Therefore, the mean should be approximately 8.12 ounces such that only 1% of the bags contain less than 8 ounces of potato chips.