To solve this problem, we need to use the properties of a normal distribution, specifically the Z-scores, which measure how far a given value is from the mean in terms of standard deviations.

Given:

• Mean ($\mu$) = 481 calories
• Standard deviation ($\sigma$) = 25 calories

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Part 1: Proportion of servings between 480 and 490 calories

We need to find the proportion of servings with calories between 480 and 490.

Step 1: Calculate the Z-scores for 480 and 490.

The formula for a Z-score is:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the calorie value
• $\mu$ is the mean
• $\sigma$ is the standard deviation

Z-score for 480:

$Z_{480} = \frac{480 - 481}{25} = \frac{-1}{25} = -0.04$

Z-score for 490:

$Z_{490} = \frac{490 - 481}{25} = \frac{9}{25} = 0.36$

Step 2: Find the corresponding probabilities for the Z-scores

We use a Z-table or a standard normal distribution calculator to find the cumulative probability for each Z-score.

• The cumulative probability for $Z = -0.04$ is approximately 0.4840
• The cumulative probability for $Z = 0.36$ is approximately 0.6406

Step 3: Find the proportion of servings between 480 and 490 calories

The proportion is the difference between the cumulative probabilities for the two Z-scores:

$P(480 < X < 490) = P(Z_{490}) - P(Z_{480}) = 0.6406 - 0.4840 = 0.1566$

So, the proportion of servings with calories between 480 and 490 is 0.1566 or 15.66%.

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Part 2: 83rd percentile of the number of calories

The 83rd percentile means we are looking for the calorie value that separates the lowest 83% of servings from the highest 17%. This corresponds to a Z-score where the cumulative probability is 0.83.

Step 1: Find the Z-score for the 83rd percentile

Using a Z-table or calculator, we find that the Z-score for the 83rd percentile is approximately 0.954.

Step 2: Convert the Z-score to a calorie value

Using the Z-score formula $Z = \frac{X - \mu}{\sigma}$, we solve for $X$:

$0.954 = \frac{X - 481}{25}$

Multiplying both sides by 25:

$0.954 \times 25 = X - 481$

$23.85 = X - 481$

Add 481 to both sides:

$X = 481 + 23.85 = 504.85$

So, the 83rd percentile of the number of calories is approximately 504.85 calories.

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Summary:

• Proportion of servings between 480 and 490 calories: 0.1566 (or 15.66%)
• 83rd percentile of the number of calories: 504.85 calories

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Would you like further clarification or details?

Here are 5 related questions you might find interesting:

1. What is the probability that a serving has less than 460 calories?
2. How many calories correspond to the 95th percentile?
3. What proportion of servings have more than 500 calories?
4. If the standard deviation increased to 30, how would that affect the 83rd percentile?
5. How would you calculate the median number of calories in this normal distribution?

Tip: Understanding how to calculate Z-scores is crucial for working with normal distributions!