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Standard deviation is a measure of how spread out the values in a dataset are. It tells us, on average, how far each value in the set is from the mean. When we say that the weight of a running shoe is normally distributed with a standard deviation of 0.5 ounces, it means most shoes are within 0.5 ounces above or below the mean weight of 12 ounces.
When adjusting the standard deviation, it affects how tightly or widely the weights are spread around the mean. If the standard deviation is small, it indicates that most of the shoe weights are clustered closely around the mean. Conversely, a larger standard deviation implies that the weights deviate more from the mean. In part (b) of the exercise, by needing shoes to meet a certain weight requirement (99.9% weighing less than 13 ounces), we find that a smaller standard deviation of about 0.324 ounces is needed. This ensures less variability in the shoe weights, keeping more of them under the desired maximum weight.

The z-score is a statistic that tells us how far, in standard deviations, a particular value is from the mean of a dataset. It's calculated using the formula: \[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the specific value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation of the dataset.
This score allows us to standardize different datasets on a common scale, making them easily comparable. In the exercise, we calculated the z-score for a shoe weighing 13 ounces when the mean weight is 12 ounces and the standard deviation is 0.5 ounces. The calculation:\[ z = \frac{13 - 12}{0.5} = 2 \]means this shoe's weight is 2 standard deviations above the mean. Utilizing z-scores, we can assess probabilities for different scenarios and make meaningful comparisons across different conditions or groups.

Probability is the measure of the likelihood that a particular event will occur. In the context of normal distribution, the probability of a specific outcome can be found using the z-score and corresponding z-table. The z-table provides the likelihood of a value being less than a specific z-score in a standard normal distribution.
For instance, to find the probability that a shoe weighs more than 13 ounces, we first calculated the z-score as 2. Then, by using the z-table, we found that the probability of a z-score less than 2 is 0.9772. Since we want the opposite (greater than 13 ounces), we take the complement:\[ P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 \]Thus, the probability that a shoe weighs more than 13 ounces is 0.0228, or 2.28%. These calculations are pivotal in determining how often certain events will likely happen based on your dataset's characteristics.

Mean, in statistics, represents the average of a dataset. By adjusting the mean, we essentially shift the entire dataset to the left or right on the number line without altering the spread. In the exercise, to ensure 99.9% of shoes weigh less than 13 ounces with a standard deviation remaining at 0.5 ounces, we calculated the required mean weight.
By utilizing the z-score of 3.090 for 99.9%, we solve:
\[ 3.090 = \frac{13 - \mu}{0.5} \]
This yields:\[ \mu = 13 - 1.545 = 11.455 \]
Thus, for the company to claim that most shoes meet the weight requirement with the original variability, the mean weight should be 11.455 ounces. Adjusting the mean ensures that the central tendency of the shoe weights aligns with desired specifications without changing the distribution's standard deviation.