krit.club logo

Statistics and Probability - Measures of dispersion: range, interquartile range, and box-and-whisker plots

Grade 9IB

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

🔑Concepts

•

The Range is the simplest measure of dispersion, representing the total spread of the data. It is calculated as the difference between the maximum value and the minimum value: Range=xmax−xminRange = x_{max} - x_{min}.

A number line showing the distance between minimum and maximum values as the Range.
•

The Interquartile Range (IQR) measures the spread of the middle 50% of the data. It is the difference between the upper quartile (Q3Q_3) and the lower quartile (Q1Q_1): IQR=Q3−Q1IQR = Q_3 - Q_1. It is less affected by outliers than the range.

A box plot section highlighting the middle 50 percent as the IQR.
•

A Box-and-Whisker Plot visually summarizes a data set using a five-number summary: Minimum, Q1Q_1, Median, Q3Q_3, and Maximum.

Complete box-and-whisker plot structure.
•

Outliers are extreme values that fall significantly outside the rest of the data. Mathematically, they are values smaller than Q1−1.5×IQRQ_1 - 1.5 \times IQR or larger than Q3+1.5×IQRQ_3 + 1.5 \times IQR.

📐Formulae

Range=xmax−xminRange = x_{max} - x_{min}

IQR=Q3−Q1IQR = Q_3 - Q_1

LowerOutlierBound=Q1−1.5×IQRLower Outlier Bound = Q_1 - 1.5 \times IQR

UpperOutlierBound=Q3+1.5×IQRUpper Outlier Bound = Q_3 + 1.5 \times IQR

PositionofMedian=n+12Position of Median = \frac{n + 1}{2}

💡Examples

Problem 1:

The marks obtained by 9 students in a quiz are: 15,12,18,10,20,15,17,9,2115, 12, 18, 10, 20, 15, 17, 9, 21. Calculate the Range and the Interquartile Range (IQR).

Solution:

  1. Order the data from least to greatest: 9,10,12,15,15,17,18,20,219, 10, 12, 15, 15, 17, 18, 20, 21
  2. Find the Range: Max−Min=21−9=12Max - Min = 21 - 9 = 12
  3. Find the Median (Q2Q_2): The middle value of 9 numbers is the 5th5^{th} position: Median=15Median = 15
  4. Find Q1Q_1: The median of the lower half (9,10,12,159, 10, 12, 15) is 10+122=11\frac{10 + 12}{2} = 11
  5. Find Q3Q_3: The median of the upper half (17,18,20,2117, 18, 20, 21) is 18+202=19\frac{18 + 20}{2} = 19
  6. Calculate IQRIQR: Q3−Q1=19−11=8Q_3 - Q_1 = 19 - 11 = 8

Explanation:

To find measures of dispersion, the data must first be ordered. The range gives the total spread (1212), while the IQR (88) focuses on the spread of the middle half of the marks, which is less affected by the highest and lowest scores.

Problem 2:

A dataset has Q1=20Q_1 = 20, Median=25Median = 25, and Q3=32Q_3 = 32. Determine if a value of 5555 would be considered an outlier in this dataset.

Solution:

  1. Calculate the IQR: IQR=Q3−Q1=32−20=12IQR = Q_3 - Q_1 = 32 - 20 = 12
  2. Calculate the multiplier for outliers: 1.5×IQR=1.5×12=181.5 \times IQR = 1.5 \times 12 = 18
  3. Calculate the Upper Bound: Q3+18=32+18=50Q_3 + 18 = 32 + 18 = 50
  4. Compare the value to the bound: 55>5055 > 50

Explanation:

Since the value 5555 is greater than the upper boundary of 5050, it is statistically classified as an outlier. In a box-and-whisker plot, the whisker would stop at the last data point within the 5050 limit, and 5555 would be plotted as a separate dot.

Problem 3:

Given the following box-and-whisker plot for the heights (in cm) of a group of plants, identify the five-number summary and calculate the Range and IQR.

Box plot with scale from 8 to 32 showing min 10, Q1 14, median 18, Q3 22, and max 30.

Solution:

  1. Minimum = 1010
  2. Q1=14Q_1 = 14
  3. Median = 1818
  4. Q3=22Q_3 = 22
  5. Maximum = 3030

Range=30−10=20Range = 30 - 10 = 20 cm IQR=22−14=8IQR = 22 - 14 = 8 cm

Explanation:

The whiskers end at the Minimum (1010) and Maximum (3030). The box starts at Q1Q_1 (1414) and ends at Q3Q_3 (2222). The line inside the box is the Median (1818).

Problem 4:

A set of data points is: 5,12,13,15,18,22,505, 12, 13, 15, 18, 22, 50. Use the 1.5×IQR1.5 \times IQR rule to determine if the value 5050 is an outlier.

Box plot showing the whisker ending at the calculated bound of 37 and a separate point at 50 representing the outlier.

Solution:

  1. Find Median: Middle value is 1515.
  2. Find Q1Q_1 (median of lower half: 5,12,135, 12, 13): Q1=12Q_1 = 12.
  3. Find Q3Q_3 (median of upper half: 18,22,5018, 22, 50): Q3=22Q_3 = 22.
  4. Calculate IQRIQR: 22−12=1022 - 12 = 10.
  5. Calculate Upper Bound: Q3+1.5×10=22+15=37Q_3 + 1.5 \times 10 = 22 + 15 = 37.
  6. Compare: 50>3750 > 37, therefore 5050 is an outlier.

Explanation:

Since 5050 exceeds the upper threshold calculated by adding 1.51.5 times the interquartile range to the third quartile, it is statistically classified as an outlier.