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To tackle problems involving normal distribution, it's essential to transform the data into a standardized form using a z-score. The z-score is crucial because it allows us to understand how far 'x' is from the mean in terms of the standard deviation. You can calculate the z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \]where

• \(x\) is the value you are interested in,
• \(\mu\) is the mean of the distribution, and
• \(\sigma\) is the standard deviation.

In our example, given \(x=30\), \(\mu=20\), and \(\sigma=3.4\), substituting these values into the formula yields \(z \approx 2.941\). This result shows that the value 30 is approximately 2.941 standard deviations above the mean.

Once you have the z-score, the next step is determining the probability of that z-score within a standard normal distribution. The standard normal distribution is a special normal distribution that has a mean of 0 and a standard deviation of 1. By converting any normal distribution into this format through z-scores, we can easily use statistical tables or calculators to find probabilities. For the z-score calculated previously (\(z = 2.941\)), we use a standard normal table to find \(P(Z \leq 2.941)\). This table indicates the probability of being below the z-score. In this case, \(P(Z \leq 2.941)\) is approximately 0.9984. To find the probability of our original event \(P(x \geq 30)\), we compute:\[ P(Z \geq 2.941) = 1 - P(Z \leq 2.941) = 1 - 0.9984 = 0.0016 \]Thus, almost all the values are below 30 on this distribution, confirming its rarity under this parameter set.

Interpreting the probability result involves expressing what it means in a real-world context. Our calculation indicates \(P(x \geq 30) \approx 0.0016\), or 0.16%. This means that in a normal distribution with a mean of 20 and a standard deviation of 3.4, there's only a 0.16% chance that a value will be 30 or more. In practical terms, this suggests that reaching or exceeding the value of 30 is an exceptionally rare event given the typical spread of data around the mean. Understanding this probability helps in assessing risk and making informed decisions based on expected outcomes within the normal distribution framework. It's a powerful application for fields like finance, quality control, and any area that involves uncertainty and variability.