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Female heights For the 261 female heights shown in the box plot in Figure \(2.16,\) the mean was 65.3 inches and the standard deviation was 3.0 inches. The shortest person in this sample had a height of 56 inches. a. Find the z-score for the height of 56 inches. b. What does the negative sign for the z-score represent? c. Is this observation a potential outlier according to the 3 standard deviation distance criterion? Explain.

Short Answer

Expert verified
The z-score is approximately -3.1, indicating a height significantly below average, and it suggests the height is a potential outlier.

Step by step solution

01

Define the Z-score formula

The Z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the observed value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In this case, \( X = 56 \), \( \mu = 65.3 \), and \( \sigma = 3.0 \).
02

Calculate the Z-score

Substitute the given values into the formula: \[ z = \frac{(56 - 65.3)}{3.0} = \frac{-9.3}{3.0} \approx -3.1 \] So, the Z-score for the height of 56 inches is approximately \(-3.1\).
03

Interpret the Negative Z-score

A negative Z-score indicates that the observed value is below the mean. Specifically, a Z-score of approximately \(-3.1\) means that the height of 56 inches is 3.1 standard deviations below the mean height.
04

Check for Outlier Status

The 3 standard deviation criterion for identifying potential outliers suggests that any value with a Z-score below \(-3\) or above \(3\) can be considered a potential outlier. Since the Z-score of \(-3.1\) is less than \(-3\), the height of 56 inches is considered a potential outlier.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variability or dispersion in a set of data values. In simpler terms, it tells us how spread out the numbers are around the mean (average). When the data points are close to the mean, the standard deviation is smaller. Conversely, if the data points are spread out over a wide range, the standard deviation is larger.

To compute the standard deviation, you can follow these steps:
  • Calculate the mean of the data set.
  • Subtract the mean from each data point to find the deviation of each point.
  • Square each deviation to eliminate negative values.
  • Find the average of these squared deviations, which is known as the variance.
  • Take the square root of the variance to get the standard deviation.
In the context of the female heights example, the standard deviation was given as 3.0 inches. This indicates that the majority of female heights vary by about 3 inches from the mean height of 65.3 inches.
Detecting Outliers with Z-Scores
Outlier detection involves identifying data points that differ significantly from others. They are data points that are either much higher or much lower than the rest.One common method to detect outliers is using Z-scores. The Z-score formula helps determine how far a data point is from the mean in units of standard deviations. Here's the formula again:

\[ z = \frac{(X - \mu)}{\sigma} \]
  • Where \(X\) is the data point.
  • \(\mu\) is the mean.
  • \(\sigma\) is the standard deviation.
A Z-score greater than 3 or less than -3 typically indicates an outlier. In the exercise for female heights, the Z-score for a height of 56 inches was -3.1. Since this is less than -3, it marks the height as a potential outlier.Outliers can have various implications in real-world scenarios, such as affecting the overall analysis or indicating errors or special cases.
Mean and Median Interpretation
Understanding the mean and median is crucial when analyzing a set of data. The **mean**, or average, is calculated by summing all data points and dividing by the count of data points. It gives a central value but can be skewed by extremely high or low values (outliers).

The **median** is the middle value when data points are arranged in order. It is less affected by outliers and provides a better sense of the dataset's center in a skewed distribution. With the female heights example, the mean was 65.3 inches. This average provides a rough estimate of the central height. However, if the data has many outliers, the mean might not fully represent the central tendency. In such cases, the median could offer better insights. If the median differs significantly from the mean, it may indicate skewness or outliers impacting the data. Balancing the insights from both statistics ensures a well-rounded view of data distribution.

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