Question 1170505: A machine fills pop bottles with a standard deviation of 25 mL no matter what the setting is for the mean. At how many mililiters should the mean be set, so that 90% of the bottles will contain at least 500 mL?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let X be the amount of liquid filled in the pop bottles. We are given:
* Standard deviation (σ) = 25 mL
* We want 90% of the bottles to contain at least 500 mL, which means P(X ≥ 500) = 0.90.
We want to find the mean (μ) such that this condition is satisfied.
**1. Convert to Standard Normal Distribution**
We can standardize the random variable X using the formula:
Z = (X - μ) / σ
where Z follows a standard normal distribution (mean 0, standard deviation 1).
We want P(X ≥ 500) = 0.90. Let's rewrite this in terms of Z:
P((X - μ) / σ ≥ (500 - μ) / 25) = 0.90
P(Z ≥ (500 - μ) / 25) = 0.90
**2. Find the Z-score**
We need to find the z-score such that the area to the right of it is 0.90. This means the area to the left of it is 1 - 0.90 = 0.10.
Using a standard normal distribution table or a calculator, we find the z-score corresponding to a cumulative probability of 0.10. This z-score is approximately -1.28.
So, (500 - μ) / 25 = -1.28.
**3. Solve for μ**
Now, we solve for μ:
500 - μ = -1.28 * 25
500 - μ = -32
μ = 500 + 32
μ = 532
**4. Answer**
The mean should be set to 532 mL.
**Therefore, the mean should be set to 532 mL so that 90% of the bottles contain at least 500 mL.**
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