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Statistics - Averages and Measures of Spread

Grade 9IGCSE

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

🔑Concepts

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The Mean is the arithmetic average of a set of numbers, calculated by dividing the sum of all values by the total count of values (nn).

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The Median is the middle value in a data set when the values are arranged in ascending or descending order. If there is an even number of values, it is the mean of the two middle numbers.

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The Mode is the value that appears most frequently in a data set. A set can be bimodal (two modes) or have no mode.

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The Range is a measure of spread calculated as the difference between the largest and smallest values: Range=xmax⁡−xmin⁡\text{Range} = x_{\max} - x_{\min}.

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For discrete frequency tables, the mean is found by ∑fx∑f\frac{\sum fx}{\sum f}, where ff is the frequency and xx is the value.

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For grouped data, the mean is estimated using the midpoint of each class interval as the xx value.

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Quartiles divide a ranked data set into four equal parts. Q1Q_1 is the lower quartile (25%25\%), Q2Q_2 is the median (50%50\%), and Q3Q_3 is the upper quartile (75%75\%).

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The Interquartile Range (IQRIQR) measures the spread of the middle 50%50\% of the data and is less affected by outliers than the range.

📐Formulae

Mean(xˉ)=∑xn\text{Mean} (\bar{x}) = \frac{\sum x}{n}

Mean from Frequency Table=∑fx∑f\text{Mean from Frequency Table} = \frac{\sum fx}{\sum f}

Range=Maximum Value−Minimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value}

Interquartile Range (IQR)=Q3−Q1\text{Interquartile Range (IQR)} = Q_3 - Q_1

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

Upper Quartile (Position)Q3=3(n+1)4\text{Upper Quartile (Position)} Q_3 = \frac{3(n+1)}{4}

💡Examples

Problem 1:

Find the mean, median, mode, and range of the following data set: 3,7,2,7,6,53, 7, 2, 7, 6, 5.

Solution:

  1. Arrange in order: 2,3,5,6,7,72, 3, 5, 6, 7, 7
  2. Mean=2+3+5+6+7+76=306=5\text{Mean} = \frac{2+3+5+6+7+7}{6} = \frac{30}{6} = 5
  3. Median=5+62=5.5\text{Median} = \frac{5+6}{2} = 5.5 (average of the 3rd and 4th values)
  4. Mode=7\text{Mode} = 7 (most frequent value)
  5. Range=7−2=5\text{Range} = 7 - 2 = 5

Explanation:

To find averages, first sort the data. Use the sum for the mean, the middle position for the median, frequency for the mode, and the gap between extremes for the range.

Problem 2:

Calculate the mean from the following frequency table: Value (xx): 2, 4, 6 Frequency (ff): 3, 5, 2

Solution:

  1. Calculate fxfx for each row: 2×3=62 \times 3 = 6 4×5=204 \times 5 = 20 6×2=126 \times 2 = 12
  2. Sum of ff: ∑f=3+5+2=10\sum f = 3 + 5 + 2 = 10
  3. Sum of fxfx: ∑fx=6+20+12=38\sum fx = 6 + 20 + 12 = 38
  4. Mean=∑fx∑f=3810=3.8\text{Mean} = \frac{\sum fx}{\sum f} = \frac{38}{10} = 3.8

Explanation:

When data is in a frequency table, multiply each value by its frequency to find the total sum, then divide by the total number of items (the sum of frequencies).

Problem 3:

Find the IQRIQR for the data set: 12,15,17,19,21,25,2812, 15, 17, 19, 21, 25, 28.

Solution:

  1. n=7n = 7
  2. Q1Q_1 position: 7+14=2\frac{7+1}{4} = 2nd value →Q1=15\rightarrow Q_1 = 15
  3. Q3Q_3 position: 3(7+1)4=6\frac{3(7+1)}{4} = 6th value →Q3=25\rightarrow Q_3 = 25
  4. IQR=Q3−Q1=25−15=10IQR = Q_3 - Q_1 = 25 - 15 = 10

Explanation:

The Interquartile Range is the difference between the upper and lower quartiles. For small data sets, use the (n+1)(n+1) formula to find the positions of the quartiles.