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Statistics and Probability - Quartiles, Percentiles, and Interquartile Range

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Quartiles divide a sorted data set into four equal parts. The three cut points are the Lower Quartile (Q1Q_1), the Median (Q2Q_2), and the Upper Quartile (Q3Q_3).

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The Interquartile Range (IQRIQR) represents the range of the middle 50%50\% of the data. It is calculated as the difference between the upper and lower quartiles: IQR=Q3−Q1IQR = Q_3 - Q_1.

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Percentiles divide the data into 100100 equal parts. For example, the kthk^{th} percentile is the value below which k%k\% of the data falls. Q1Q_1 corresponds to the 25th25^{th} percentile, Q2Q_2 to the 50th50^{th} percentile, and Q3Q_3 to the 75th75^{th} percentile.

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The Five-Number Summary consists of the Minimum value, Q1Q_1, Q2Q_2 (Median), Q3Q_3, and the Maximum value. This summary is used to construct a Box-and-Whisker Plot.

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Outliers are often defined as values that fall more than 1.5×IQR1.5 \times IQR below Q1Q_1 or more than 1.5×IQR1.5 \times IQR above Q3Q_3.

📐Formulae

Position of Q1=14(n+1)\text{Position of } Q_1 = \frac{1}{4}(n + 1) tokens

Position of Q2 (Median)=12(n+1)\text{Position of } Q_2 \text{ (Median)} = \frac{1}{2}(n + 1) tokens

Position of Q3=34(n+1)\text{Position of } Q_3 = \frac{3}{4}(n + 1) tokens

IQR=Q3−Q1IQR = Q_3 - Q_1

Lower Outlier Boundary=Q1−1.5×IQR\text{Lower Outlier Boundary} = Q_1 - 1.5 \times IQR

Upper Outlier Boundary=Q3+1.5×IQR\text{Upper Outlier Boundary} = Q_3 + 1.5 \times IQR

💡Examples

Problem 1:

Find the Q1Q_1, Q3Q_3, and IQRIQR for the following set of test scores: 12,15,17,20,22,25,2812, 15, 17, 20, 22, 25, 28.

Solution:

  1. Arrange the data in ascending order: 12,15,17,20,22,25,2812, 15, 17, 20, 22, 25, 28.
  2. Find the number of terms: n=7n = 7.
  3. Find Q2Q_2 (Median): The middle term is 2020.
  4. Find Q1Q_1: The median of the lower half (12,15,1712, 15, 17) is 1515.
  5. Find Q3Q_3: The median of the upper half (22,25,2822, 25, 28) is 2525.
  6. Calculate IQRIQR: IQR=Q3−Q1=25−15=10IQR = Q_3 - Q_1 = 25 - 15 = 10.

Explanation:

Since nn is odd, the median is the center value. Q1Q_1 and Q3Q_3 are found by taking the middle values of the lower and upper subsets respectively.

Problem 2:

In a dataset of 200200 students, a student's score is at the 85th85^{th} percentile. How many students scored lower than this student?

Solution:

  1. Identify the percentage: 85%85\%.
  2. Identify the total number of students: n=200n = 200.
  3. Calculate the number of students: Number of students=85100×200=170\text{Number of students} = \frac{85}{100} \times 200 = 170.

Explanation:

The percentile rank indicates the percentage of scores that fall below a specific value. Therefore, 85%85\% of the 200200 students scored lower.

Problem 3:

Determine if there are any outliers in the following data: 2,22,24,26,28,30,502, 22, 24, 26, 28, 30, 50, given Q1=22Q_1 = 22 and Q3=30Q_3 = 30.

Solution:

  1. Calculate IQRIQR: IQR=30−22=8IQR = 30 - 22 = 8.
  2. Calculate the lower boundary: Q1−1.5×IQR=22−(1.5×8)=22−12=10Q_1 - 1.5 \times IQR = 22 - (1.5 \times 8) = 22 - 12 = 10.
  3. Calculate the upper boundary: Q3+1.5×IQR=30+(1.5×8)=30+12=42Q_3 + 1.5 \times IQR = 30 + (1.5 \times 8) = 30 + 12 = 42.
  4. Check for values outside [10,42][10, 42].
  5. The value 22 is less than 1010, and 5050 is greater than 4242.

Explanation:

Using the 1.5×IQR1.5 \times IQR rule, both 22 and 5050 are classified as outliers because they fall outside the calculated boundaries.