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A probability distribution describes how the values of a random variable are spread or distributed. In the context of this problem, the random variable is the fracture strength of tempered glass. The population average (mean) is given as 14, and the variability around this mean is described by the standard deviation of 2.

When discussing probability distributions in statistics, a normal distribution is often used, especially because of its convenient properties (the famous bell-shaped curve). This is relevant when considering larger sample sizes because of the Central Limit Theorem, which states that the means of sufficiently large samples will approximate a normal distribution, regardless of the population's distribution shape. This allows statisticians to predict probabilities related to sample means with more reliability.

In this exercise, the mean fracture strength is part of a **normal distribution** with parameters \( \mu = 14 \) and \( \sigma = 2 \). Utilizing this model is key for analyzing and making probabilistic predictions about fracture strength in the sample of 100 pieces of glass.

A confidence interval provides a range within which we expect the true population parameter (like the mean) to fall, with a certain level of confidence. In real-world terms, if we repeat the sample process multiple times, the true mean will fall within this calculated interval in a specified percentage of cases (typically 95%, in this scenario).

To compute a confidence interval, we need the sample mean, standard deviation, and the sample size, along with the desired level of confidence. Common choices are **90%**, **95%**, and **99%** intervals. These correspond to specific Z-values (from the standard normal distribution), such as **1.645**, **1.96**, and **2.576** for one-tailed intervals, respectively.

• For example, in this exercise, a **95% confidence interval** is calculated around a mean of 14, with a standard deviation adjusted for the sample size (\( \frac{2}{\sqrt{100}} = 0.2\)).
• The interval is determined by multiplying the Z-value (1.96 for 95% confidence) by the standard deviation.
• This yields an interval: \((14 - 1.96 \times 0.2, 14 + 1.96 \times 0.2)\), or approximately \((13.608, 14.392)\).

We're saying we're 95% confident that the true average fracture strength lies between these values.

A Z-score is a statistical measurement describing a value's relationship to the mean of a group of values, expressed in terms of standard deviations from the mean. For instance, a Z-score of 2.5 signifies that the data point is 2.5 standard deviations above the mean.

In the context of the exercise, calculating the **Z-score** helps us determine how unusual or common certain fracture strengths are within the sample. To find a Z-score, the formula is: \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the value in question (such as 14.5 in part a), \(\mu\) is the mean, and \(\sigma\) is the standard deviation (adjusted for sample size if needed).

• This calculation transforms the fracture strength data to fit within a standard normal distribution, enabling the use of statistical tables to find the associated probabilities.
• In Part (a), for instance, the Z-score for 14.5 is calculated as \(Z = \frac{14.5 - 14}{0.2} = 2.5\).
• This tells us that a sample mean of 14.5 is 2.5 standard deviations above the population mean of 14.

The Z-score helps in finding probabilities using standard normal distribution tables, thus playing a crucial role in both probability and confidence interval calculations.