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The concept of mean and standard deviation is fundamental to understand a normal distribution. In simple terms, the mean is the average of all the data points in a set. It gives us an idea of the central value around which the data points are spread. In Ken's case, the mean (\( \mu \)) is 25 minutes, indicating that, on average, it takes him 25 minutes to drive to work.

Standard deviation, denoted as \( \sigma \), measures how spread out the numbers in a data set are. A small standard deviation means that the values are close to the mean, whereas a larger standard deviation means the values are more spread out. For Ken, the time varies about 4.5 minutes from the mean.

Understanding these two parameters helps you see how often Ken's travel time will fall within certain ranges, crucial for gauging punctuality.

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. Formulaically, it helps you understand how far away a particular score is from the mean in terms of standard deviations. It is calculated as:

\[ z = \frac{x - \mu}{\sigma} \]

Where:

• \( x \) is the value of interest.
• \( \mu \) is the mean of the distribution.
• \( \sigma \) is the standard deviation.

For Ken, to find how often he takes more than 35 minutes, the Z-score is computed as:
\[ z = \frac{35 - 25}{4.5} \approx 2.22 \]
A Z-score of 2.22 indicates that 35 minutes is 2.22 standard deviations above the average driving time.

Once you've calculated the Z-score, you can determine the probability of an event by referencing the standard normal distribution (Z-table). This table shows how likely a value is to appear in a normal distribution.

Using Ken's Z-score of approximately 2.22, you consult the Z-table, finding a numerical value of 0.9861. This value means that 98.61% of the instances occur at or below 35 minutes. In other words, it shows the percentage of time Ken drives to work in under 35 minutes.

This insight is helpful for predicting the likelihood of events falling within a certain range in normally distributed datasets.

To determine how often Ken is late, you need the proportion of time that exceeds 35 minutes. Since the Z-table tells us the probability of Ken arriving within 35 minutes is 98.61%, you find the probability of being late by subtracting this from 1:

\[ 1 - 0.9861 = 0.0139 \]

This calculation reveals that Ken is late 1.39% of the time. This percentage is critical for knowing how much time Ken should consider budgeting more prudently, ensuring punctuality when a timely arrival is essential.

Such calculations using normal distribution provide a powerful way to plan for regularity and predict timing-related outcomes based on past performances.