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The normal distribution is a vital concept in statistics that forms a "bell-shaped" curve when plotted on a graph. It is one of the most common probability distributions in the natural and social sciences. This distribution is defined by two parameters: the mean (average) and the standard deviation (a measure of spread).

A normal distribution is symmetric around its mean, which means that data near the mean are more frequent in occurrence than data far from the mean. In our exercise, the diameters of the bearings are normally distributed with a mean of 1.5 mm and a standard deviation of 0.025 mm. This implies that the most common diameters center around the mean, but there is a certain spread that allows for diameters to be both smaller and larger than the mean.

Understanding the normal distribution helps us use other statistical tools, like the Z-score, to find probabilities, especially when looking for values that exceed or fall below a certain range.

The Z-score is a statistical tool that measures how far a data point is from the mean of a distribution, in units of standard deviation. It tells you how many standard deviations an element is from the mean. The formula for calculating a Z-score is:

\[ Z = \frac{X - \mu}{\sigma} \]

Where:

• \[ X \] is the value of the random variable.
• \[ \mu \] is the mean of the distribution.
• \[ \sigma \] is the standard deviation of the distribution.

For the exercise, calculating the Z-score for a diameter of 1.6 mm is crucial because it tells us how much this diameter deviates from the mean of 1.5 mm. When we calculated the Z-score to be 4, it indicates that 1.6 mm is 4 standard deviations above the mean. This large Z-score suggests that a bearing diameter of 1.6 mm is very uncommon, given our normal distribution parameters.

Z-scores can be used to find probabilities in a normal distribution by checking how likely it is to observe such extremes.

A random variable is a numerical outcome of a random process. We use random variables to quantify and represent real-world scenarios in terms of statistics and probability. In the exercise provided, the diameter of a single bearing is considered a random variable, denoted by \( X \).

The random variable \( X \) is described here by a normal distribution with a given mean and standard deviation, \( X \sim N(1.5, 0.025^2) \). Since the diameters are independent, each bearing represents a separate realization of the random variable \( X \). This means that while each bearing is affected by the overall distribution, any specific outcome, like our concern of diameters exceeding 1.6 mm, can be considered individually.

Understanding random variables helps us navigate through probability questions and make statistical predictions based on given data conditions.

The complement rule is a fundamental principle in probability that helps in finding the probability of the occurrence of an event by considering the non-occurrence of the opposite event. It can be expressed as:

\[ P( ext{Event}) = 1 - P( ext{Complement of Event}) \]

In the exercise, we need to determine the probability of at least one of the bearings having a diameter exceeding 1.6 mm. It’s easier to first find the probability that none of them exceeds 1.6 mm and then use the complement rule. Here’s how it works:

• We calculate the probability that an individual bearing will not exceed 1.6 mm.
• Since the bearings are independent, raise this probability to the power of 10 (for 10 bearings).
• The complement rule is then used to find that at least one exceeds the diameter: \( 1 - (1-P( ext{Exceeds})^{10}) \).

This technique greatly simplifies the calculation of probabilities for events involving multiple trials, making it a powerful tool in statistical analysis.