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In a uniform distribution, every value within a specific interval is equally likely to occur. When dealing with uniform distribution, calculating the mean (average) and variance can help summarize and understand the data spread.
The mean of a uniform distribution on the interval \(a, b\) is calculated using the formula: \\[ \mu = \frac{a + b}{2} \] This formula finds the midpoint of the interval where the average value of the distribution lies.
For our exercise with a depth interval of \(7.5, 20\), plugging in the values we have: \[ \mu = \frac{7.5 + 20}{2} = 13.75 \] This means that, on average, the bioturbation layer depth is 13.75 cm.
Variance, on the other hand, measures how spread out the values are from the mean. It tells us about the distribution's variability. For a uniform distribution, variance \(\sigma^2\) is determined by: \[ \sigma^2 = \frac{(b-a)^2}{12} \] Substituting the depth values, we find: \[ \sigma^2 = \frac{(20 - 7.5)^2}{12} = 13.02 \] This shows there's a moderate spread of depth values around the mean.

The cumulative distribution function (CDF) helps us understand the probability that a random variable takes on a value less than or equal to a particular number. For uniform distribution, the CDF is a linear function because each interval is equally probable.
For a uniform distribution within \(a, b\), the formula for the CDF is:

• \( F(x) = 0 \, \text{for} \; x < a \)
• \( F(x) = \frac{x-a}{b-a} \, \text{for} \; a \leq x \leq b \)
• \( F(x) = 1 \, \text{for} \; x > b \)

In this problem, substituting \(a = 7.5\) and \(b = 20\), the CDF becomes: \\[ F(x) = \frac{x-7.5}{12.5}, \text{for} \; 7.5 \leq x \leq 20 \]
This function provides probabilities of the random variable depth being less than or equal to \(x\). For any depth value within \(7.5\, \text{and}\, 20\), simply input into the formula to get the cumulative probability.

Probability calculations are essential for determining the likelihood of specific outcomes. With a uniform distribution, probabilities are derived using the CDF. Let's see how these calculations are applied.
To find the probability of the depth being at most 10 cm, we use \(F(x)\) from the CDF at \(x = 10\): \\[ P(X \leq 10) = F(10) = \frac{10-7.5}{12.5} = 0.2 \] So, there's a 20% chance the depth is 10 cm or less.
To calculate the probability of depth between 10 cm and 15 cm, subtract the CDF value at 10 from that at 15:

• \(F(15) = \frac{15-7.5}{12.5} = 0.6 \)
• \(P(10 < X < 15) = F(15) - F(10) = 0.6 - 0.2 = 0.4 \)

Thus, there's a 40% likelihood of the depth being between these values.
These calculations illustrate how uniform distribution can predict ranges and thresholds.

Standard deviation is a crucial statistic that quantifies the amount of variation in a set of data values. In a uniform distribution, it helps you understand how data points differ from the mean.
For this problem, the standard deviation \(\sigma\) is the square root of the variance: \\[ \sigma = \sqrt{13.02} = 3.60 \] The standard deviation tells us, on average, how far values deviate from the mean (13.75 cm).
To find probabilities of the depth staying within one standard deviation of the mean, the range is calculated:

• First standard deviation: \(13.75 \pm 3.60 = 10.15 \, \text{to} \, 17.35 \)
• Probability within 1 standard deviation: \[ P(10.15 < X < 17.35) = F(17.35) - F(10.15) = 0.784 - 0.212 = 0.572 \] That means a 57.2% chance of being within this range.
  
• Second standard deviation: \(13.75 \pm 7.20 = 6.55 \, \text{to} \, 20.95\)
• Probability within 2 standard deviations: Since the entire interval \(7.5, 20\) lies within this range, the probability is 1, meaning 100% certainty.
  

Understanding these probabilities helps to see how much variation exists around the mean and assess intervals of data in practical scenarios.