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In any given set of data, two key measurements are often considered: the mean and the standard deviation. The mean represents the average value, while the standard deviation indicates how much variation exists from this mean. In the context of our wire resistance problem, the mean resistance is given as \( \mu = 0.13 \) ohms. This is the central value around which all other resistances are expected to cluster.

The standard deviation is \( \sigma = 0.005 \) ohms, which signifies that most of the wire resistances lie within \( 0.005 \) ohms of the mean. A smaller standard deviation implies that the resistances are more closely packed around the mean, while a larger one would suggest more spread out resistances. Together, these parameters help describe the distribution of wire resistances as a normal distribution, providing a powerful way to analyze probability and variability in a natural and intuitive manner.

The normal distribution is a continuous probability distribution that is symmetrical around its mean, and it is often referred to as the "bell curve" due to its shape. When analyzing data on a normal distribution, one useful tool is to convert it into a standard normal distribution. This is done by transforming the data using z-scores. A standard normal distribution has a mean of 0 and a standard deviation of 1, simplifying calculations and making it easier to compare different sets of data.

In our scenario, the wire resistances are normally distributed, so converting these into a standard normal distribution allows us to find the likelihood of specific outcomes using standard z-tables or calculators. This is essential when determining how likely a particular measurement, like resistance falling within certain specifications, will occur. It's a handy way of translating real-world data into a more uniform and understandable framework for analysis.

Calculating probability within a normal distribution often involves finding the area under the curve, bounded by specific limits. For our wires, we needed to calculate the probability that a wire's resistance would fall between \( 0.12 \) ohms and \( 0.14 \) ohms.

By converting these values into z-scores, as previously discussed, we can find that the probability of a single wire meeting the desired specification is the area within the z-scores of \(-2\) and \(2\) on a standard normal distribution. Using a z-table, this area corresponds to a probability of approximately 0.9544, indicating that there's a 95.44% chance that any randomly chosen wire will have a resistance within this range.

This probability can then be extended to assess more complex scenarios, such as determining the likelihood that multiple events occur, like all four wires in a computer system meeting the specifications.

Z-scores are a type of standard score that indicate how many standard deviations an element is from the mean. Z-scores are calculated using the formula: \[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the context of the wire resistances, to find out whether a wire meets the specifications given by 0.12 and 0.14 ohms, each resistance value was converted to a z-score. This was achieved by placing the actual resistance values into the equation. By doing so:

• For \( x = 0.12 \), we got a z-score of \(-2\).
• For \( x = 0.14 \), we got a z-score of \(2\).

These z-scores help us determine the position of those values in a standard normal distribution. The scores essentially standardize different data points, which is crucial in determining probabilities and comparisons across different datasets.