IP AMaths Notes (Upper Sec, Year 3-4): 20) Statistics — Standard Deviation

Standard deviation quantifies spread. Handle both raw datasets and grouped frequencies.

Formulae

• Mean: \( \bar{x} = \dfrac{\sum x}{n} \).
• Variance (population): \( \sigma^2 = \dfrac{\sum (x - \bar{x})^2}{n} \).
• For grouped data with midpoints \( m_i \) and frequencies \( f_i \):
  • \( \bar{x} = \dfrac{\sum f_i m_i}{\sum f_i} \).
  • \( \sigma^2 = \dfrac{\sum f_i (m_i - \bar{x})^2}{\sum f_i} \).

Worked example 1 — Raw data

Scores: \( 4, 6, 8, 10, 12 \).

1. Mean: \( \bar{x} = \dfrac{4 + 6 + 8 + 10 + 12}{5} = 8 \).
2. Compute squared deviations: \( (4 - 8)^2 = 16 \), \( (6 - 8)^2 = 4 \), \( (8 - 8)^2 = 0 \), \( (10 - 8)^2 = 4 \), \( (12 - 8)^2 = 16 \).
3. Variance: \( \sigma^2 = \dfrac{16 + 4 + 0 + 4 + 16}{5} = 8 \).
4. Standard deviation: \( \sigma = \sqrt{8} = 2\sqrt{2} \).

Worked example 2 — Grouped data

Class intervals \( 0 \leq x < 5 \), \( 5 \leq x < 10 \), \( 10 \leq x < 15 \) with frequencies \( 3, 5, 2 \) respectively.

1. Midpoints: \( m_1 = 2.5 \), \( m_2 = 7.5 \), \( m_3 = 12.5 \).
2. Mean: \( \bar{x} = \dfrac{3(2.5) + 5(7.5) + 2(12.5)}{3 + 5 + 2} = \dfrac{7.5 + 37.5 + 25}{10} = 7.0 \).
3. Variance numerator: \( 3(2.5 - 7)^2 + 5(7.5 - 7)^2 + 2(12.5 - 7)^2 = 3(20.25) + 5(0.25) + 2(30.25) = 60.75 + 1.25 + 60.5 = 122.5 \).
4. Variance: \( \sigma^2 = \dfrac{122.5}{10} = 12.25 \).
5. Standard deviation: \( \sigma = \sqrt{12.25} = 3.5 \).

Try this

Given grouped data with class width \( 4 \) and frequencies \( 2, 7, 9, 4 \), construct the midpoints, compute \( \bar{x} \), and evaluate \( \sigma \) accurately.