Problem 6 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Given is the following information about a histogram: $$ \begin{array}{cc} \hline \hline \text { Bin } & \text { Height } \\ \hline(0,2] & 0.245 \\ (2,4] & 0.130 \\ (4,7] & 0.050 \\ (7,11] & 0.020 \\ (11,15] & 0.005 \\ \hline \hline \end{array} $$ Compute the value of the empirical distribution function in the point \(t=7\).

Step-by-Step Solution

Answer

The empirical distribution function at \( t = 7 \) is 0.425.

Step 1: Identify Relevant Bins

First, we need to identify which bins have a midpoint less than or equal to the given point \( t = 7 \). The bins spanning the intervals \((0,2]\), \((2,4]\), and \((4,7]\) have their midpoints less than or equal to \(7\).

Step 2: Add Heights of These Bins

The next step is to add the heights of the bins identified in Step 1. Specifically, we sum the heights \(0.245\), \(0.130\), and \(0.050\) to calculate the empirical distribution function at \(t = 7\).

Step 3: Calculate the Empirical Distribution Function

Add the values from step 2: \[0.245 + 0.130 + 0.050 = 0.425\]Thus, the empirical distribution function at \(t = 7\) is \(0.425\).

Key Concepts

Histogram Interpretation

Histograms are a fundamental tool in statistics for visualizing the distribution of data. They allow us to see the frequency of data points within specified ranges, called bins or intervals. In a histogram, each bin is represented by a bar, where the height of the bar indicates the frequency of data points within that interval.
The intervals in a histogram can be adjusted based on the data distribution. In our given example, the bins are:

• (0,2], (2,4], (4,7], (7,11], (11,15]

The height of the bar (or frequency) is essential for understanding the distribution. In this context, these are not just simple frequencies but represent probabilities or proportions of the sample data falling into those intervals, summing up to a total height of 1. Interpretations include:

• Higher bars indicate more data points fall within that interval.
• The sum of all the heights should equal 1, representing the total probability.

This provides a clear visual representation to analyze patterns, like skewness or modality in data.

Midpoint Calculation

A crucial concept when dealing with histograms is calculating the midpoint of each bin. The midpoint serves as a representative value of the interval, helping to understand where the bulk of the data within that interval is centered.
The midpoint of a bin can be calculated using the formula:\[ \text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \] In our problem:

• For the bin (0,2], the midpoint is \((0+2)/2 = 1\)
• For the bin (2,4], the midpoint is \((2+4)/2 = 3\)
• For the bin (4,7], the midpoint is \((4+7)/2 = 5.5\)

Understanding and calculating these midpoints is important as they help determine which bins to consider when computing the empirical distribution function at a specific point, such as when the point of interest \(t=7\). This is because it helps us identify the relevant intervals that affect our calculations.

Step-by-Step Statistics

The step-by-step approach in statistics helps break down complex problems into manageable parts. Let's explore it using the computation of the empirical distribution function at \(t=7\) as an example.
First, identify the relevant bins by finding which midpoints are less than or equal to \(7\). This involves understanding the data's range and choosing only those intervals that contribute directly to the calculation. In this problem, the relevant intervals are

• (0,2], (2,4], and (4,7]

Step two requires adding the heights of these identified bins. Calculating the sum of the specified heights helps in the formation of the empirical distribution function. For our given data, that means:

• Adding 0.245, 0.130, and 0.050

Finally, compute the total to derive the empirical distribution function at the point specified, which is \(t=7\) in this case: \[0.245 + 0.130 + 0.050 = 0.425\]Thus, this step-by-step breakdown not only simplifies the calculation process but also enhances comprehension by clarifying each task's role in achieving the final result.