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Statistics and Probability - Quartiles and percentiles

Grade 10IB

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

🔑Concepts

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

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The Lower Quartile (Q1Q_1) represents the 25th percentile, meaning 25% of the data lies below this value.

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The Median (Q2Q_2) represents the 50th percentile, dividing the data set in half.

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The Upper Quartile (Q3Q_3) represents the 75th percentile, meaning 75% of the data lies below this value.

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The Interquartile Range (IQRIQR) is the difference between the upper and lower quartiles and represents the spread of the middle 50% of the data.

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Percentiles divide the data into 100 equal parts. The kthk^{th} percentile is the value below which kk percent of the data falls.

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The five-number summary consists of the Minimum value, Q1Q_1, Median, Q3Q_3, and the Maximum value. This is visually represented using a Box and Whisker Plot.

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Outliers are extreme values. A common rule is that a value is an outlier if it is less than Q1−1.5×IQRQ_1 - 1.5 \times IQR or greater than Q3+1.5×IQRQ_3 + 1.5 \times IQR.

📐Formulae

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

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

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

IQR=Q3−Q1IQR = Q_3 - Q_1

Position of Pk=k100(n+1)Position\ of\ P_k = \frac{k}{100}(n + 1)

Lower Outlier Boundary=Q1−1.5×IQRLower\ Outlier\ Boundary = Q_1 - 1.5 \times IQR

Upper Outlier Boundary=Q3+1.5×IQRUpper\ Outlier\ Boundary = Q_3 + 1.5 \times IQR

💡Examples

Problem 1:

Given the data set: 3,7,8,5,12,14,21,15,183, 7, 8, 5, 12, 14, 21, 15, 18, find the Q1Q_1, Q2Q_2, Q3Q_3, and the IQRIQR.

Solution:

  1. Sort the data: 3,5,7,8,12,14,15,18,213, 5, 7, 8, 12, 14, 15, 18, 21.
  2. Number of terms n=9n = 9.
  3. Median (Q2Q_2) is the 9+12=5th\frac{9+1}{2} = 5^{th} term: Q2=12Q_2 = 12.
  4. Q1Q_1 is the median of the lower half (3,5,7,83, 5, 7, 8): 5+72=6\frac{5+7}{2} = 6.
  5. Q3Q_3 is the median of the upper half (14,15,18,2114, 15, 18, 21): 15+182=16.5\frac{15+18}{2} = 16.5.
  6. IQR=Q3−Q1=16.5−6=10.5IQR = Q_3 - Q_1 = 16.5 - 6 = 10.5.

Explanation:

First, the data must be ordered. Since nn is odd, the median is the middle term. To find Q1Q_1 and Q3Q_3, we find the medians of the lower and upper halves of the data respectively. The IQRIQR shows the range within which the middle 50% of values lie.

Problem 2:

In a class of 40 students, a student's score is at the 80th percentile. How many students scored lower than or equal to this student?

Solution:

Number of students=80100×40=32Number\ of\ students = \frac{80}{100} \times 40 = 32

Explanation:

The percentile rank indicates the percentage of scores that fall at or below a specific value. To find the number of students, multiply the total count by the percentile decimal (0.800.80).

Problem 3:

Identify if there are any outliers in the following data set: 10,50,52,55,58,60,9510, 50, 52, 55, 58, 60, 95. Given Q1=50Q_1 = 50 and Q3=60Q_3 = 60.

Solution:

  1. Calculate IQRIQR: IQR=60−50=10IQR = 60 - 50 = 10
  2. Calculate boundaries: Lower Boundary: 50−(1.5×10)=50−15=3550 - (1.5 \times 10) = 50 - 15 = 35 Upper Boundary: 60+(1.5×10)=60+15=7560 + (1.5 \times 10) = 60 + 15 = 75
  3. Compare data: 10<3510 < 35 (Outlier) and 95>7595 > 75 (Outlier).

Explanation:

We use the 1.5×IQR1.5 \times IQR rule. Any data point outside the range [35,75][35, 75] is considered an outlier. In this set, both 1010 and 9595 are outliers.