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To understand the normal distribution probability, we must delve into z-score calculation first. The z-score, also known as the standard score, measures how many standard deviations an element is from the mean. This process helps us understand where a specific observation lies within a normal distribution.

The formula for calculating a z-score is simple: \(z = \frac{x - \text{mean}}{\text{standard deviation}}\).
Here, \(x\) is the value you want to convert, the \(\text{mean}\) is the average of the dataset, and \(\text{standard deviation}\) measures the dispersion of the dataset.
For instance, if the mean height of females is 22.7 inches with a standard deviation of 1.0 inch, and you want to find the z-score for a height of 24.0 inches, you substitute these values into the formula: \(z = \frac{24.0 - 22.7}{1.0} = 1.3\).
This means the height of 24.0 inches is \(1.3\) standard deviations above the mean.

Standard normal transformation is a vital concept when dealing with normal distribution probabilities. It usually involves converting a normal distribution to a standard normal distribution where the mean is zero and the standard deviation is one. This process is often termed 'standardizing'.

By transforming the data to the standard normal distribution, we make use of the z-table (or standard normal table) efficiently. As demonstrated in the original exercise, we transform an observed value \(24.0\) to a z-score \(1.3\). The transformed value is easier to work with for finding exact probabilities and comparing different datasets.

The formula to achieve this conversion is again: \(z = \frac{x - \text{mean}}{\text{standard deviation}}\).
This transformation aligns the data on a common scale, allowing seamless extrapolation of probabilities from the z-table. Remember, this transformation simplifies the pursuit of probability metrics across varied distributions.

Once we calculate the z-score, the next step is to find the corresponding probability using the z-table. A z-table helps us work with the standard normal distribution and find the area (probability) to the left of any given z-score.

For example, from our previous calculation, the z-score of 1.3 corresponds to a probability of \(0.9032\). This value represents the probability that a randomly selected female will have a back-to-knee length less than 24.0 inches.
To find the probability of a value being greater than a specific z-score, we subtract this value from 1:

\(P(X > 24.0) = 1 - P(Z < 1.3)\)
\(P(X > 24.0) = 1 - 0.9032 = 0.0968\).
Therefore, there's about a \(9.68\)% chance that a randomly selected female will have a back-to-knee length greater than 24.0 inches. Working with the z-table empowers us to derive meaningful probabilities with straightforward arithmetic, making data insights more accessible.