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Calculating percentiles is a useful tool when analyzing normal distributions, especially when attempting to understand how data values rank relative to one another.
In our exercise, we worked with a normal distribution to find specific percentiles of defect lengths on a steel pipe.

Percentiles give you the value below which a certain percentage of the data falls.

• For example, the 75th percentile is the value below which 75% of the data lies.
• Conversely, the 25% of data that is above this value provides a boundary for the upper quartile.

To find a specific percentile, you look up the corresponding z-value in a standard normal distribution table or use statistical software.Let's see how this applies. For the 75th percentile, the z-value we found was approximately 0.6745. Subsequently, using the formula \( X = \mu + z \sigma \), where \( \mu = 30 \) and \( \sigma = 7.8 \), gave us the specific defect length where 75% of all defects are shorter. This turns out to be about 35.26 mm.
This method helps provide a clear understanding of the diversity of lengths in the dataset.

Probability calculations in the context of normal distributions allow us to determine the likelihood of a random variable falling within a certain range.

Consider the example where we're interested in the probability that a corrosion defect length is at most 20 mm.By transforming our normal distribution into a standard normal distribution, we can easily use the cumulative distribution function (CDF) associated with normal distributions.

The formula to find the standard normal variable (z-score) is \( Z = \frac{X - \mu}{\sigma} \).

• Here, \( X \) is the value we're interested in (20 mm), \( \mu = 30 \) is the mean, and \( \sigma = 7.8 \) is the standard deviation.
• This gives us \( Z = \frac{20 - 30}{7.8} \approx -1.28 \).

Using a standard normal distribution table or calculator, you find the probability that the defect length is less than or equal to 20 mm is approximately 0.1003.
This interpretation is crucial for understanding how likely certain defects occur within the dataset.

The standard deviation is a measure of how spread out the numbers in a data set are around the mean.

In the context of our exercise, the standard deviation plays a key role in interpreting the normal distribution of defect lengths.

The standard deviation \( \sigma = 7.8 \) mm indicates how much variation or "dispersion" there is from average defect lengths. More specifically:

• A smaller standard deviation would imply that most defect lengths are close to the mean.
• A larger standard deviation would denote that defect lengths are spread out over a wider range of values.

Understanding standard deviation is vital as it helps in calculating the z-scores, which in turn, facilitate the probability calculations and percentile determinations in a normal distribution.A solid grasp of standard deviation allows for better predictions about variability in data, which is critical in applications like material quality assessments.