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In the problem, we are dealing with a type of statistical distribution called the Binomial Distribution. Both scenarios—trees planted in sandy soil and those in clay soil—are examples of binomially distributed cases. A binomial distribution is used when we have a fixed number of independent trials, each with two possible outcomes: success or failure. In our exercise, survival of a tree represents a "success," while not surviving is a "failure."

To formally define a binomial distribution, we use parameters like the number of trials \(n\) and the probability of success \(p\). For the sandy soil, let's denote the number of trees as \(n = 50\) and the survival probability as \(p = 0.7\). This is represented as \( X \sim \text{Binomial}(n=50, p=0.7) \). Similarly, for the clay soil, \( Y \sim \text{Binomial}(n=50, p=0.6) \).

Helpful features of a binomial distribution include:

• Each trial is independent, meaning the outcome of one doesn't affect another.
• The probability of success \(p\) stays constant across trials.
• The random variable, like \( X \) or \( Y \), counts the number of successes in \(n\) trials.

This foundational understanding sets the stage for converting our binomial distribution into a normal distribution using another statistical principle called the Central Limit Theorem.

Our journey from binomial to normal distribution is fueled by the Central Limit Theorem (CLT). The CLT states that as the number of trials \(n\) becomes large, the probability distribution of the sum (or average) of random variables, each independently drawn from a random distribution, converges towards a normal distribution. This is especially useful when \(n\) is large, like our scenario with 50 trees.

This approximation allows us to swap our binomial distributions for normal distributions. Here's how it works: for a binomial distribution \(X \sim \text{Binomial}(n, p)\), the mean \(\mu\) and variance \(\sigma^2\) of the equivalent normal distribution are calculated using:

\[ \mu = np \]
\[ \sigma^2 = np(1-p) \]

For sandy soil, the approximate normal distribution is \( X \sim N(\mu_x=35, \sigma_x^2=10.5) \). For clay soil, it becomes \( Y \sim N(\mu_y=30, \sigma_y^2=12) \).

The usefulness lies in its ability to enable calculations of probabilities related to the normal distribution, such as the likelihood that \(X-Y\) falls within a specified range. This approximation simplifies the process of solving real-world problems.

Z-Scores are the key to unlocking probabilities with a normal distribution. A Z-score measures how many standard deviations a data point is from the mean. By converting any normal distribution to Z-scores, we can use standard normal distribution tables (which are readily available) to find probabilities.

Let’s illustrate this with our exercise. We want to calculate the probability that \(-5 \leq X-Y \leq 5\). First, understand that our \(X-Y\) distribution has a mean \(\mu_{X-Y} = 5\) and a variance \(\sigma_{X-Y}^2 = 22.5\), making \(X-Y \sim N(5, 22.5)\).

To standardize, we apply the Z-score formula:

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

Applying to our interval, we compute for the boundaries:\

• For lower boundary: \( Z_1 = \frac{-5-5}{\sqrt{22.5}} \approx -2.11 \)
• For upper boundary: \( Z_2 = \frac{5-5}{\sqrt{22.5}} = 0 \)

These Z-scores transform the problem to finding \(P(Z_1 \leq Z \leq Z_2)\). This is achieved using a Z-table where \( P(Z \leq -2.11) \approx 0.0174\) and \( P(Z \leq 0) = 0.5\). Thus, the probability \( P(-5 \leq X-Y \leq 5) = 0.5 - 0.0174 = 0.4826 \). Z-scores simplify understanding and calculating probabilities by aligning diverse situations back to a common framework: the standard normal distribution.