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Statistics - Standard deviation of a continuous frequency distribution

Grade 11CBSE

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

🔑Concepts

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A continuous frequency distribution consists of data grouped into class intervals (e.g., 0−100-10, 10−2010-20).

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The first step in calculating the standard deviation is to find the mid-point (xix_i) of each class interval using the formula: xi=Lower Limit+Upper Limit2x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}.

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Standard deviation (represented by the Greek letter σ\sigma) is the square root of the variance (VV or σ2\sigma^2).

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For large values of xix_i, we use the 'Step-deviation Method' to simplify calculations by introducing a new variable ui=xi−Ahu_i = \frac{x_i - A}{h}, where AA is the assumed mean and hh is the class width.

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The sum of frequencies is denoted by N=∑fiN = \sum f_i.

📐Formulae

Mean (xˉ)=∑fixiN\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{N}

Standard Deviation (Direct Method): σ=∑fi(xi−xˉ)2N\text{Standard Deviation (Direct Method): } \sigma = \sqrt{\frac{\sum f_i(x_i - \bar{x})^2}{N}}

Standard Deviation (Shortcut Method): σ=1NN∑fixi2−(∑fixi)2\text{Standard Deviation (Shortcut Method): } \sigma = \frac{1}{N}\sqrt{N\sum f_i x_i^2 - (\sum f_i x_i)^2}

Step-deviation Method: σ=hNN∑fiui2−(∑fiui)2\text{Step-deviation Method: } \sigma = \frac{h}{N}\sqrt{N\sum f_i u_i^2 - (\sum f_i u_i)^2}

where ui=xi−Ah\text{where } u_i = \frac{x_i - A}{h}

💡Examples

Problem 1:

Calculate the standard deviation for the following distribution: Classes: 0−10,10−20,20−300-10, 10-20, 20-30 Frequencies: 2,3,52, 3, 5

Solution:

  1. Find mid-points (xix_i): 5,15,255, 15, 25.
  2. Calculate N=∑fi=2+3+5=10N = \sum f_i = 2 + 3 + 5 = 10.
  3. Calculate Mean (xˉ\bar{x}): xˉ=(2×5)+(3×15)+(5×25)10=10+45+12510=18010=18\bar{x} = \frac{(2 \times 5) + (3 \times 15) + (5 \times 25)}{10} = \frac{10 + 45 + 125}{10} = \frac{180}{10} = 18
  4. Calculate deviations (xi−xˉ)(x_i - \bar{x}) and their squares:
  • For xi=5:(5−18)=−13,(−13)2=169x_i = 5: (5 - 18) = -13, (-13)^2 = 169
  • For xi=15:(15−18)=−3,(−3)2=9x_i = 15: (15 - 18) = -3, (-3)^2 = 9
  • For xi=25:(25−18)=7,(7)2=49x_i = 25: (25 - 18) = 7, (7)^2 = 49
  1. Calculate ∑fi(xi−xˉ)2\sum f_i(x_i - \bar{x})^2: 2×169=3383×9=275×49=245Total=610\begin{array}{r} 2 \times 169 = 338 \\ 3 \times 9 = 27 \\ 5 \times 49 = 245 \\ \hline \text{Total} = 610 \end{array}
  2. Calculate σ\sigma: σ=61010=61≈7.81\sigma = \sqrt{\frac{610}{10}} = \sqrt{61} \approx 7.81

Explanation:

We first identify the class marks, then find the arithmetic mean. Using the mean, we find the squared deviations for each class, multiply them by their respective frequencies, and finally take the square root of the average squared deviation.