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The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It's measured in terms of standard deviations from the mean. In our exercise, we are given a mean wearout time of 200 hours and a standard deviation of 10 hours. To find out how much 220 hours deviates from the mean, we use the z-score formula:

\[ z = \frac{x - \mu}{\sigma} \]
Here, \( x \) is the value we're interested in (220 hours), \( \mu \) is the mean (200 hours), and \( \sigma \) is the standard deviation (10 hours). Plugging in the values:

\[ z = \frac{220 - 200}{10} = 2 \]
So, a wearout time of 220 hours is 2 standard deviations above the mean. This makes it easier to determine the probability associated with this value using a standard normal distribution table (z-table).

Standard deviation is a measure of the amount of variation or dispersion of a set of values. In more practical terms, it tells us how much individual values in a dataset differ from the mean. A small standard deviation means that most numbers are close to the average, while a large standard deviation indicates a wider range of values.

In our example, the standard deviation of the wearout times is 10 hours. This indicates that most of the times are within 10 hours of the mean (200 hours). To visualize it, consider:

• 1 standard deviation from the mean will cover about 68% of all values.
• 2 standard deviations will cover about 95% of all values.
• 3 standard deviations will encompass about 99.7% of all values.

So, when we calculated the z-score and found it to be 2, we were able to conclude that 220 hours is relatively far from the mean in statistical terms.

Probability is a measure of how likely an event is to occur. It's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the context of our problem, we are looking for the probability that a machine has a wearout time of more than 220 hours.

To find this probability, we first calculate the z-score, which was 2. Next, we use the z-table to find the probability associated with this z-score. The z-table gives the probability that a value is less than the given z-score. For a z-score of 2, this probability is 0.9772, or 97.72%.

However, we needed the probability that the wearout time is more than 220 hours. So, we take the complement of 0.9772, because:

\[ P(X > 220) = 1 - P(X < 220) = 1 - 0.9772 = 0.0228 \]
Thus, the probability that a machine has a wearout time of more than 220 hours is 0.0228, or 2.28%. This low probability indicates that such an event is relatively rare, given the normal distribution of wearout times.