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The binomial distribution is often used to model scenarios where you have a fixed number of trials and two outcomes, such as becoming ill or not from bacteria exposure. However, calculating probabilities directly from the binomial distribution can be complex for large numbers like 1000 trials. This is where the normal approximation becomes particularly useful.

When the sample size is sufficiently large, the binomial distribution can be approximated by a normal distribution. This is applicable to our problem because we are dealing with a scenario where 1000 individuals are exposed to a bacteria with a 20% infection rate. To use the normal approximation, we calculate the mean \( \mu \) and the standard deviation \( \sigma \) of the binomial distribution.

From our problem, the mean \( \mu \) is calculated as \( n \times p = 1000 \times 0.2 = 200 \), and the standard deviation \( \sigma \) is \( \sqrt{n \times p \times (1-p)} = \sqrt{1000 \times 0.2 \times 0.8} \approx 12.65 \). With these parameters, we can transform our binomial distribution into a normal one, making it easier to compute probabilities.

Continuity correction is an essential step when using the normal approximation for a binomial distribution. This is because the binomial distribution is discrete, while the normal distribution is continuous.

When converting a discrete variable to a continuous one, we adjust our calculations slightly, known as the continuity correction. For example, if we want to calculate the probability of more than 225 people becoming ill, we actually calculate \( P(X > 225.5) \). This small adjustment of adding or subtracting 0.5 is necessary for accuracy.

In problems where you're dealing with exact values like 'between 175 and 225', you would apply corrections such as \( P(174.5 < X < 225.5) \). These corrections help align the intervals to better fit a continuous distribution, improving the accuracy of your probability calculations.

The z-score is a pivotal concept when utilizing the normal approximation. It allows you to transform normal distribution probabilities into standard normal distribution probabilities, which can easily be found in z-tables.

To calculate a z-score, you use the formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the value you're interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the bacteria problem, to determine the likelihood of more than 225 people becoming ill, we compute the z-score for 225.5 using this formula, resulting in \( z \approx 2.01 \).

Once you have the z-score, you can look up standard normal distribution tables to find the corresponding probability. A similar process is used for determining probabilities between values or finding a specific value with given probabilities, as shown in the exercise's solutions.

After understanding how to apply normal approximation, continuity correction, and compute z-scores, you can calculate probabilities. You can determine the likelihood of observing certain outcomes in a binomially-distributed scenario using these steps.

For instance, to find the probability of more than 225 people being ill, we calculate \( P(Z > 2.01) \) using the z-score of 2.01. From z-tables, \( P(Z > 2.01) \approx 0.0222 \), indicating there's about a 2.22% chance this occurs.

Similarly, for determining the probability of outcomes falling between two values, like between 175 and 225 ill people, we find the relevant z-scores, then use these to calculate the cumulative probabilities. Subtracting these probabilities gives the desired result.

Probability calculations using these methods increase precision, especially for large sample sizes, making them indispensable in simplifying complex binomial problems.