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When talking about weight distribution in a normal distribution, we refer to how data spreads around the mean. Specifically, a normal distribution is symmetric about its mean, creating a bell-shaped curve. For parcels sent in the given example, their weights are distributed with a mean of 12 pounds and a standard deviation of 3.5 pounds. This means most parcels' weights cluster around 12 pounds.
Weight distribution can affect business decisions, such as establishing weight boundaries. If the weight distribution of parcels is well understood, a parcel service can make informed decisions about surcharge thresholds to make sure that most parcels will not incur extra costs.

The Z-Score is a key concept in statistics, especially in the context of normal distributions. It measures the number of standard deviations a data point is from the mean.
In the exercise provided, a Z-Score was used to find the point where 99% of parcels fall below a specific weight. This is done by using a Z-Score table to find that the 99th percentile corresponds to a Z-Score of about 2.33.
This signifies that the weight at this score is 2.33 standard deviations above the mean. The Z-Score formula used is:

• \[ Z = \frac{X - \text{mean}}{\text{standard deviation}} \]
• Where you substitute the known mean and standard deviation to find the weight at this Z-Score.

Percentiles are values in a data set that divide the data into 100 equal parts. Each percentile represents a particular threshold value under which a certain percentage of data falls.
In the context of this problem, we are looking for the 99th percentile. This means we want to find a weight at which 99% of all parcels are below that weight. This is important when setting surcharge benchmarks, ensuring that only the heaviest parcels incur surcharges.
Percentiles give businesses a way to interpret and convey the spread of their data in terms that can be easily understood and applied to decision-making.

Standard deviation is a measure of how spread out the numbers are in a data set. In a normal distribution, around 68% of data falls within one standard deviation of the mean, 95% falls within two, and 99.7% falls within three.
The given parcel example has a standard deviation of 3.5 pounds, illustrating that most parcel weights vary by 3.5 pounds from the average (12 pounds).
Understanding standard deviation allows the parcel service to predict and evaluate options like surcharge levels effectively since they can quantify the usual variability in parcel weights.
In practical terms, knowing the standard deviation helps ensure that the majority of parcels incur a fair postage fee, thus aligning operational costs with price structures.