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Understanding the Z-score can be really helpful when dealing with normal distributions. It's a way of expressing how far away a specific measurement is from the mean. By calculating a Z-score, we can determine how unusual or usual a particular value is, relative to the distribution's mean.

The Z-score can be found using the formula: \[ Z = \frac{X - \mu}{\sigma} \], where:

• \( X \) is the value of interest (e.g., dot diameter)
• \( \mu \) is the mean of the distribution
• \( \sigma \) is the standard deviation

For example, if the dot diameter is \(0.0026\), with a mean \(\mu = 0.002\) and standard deviation \(\sigma = 0.0004\), the Z-score is \(1.5\). This means that \(0.0026\) is 1.5 standard deviations above the mean.

Z-scores help us understand whether a value is typically expected or more rare based on the context of a normal distribution.

After finding a Z-score, you can easily calculate various probabilities. This involves using Z-score tables (also known as standard normal distribution tables) to find out how likely it is for a value to occur in a normal distribution.

For instance, if you want to know the probability of a dot diameter exceeding \(0.0026\) and calculate that the Z-score is \(1.5\), you will check the table for \(P(Z < 1.5)\). The Z-table gives us \(0.9332\), meaning there's a 93.32% chance a value is below 1.5 standard deviations from the mean. To find the probability of a diameter greater than \(0.0026\), we subtract this from 1 (as probabilities sum up to 1). Thus, \(1 - 0.9332 = 0.0668\), or 6.68% chance.

Similarly, to find the probability between two diameters, you compute Z-scores for both and then find respective probabilities from the Z-table, subtracting to get the desired probability. Such calculations open up understanding of data distribution and frequency of different outcomes.

The standard deviation is a key concept in statistics and normal distribution. It measures the spread or how much variation there is from the average (mean). In a normal distribution, smaller standard deviations mean that the values are closer to the mean, while larger ones indicate more spread.

When adjusting the standard deviation to achieve certain probabilities, as in exercise part (c), we're tweaking how spread out we expect our data to be to fit a new criterion. Let's say we need a probability of 0.995 for a specific range (like from 0.0014 to 0.0026). We adjust our standard deviation using the known value of Z, here approximately \(-2.575\) to \(2.575\). Solving the equation \[ \sigma' = \frac{X_2 - X_1}{2 \times (2.575)} \] where \(X_2\) and \(X_1\) are the limits of the interval gives us the new spread needed.

In our example, this calculation returns \(0.000233\), meaning the new standard deviation should be \(0.000233\) inches to achieve the desired probability coverage for the specified range. Understanding standard deviation allows us to comprehend not just individual events, but the behavior of entire data sets.