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Statistics - Mean deviation for ungrouped data

Grade 11CBSE

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

🔑Concepts

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Dispersion refers to the scattering or spread of data around a central value. Mean Deviation is a measure of dispersion that considers the average of absolute differences from a central tendency.

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Mean Deviation for ungrouped data can be calculated about two central values: the Mean (xˉ\bar{x}) or the Median (MM).

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The absolute value symbol ∣xi−a∣|x_i - a| is used to ensure that all deviations are treated as positive distances, preventing positive and negative deviations from cancelling each other out.

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To find Mean Deviation about the Mean, first calculate the arithmetic mean xˉ\bar{x}, then find the sum of absolute differences ∣xi−xˉ∣|x_i - \bar{x}|, and divide by the total number of observations nn.

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To find Mean Deviation about the Median, first arrange the data in ascending order to find the median MM, then calculate the sum of absolute differences ∣xi−M∣|x_i - M|, and divide by nn.

📐Formulae

xˉ=∑i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}

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

M={(n+12)th observationif n is odd(n2)th obs+(n2+1)th obs2if n is evenM = \begin{cases} \left(\frac{n+1}{2}\right)^{th} \text{ observation} & \text{if } n \text{ is odd} \\ \frac{(\frac{n}{2})^{th} \text{ obs} + (\frac{n}{2}+1)^{th} \text{ obs}}{2} & \text{if } n \text{ is even} \end{cases}

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

💡Examples

Problem 1:

Find the mean deviation about the mean for the following data: 6,7,10,12,13,4,8,126, 7, 10, 12, 13, 4, 8, 12.

Solution:

  1. Find the Mean xˉ\bar{x}: n=8n = 8 xˉ=6+7+10+12+13+4+8+128=728=9\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9
  2. Calculate absolute deviations ∣xi−9∣|x_i - 9|: ∣6−9∣=3,∣7−9∣=2,∣10−9∣=1,∣12−9∣=3,∣13−9∣=4,∣4−9∣=5,∣8−9∣=1,∣12−9∣=3|6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3
  3. Sum of absolute deviations: 3213451+322\begin{array}{r} 3 \\ 2 \\ 1 \\ 3 \\ 4 \\ 5 \\ 1 \\ + 3 \\ \hline 22 \end{array}
  4. M.D.(xˉ)=228=2.75M.D.(\bar{x}) = \frac{22}{8} = 2.75

Explanation:

We first calculated the arithmetic mean of the 8 values. Then, we found the distance of each value from the mean (ignoring the sign) and averaged those distances.

Problem 2:

Find the mean deviation about the median for the data: 3,9,5,3,12,10,18,4,7,19,213, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Solution:

  1. Arrange data in ascending order: 3,3,4,5,7,9,10,12,18,19,213, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
  2. Find Median (MM): n=11n = 11 (odd) M=(11+12)th term=6th term=9M = \left(\frac{11+1}{2}\right)^{th} \text{ term} = 6^{th} \text{ term} = 9
  3. Calculate absolute deviations ∣xi−9∣|x_i - 9|: ∣3−9∣=6,∣3−9∣=6,∣4−9∣=5,∣5−9∣=4,∣7−9∣=2,∣9−9∣=0,∣10−9∣=1,∣12−9∣=3,∣18−9∣=9,∣19−9∣=10,∣21−9∣=12|3-9|=6, |3-9|=6, |4-9|=5, |5-9|=4, |7-9|=2, |9-9|=0, |10-9|=1, |12-9|=3, |18-9|=9, |19-9|=10, |21-9|=12
  4. Sum of deviations: 6+6+5+4+2+0+1+3+9+10+12=586+6+5+4+2+0+1+3+9+10+12 = 58
  5. M.D.(M)=5811≈5.27M.D.(M) = \frac{58}{11} \approx 5.27

Explanation:

Since the number of observations is odd, the median is the middle-most value after sorting. We then calculated the average of the absolute differences between each data point and this median.