krit.club logo

Statistics - Mean deviation for grouped data

Grade 11CBSE

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

🔑Concepts

•

Grouped data is classified into two types: Discrete Frequency Distribution and Continuous Frequency Distribution.

•

For a Discrete Frequency Distribution, the data consists of nn distinct values x1,x2,…,xnx_1, x_2, \dots, x_n occurring with frequencies f1,f2,…,fnf_1, f_2, \dots, f_n respectively.

•

For a Continuous Frequency Distribution, data is given in class intervals. The mid-point of each class (class mark), denoted by xix_i, represents the observations.

•

Mean Deviation about Mean (M.D.(xˉ)M.D.(\bar{x})) measures the average of the absolute differences between each observation xix_i and the arithmetic mean xˉ\bar{x}, weighted by their frequencies fif_i.

•

Mean Deviation about Median (M.D.(M)M.D.(M)) measures the average of the absolute differences between each observation xix_i and the median MM, weighted by their frequencies fif_i.

•

In continuous distributions, the median is calculated using the formula L+(N2−Cf)×hL + \left( \frac{\frac{N}{2} - C}{f} \right) \times h, where LL is the lower limit of the median class, CC is the cumulative frequency of the preceding class, ff is the frequency of the median class, and hh is the class width.

📐Formulae

xˉ=∑i=1nfixi∑i=1nfi=1N∑i=1nfixi\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} = \frac{1}{N} \sum_{i=1}^{n} f_i x_i

M.D.(xˉ)=∑i=1nfi∣xi−xˉ∣NM.D.(\bar{x}) = \frac{\sum_{i=1}^{n} f_i |x_i - \bar{x}|}{N}

M=L+(N2−Cf)×hM = L + \left( \frac{\frac{N}{2} - C}{f} \right) \times h

M.D.(M)=∑i=1nfi∣xi−M∣NM.D.(M) = \frac{\sum_{i=1}^{n} f_i |x_i - M|}{N}

💡Examples

Problem 1:

Find the mean deviation about the mean for the following discrete frequency distribution: xix_i: 2, 5, 6, 8, 10 fif_i: 2, 8, 10, 7, 8

Solution:

  1. Calculate N=∑fi=2+8+10+7+8=35N = \sum f_i = 2 + 8 + 10 + 7 + 8 = 35.
  2. Calculate ∑fixi=(2×2)+(5×8)+(6×10)+(8×7)+(10×8)=4+40+60+56+80=240\sum f_i x_i = (2 \times 2) + (5 \times 8) + (6 \times 10) + (8 \times 7) + (10 \times 8) = 4 + 40 + 60 + 56 + 80 = 240.
  3. Mean xˉ=24035≈6.86\bar{x} = \frac{240}{35} \approx 6.86.
  4. Calculate absolute deviations ∣xi−xˉ∣|x_i - \bar{x}|: ∣2−6.86∣=4.86|2 - 6.86| = 4.86, ∣5−6.86∣=1.86|5 - 6.86| = 1.86, ∣6−6.86∣=0.86|6 - 6.86| = 0.86, ∣8−6.86∣=1.14|8 - 6.86| = 1.14, ∣10−6.86∣=3.14|10 - 6.86| = 3.14.
  5. Calculate ∑fi∣xi−xˉ∣=(2×4.86)+(8×1.86)+(10×0.86)+(7×1.14)+(8×3.14)=9.72+14.88+8.6+7.98+25.12=66.3\sum f_i |x_i - \bar{x}| = (2 \times 4.86) + (8 \times 1.86) + (10 \times 0.86) + (7 \times 1.14) + (8 \times 3.14) = 9.72 + 14.88 + 8.6 + 7.98 + 25.12 = 66.3.
  6. M.D.(xˉ)=66.335≈1.89M.D.(\bar{x}) = \frac{66.3}{35} \approx 1.89.

Explanation:

We first find the weighted mean of the observations. Then, we find the absolute difference of each observation from this mean, multiply by the respective frequency, and divide the sum of these products by the total frequency NN.

Problem 2:

Calculate the total frequency NN and the mean xˉ\bar{x} for a set where the sum of fixif_i x_i is calculated as follows: 120450300+1301000\begin{array}{r} 120 \\ 450 \\ 300 \\ + 130 \\ \hline 1000 \end{array} Given N=50N = 50.

Solution:

Sum of products ∑fixi=1000\sum f_i x_i = 1000. Total frequency N=50N = 50. Mean xˉ=∑fixiN=100050=20\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1000}{50} = 20.

Explanation:

The vertical addition shows the sum of fixif_i x_i is 10001000. Using the mean formula for grouped data, we divide this by the total frequency N=50N=50 to get 2020.