
The amount of cereal in a box in normally distributed with a mean of 20 ounces.  If the packager is required have to at least 97.5% of the cereal with 17 or more ounces of cereal, what is the standard deviation for the amount of cereal in a box?

We are asked to find the **standard deviation** of a normally distributed random variable representing the amount of cereal in a box. Let's break this down step-by-step.

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### Step 1: Extract the problem information
- **Mean** (\(\mu\)) = 20 ounces
- **Probability Requirement**: 97.5% of the cereal should have **17 or more ounces**. This means the lower tail (cereal less than 17 ounces) should contain only 2.5% of the distribution.
- **Goal**: Find the **standard deviation** (\(\sigma\)).

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### Step 2: Use the Z-score formula
The cumulative probability for 17 ounces corresponds to the **2.5th percentile** (lower tail), which has a **Z-score** value of approximately:

\[
Z = -1.96
\]

This value is obtained from standard normal distribution tables.

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### Step 3: Set up the Z-score equation
The formula for the Z-score is:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \(Z\) = Z-score for 17 ounces (\(-1.96\))
- \(X\) = 17 ounces
- \(\mu\) = 20 ounces
- \(\sigma\) = standard deviation (what we are solving for)

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### Step 4: Solve for \(\sigma\)
Substitute the known values into the Z-score equation:

\[
-1.96 = \frac{17 - 20}{\sigma}
\]

Simplify:

\[
-1.96 = \frac{-3}{\sigma}
\]

Now, solve for \(\sigma\):

\[
\sigma = \frac{3}{1.96}
\]

\[
\sigma \approx 1.53
\]

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### Final Answer
The standard deviation \(\sigma\) is approximately:

\[
\sigma \approx 1.53 \, \text{ounces}
\]

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Would you like further details or another example? 

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### Related Questions:
1. How is the Z-score table used to calculate probabilities in normal distributions?
2. What does it mean if a data point lies more than 3 standard deviations from the mean?
3. How would the standard deviation change if 99% of the cereal needed to weigh 17 ounces or more?
4. What is the cumulative distribution function (CDF) in normal distributions?
5. How can I determine the mean if only the Z-score and standard deviation are known?

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**Tip:** Always remember that Z-scores are useful for transforming normal distributions to the standard normal distribution, which simplifies probability calculations.

The amount of cereal in a box is normally distributed with a mean of 20 ounces. If the packager is required to have at least 97.5% of the cereal with 17 or more ounces of cereal, what is the standard deviation for the amount of cereal in a box?

Determine the standard deviation for a normally distributed cereal weight, where the mean is 20 ounces and 97.5% of the boxes must contain at least 17 ounces. This solution involves key probability and Z-score concepts, suitable for high school students in Grades 10-12.

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