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The mean of a uniformly distributed variable describes the central tendency of the data, which in our case is the amount of cola per can. For a uniform distribution, finding the mean is straightforward: you simply take the average of the smallest and largest values in the distribution.
In this problem, the smallest amount of cola a can could contain is 11.96 ounces while the largest is 12.05 ounces. To find the mean, we calculate the average of these two numbers:

• The formula for the mean is: \( \mu = \frac{a + b}{2} \)
• Substituting the given values results in: \( \mu = \frac{11.96 + 12.05}{2} = 11.995 \text{ ounces} \)

This value of 11.995 ounces is our expected average amount of cola per can.

Standard deviation is a measure that helps us understand the amount of variation or dispersion in a set of values. In a uniform distribution, it tells us how spread out the cola amounts are around the mean value.
For a uniform distribution, the formula to calculate the standard deviation is:

• \( \sigma = \sqrt{\frac{(b-a)^2}{12}} \)
• Here, \( b \) is the maximum value (12.05 ounces), and \( a \) is the minimum value (11.96 ounces).

Let's plug in the numbers:

• \( \sigma = \sqrt{\frac{(12.05 - 11.96)^2}{12}} = \sqrt{\frac{0.0081}{12}} \)
• This simplifies to approximately \( \sigma \approx 0.0262 \text{ ounces} \)

This result indicates the typical deviation of the amount of cola in each can from the mean.

In probability calculations, especially with uniform distributions, we often want to find the likelihood of a variable falling within a certain range. Let’s look at a few specific probability calculations for this cola distribution.For the probability of selecting a can that has less than 12 ounces:

• We calculate the fraction of the range below 12: \( P(X < 12) = \frac{12 - 11.96}{12.05 - 11.96} \)
• This results in \( \frac{0.04}{0.09} = 0.4444 \)

For the probability of more than 11.98 ounces:

• We calculate: \( P(X > 11.98) = \frac{12.05 - 11.98}{12.05 - 11.96} \)
• This results in \( \frac{0.07}{0.09} \approx 0.7778 \)

And, for more than 11.00 ounces:

• Since this amount is below the lower limit of our distribution, every can has over 11.00 ounces, so \( P(X > 11.00) = 1 \)

These calculations tell us how likely it is for a can to contain an amount within a specified range.

Understanding how the solution process unfolds can make the concepts much clearer. Step-by-step problem solving helps deconstruct complex ideas into manageable bits.
Here’s how the exercise solution breaks down: 1. **Mean Calculation**: We address the average amount of cola using the endpoints of the distribution. Find this via the formula for the mean of a uniform distribution. 2. **Standard Deviation**: This step illustrates the consistency of cola amounts across cans by applying the uniform distribution's standard deviation formula. 3. **Probability for Less than 12 Ounces**: This involves figuring out the proportion of cans with less than 12 ounces by determining what portion of the distribution falls below this threshold. 4. **Probability for More than 11.98 Ounces**: Here, you're determining what fraction of all cans exceeds 11.98 ounces. 5. **Probability for More than 11.00 Ounces**: Finally, confirming that any can has more than 11.00 ounces because it falls entirely within the distribution range. Working through these steps allows you to deeply grasp each element of the solution, preparing you to tackle similar problems effortlessly in the future.