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The Z-score is a vital concept in statistics, especially when dealing with normal distributions. It helps us understand how far away a specific data point is from the mean, expressed in terms of standard deviations. In simpler terms, a Z-score tells us how many standard deviations a data point is from the average.

To calculate the Z-score, use the formula:

• \[Z = \frac{X - \mu}{\sigma}\]

where:

• \(X\) is the data point you are investigating,
• \(\mu\) is the mean of the distribution,
• \(\sigma\) is the standard deviation of the distribution.

By standardizing the values this way, we can compare them easily to a standard normal distribution to find probabilities. For our problem, when we found the Z-score for an envelope weighing 2 grams, we calculated it as 2.0, meaning it is 2 standard deviations above the mean weight.

The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. This distribution is instrumental because it allows us to use Z-scores to find probabilities and compare different sets of data.

The idea is that any normal distribution can be transformed into the standard normal distribution through Z-scores. Using the standard normal distribution table, also known as the Z-table, we can easily determine the probability of a random variable falling within a particular range of standard deviations.

In our context, once we calculated the Z-score of 2.0, we referred to the Z-table to understand the probability of finding an envelope weighing up to 2 grams. The value from the table, 0.9772, told us the likelihood of such an envelope being less than or equal to 2 grams.

Calculating probabilities within a normal distribution involves understanding the area under the curve of the distribution graph. Essentially, this area corresponds to the likelihood of a data point falling within a certain range.

After obtaining a Z-score, we look it up on a Z-table, which provides the probability of a data point being less than or equal to that Z-score value. However, sometimes, as with our problem, what's needed is the complementary probability - that is, the probability of a data point exceeding a specific value.

• For example, if a Z-table gives \( P(Z \leq 2) \approx 0.9772 \), then \( P(Z > 2)\) will be the complement of this probability.

This is calculated as:

• \[P(X > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228\]

This result tells us that there's a 2.28% chance of an envelope weighing over 2 grams. Therefore, in a batch of 1000 envelopes, about 23 are expected to exceed this weight.