Math Problem Statement

The following data represent the dividend yields​ (in percent) of a random sample of 28 publicly traded stocks. 2.31 0.15 2.82 0.53 0.02 2.61 1.4 0.76 0 1.94 0 0.84 3.19 0.48 0.38 3.54 1.08 0.12 2.14 3.07 0 0.1 0 0.8 2.98 0.99 0.64 2.08 ​(a) Compute the​ five-number summary.

Solution

To compute the five-number summary of the dividend yields, we need to find the following values:

  1. Minimum: The smallest value in the dataset.
  2. First quartile (Q1): The value at the 25th percentile of the dataset.
  3. Median (Q2): The value at the 50th percentile of the dataset.
  4. Third quartile (Q3): The value at the 75th percentile of the dataset.
  5. Maximum: The largest value in the dataset.

Step-by-Step Calculation:

  1. Arrange the data in ascending order: 0,0,0,0,0.02,0.1,0.12,0.15,0.38,0.48,0.53,0.64,0.76,0.8,0.84,0.99,1.08,1.4,1.94,2.08,2.14,2.31,2.61,2.82,2.98,3.07,3.19,3.540, 0, 0, 0, 0.02, 0.1, 0.12, 0.15, 0.38, 0.48, 0.53, 0.64, 0.76, 0.8, 0.84, 0.99, 1.08, 1.4, 1.94, 2.08, 2.14, 2.31, 2.61, 2.82, 2.98, 3.07, 3.19, 3.54

  2. Minimum: The smallest value is 00.

  3. First quartile (Q1): To find the first quartile, we locate the value at the 25th percentile. For a dataset of 28 numbers: Q1=Value at position(28+14)=Value at position 7.25Value between 7th and 8th values.Q1 = \text{Value at position} \left(\frac{28+1}{4}\right) = \text{Value at position 7.25} \approx \text{Value between 7th and 8th values.} The 7th and 8th values are 0.120.12 and 0.150.15, so: Q10.12+0.152=0.135.Q1 \approx \frac{0.12 + 0.15}{2} = 0.135.

  4. Median (Q2): The median is the value at the 50th percentile: Q2=Value at position(28+12)=Value at position 14.5.Q2 = \text{Value at position} \left(\frac{28+1}{2}\right) = \text{Value at position 14.5}. The 14th and 15th values are 0.80.8 and 0.840.84, so: Q20.8+0.842=0.82.Q2 \approx \frac{0.8 + 0.84}{2} = 0.82.

  5. Third quartile (Q3): To find the third quartile, locate the value at the 75th percentile: Q3=Value at position(3×28+14)=Value at position 21.75.Q3 = \text{Value at position} \left(3 \times \frac{28+1}{4}\right) = \text{Value at position 21.75}. The 21st and 22nd values are 2.142.14 and 2.312.31, so: Q32.14+2.312=2.225.Q3 \approx \frac{2.14 + 2.31}{2} = 2.225.

  6. Maximum: The largest value is 3.543.54.

The Five-Number Summary:

  • Minimum: 00
  • First quartile (Q1): 0.1350.135
  • Median (Q2): 0.820.82
  • Third quartile (Q3): 2.2252.225
  • Maximum: 3.543.54

Do you want more details or have any questions?

Here are 5 related questions for further exploration:

  1. How would the interquartile range (IQR) be calculated from this dataset?
  2. What does the shape of the data distribution suggest based on this five-number summary?
  3. How could outliers be detected using this five-number summary?
  4. What impact would additional data points have on the quartiles and median?
  5. How can the five-number summary be visually represented?

Tip: The interquartile range (IQR) is a useful measure for detecting the spread of the middle 50% of data points in a dataset.

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Math Problem Analysis

Mathematical Concepts

Descriptive Statistics
Five-number Summary
Percentiles
Data Distribution

Formulas

First Quartile (Q1) = (n+1)/4th position
Median (Q2) = (n+1)/2th position
Third Quartile (Q3) = 3*(n+1)/4th position
Interquartile Range (IQR) = Q3 - Q1

Theorems

Quartiles and Percentiles Theorem
Median as the 50th percentile
Minimum and Maximum definitions in statistics

Suitable Grade Level

Grades 9-12