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.