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A z-score gives us an understanding of how far a data point is from the mean in terms of standard deviations. It's a useful tool in statistics, especially when working with normal distributions. To calculate a z-score, take the value you're examining, subtract the mean of the distribution, and divide the result by the standard deviation. This formula is: \[ z = \frac{X - \mu}{\sigma} \]where:

• \( X \) is the data value,
• \( \mu \) is the mean,
• \( \sigma \) is the standard deviation.

Z-scores allow you to compare different data points from the same or different distributions by standardizing them. They also help when consulting standard normal distribution tables to find related probabilities. For instance, with a yield strength value of 40 and a mean of 43 along with a standard deviation of 4.5, your z-score is approximately -0.67, indicating that 40 is 0.67 standard deviations below the mean.

Percentiles show the relative ranking of data points in a set. The nth percentile of a dataset is the value below which n% of the data falls. In this context, it is particularly used to determine what value separates a certain percentage of observations.
For practical use, percentiles are often expressed in terms of z-scores when dealing with normal distributions. For example, when asked to find the yield strength that separates the strongest 75%, you're looking for the 25th percentile.
The approach involves using a z-score that corresponds to the desired cumulative probability from the standard normal distribution. We found that a z-score of 0.674 separates the lowest 75% from the highest 25% in a normal distribution. Calculating backwards from the z-score gives the physical yield strength value which, in this problem, turned out to be approximately 46.03 ksi.

Probability determination in the context of normal distributions involves finding the likelihood of a random variable falling within a certain range.
With continuous variables, such as yield strength, we use the cumulative distribution function (CDF) and z-scores to calculate these probabilities. For example, to find the probability that a yield strength is at most 40, we calculated the z-score and used it to find the cumulative probability from the standard normal distribution table. This told us that 25.14% of observations are at most 40 ksi.
Similarly, for a yield strength greater than 60, a different z-score calculation gives a very low probability, indicating it is extremely unlikely for values so far from the mean in this distribution. Using the tables, the probability is approximately zero, showing that 60 is out in the tail of the distribution.

A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In statistical contexts, it is often used to model uncertainties. Random variables can take on different types, including continuous or discrete.
For normal distributions, these variables can take any value on the entire number line. The yield strength of the A 36 grade steel in our problem is an example of a continuous random variable that follows a normal distribution. It has a mean (average value) and a standard deviation (how spread out the values are from the mean).

• The mean, \( \mu \), is central to the distribution where most values cluster.
• The standard deviation, \( \sigma \), indicates spread; larger values mean more spread.

Understanding random variables in the context of normal distribution helps in making probabilistic predictions, analyzing data variation, and determining likelihoods of various outcomes in a set of data.