Mean deviation

Mean deviation or the average deviation is the average of the deviation of the item from the measure of central tendency i.e. mean, median or mode. Mean deviation, however, gives best results when deviations are taken from the median. While computing the value of mean deviation, positive and negative signs are ignored, i.e., all value are treated as positive.

Theoretically, deviations can be taken from any of the three average mentioned above but in actual practice mean deviation is calculated either from mean or from the median. Between mean and median, the latter is supposed to be better than the former, because the sum of the deviations from the median is less than the sum of the deviations from the mean. Therefore, the value of the mean deviation from the median is always less than the value calculated from mean. So, as a choice between mean and median theoretically, the median should be preferred. But because of wide applications of mean in statistics, in practice, mean deviation is generally computed from mean. In aggregating deviations, the algebraic signs are not taken into account. There is no mathematical justification for ignoring the signs of the deviation. However, simple reasons may be that if we do not ignore the signs of deviation, then the mean deviation from mean becomes zero. Since the main objective of a measure of dispersion is to study the scatter of the given observations from a central value, the ignoring of the deviation does not matter much. However, leaving of plus and minus signs render mean deviation incapable of further algebraic treatment.

formula for mean deviation
formula for mean deviation

Calculation of Mean Deviation

Individual Series

Let X1, X2, X3, ...... Xn be the N-variate value and \(\overline {X}\) and Md be the ar thematic mean and median of the series. The mean deviation is defined by the following formula:

M.D. from mean = \(\frac{\sum|X - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum|X - Md|}{N}\)

.
.

Discrete Series

Let X1, X2, X3, ....... Xn be the variate value and f1, f2, f3, ........... fn be their corresponding frequencies respectively. If \(\overline {X}\) and Md are arithmetic mean and median of the series, then mean deviation is defined by the following formula:

M.D. from mean = \(\frac {\sum{f}|X - \overline{X}|}{\sum{f}}\) = \(\frac {\sum{f}|X - \overline{X}|}{N}\)

M.D. from median = \(\frac {\sum{f}|X - Md|}{\sum{f}}\) = \(\frac {\sum{f}|X - Md|}{N}\)

where, N = \(\sum{f}\) = total frequency

Continuous Series

Let X1, X2, X3, ...... Xnbe the mid-value of corresponding classes and f1, f2, f3, ........ fn be their corresponding frequencies. If \(\overline {X}\) and Md are arithmetic mean and median, then mean deviation is defined by the following formula:

M.D. from mean = \(\frac{\sum{f}|m - \overline{X}|}{\sum{f}}\) = \(\frac{\sum{f}|m - \overline{X}|}{N}\)

M.D. from median = \(\frac{\sum{f}|m - Md|}{\sum{f}}\) = \(\frac{\sum{f}|m - Md|}{N}\)

where, N = \(\sum{f}\) = total frequency

Mean deviation is an absolute measure of dispersion. For compairing two or more series having different unit, we calculate coefficient of mean deviation. The relative measure of dispersion based on mean deviation is called coefficient of mean deviation.

Coefficient of M.D. from mean = \(\frac {M.D. from\;mean}{Mean}\)

Coefficient of M.D. from median = \(\frac {M.D. from\;median}{Median}\)

Merits and Demerits of Mean Deviation

Merits

  1. Mean deviation is based on all the observations and is thus definitely a better measure of dispersion than the range and quartile deviation.
  2. Mean deviation is rigidly defined and is easy to understand and calculate.
  3. Mean deviation is rigidly understood. It is the average of the deviations from a measure of central tendency.
  4. As compared with standard deviation, it is less affected by extreme observations.
  5. Since mean deviation is based on the deviation about an average, it provides a better measure for comparison about the formation of different distributions.

Demerits

  1. The strongest objection against mean deviation is that while computing its value we take the absolute value of the deviations about an average and ignore the signs of the deviations.
  2. The step of ignoring the signs of the deviations is mathematically unsound and illogical. It creates artificially and renders mean deviation useless for further mathematical treatment.

Coefficient of M.D = (M.D) ÷ Median


Calculation of mean deviation from mean

Wages (Rs) X D = X - \(\overline X\) \(\begin {vmatrix}D\\ \end {vmatrix}\)
90 -18 18
100 -8 8
110 2 2
115 7 7
125 17 17
\(\sum\)X = 540 \(\sum\)\(\begin {vmatrix}D\\ \end {vmatrix}\) = 52

Mean (\(\overline X\)) = \(\frac {\sum X}N\) = \(\frac {540}5\) = 108

MeanDeviation from mean = \(\frac {\sum\begin {vmatrix}D\\ \end {vmatrix}}N\) = \(\frac {52}5\) = 10.4Ans

The given data in ascending order:

100. 150, 200, 250, 300, 350, 400

N = 7

\begin{align*} Median\; (M_d) &= (\frac {N + 1}2)^{th} term\\ &= (\frac {7 + 1}2)^{th}) term\\ &= (\frac 82)^{th} term\\ &= 4^{th} term\\ &= 250\\ \end{align*}

Calculating the mean deviation from median

x x - mdn. \(\begin{vmatrix} x - mdn\\ \end{vmatrix}\)
100 100 - 250 = -150 150
150 150 - 250 = -100 100
200 200 - 250 = -50 50
250 250 - 250 = 0 0
300 300 - 250 = 50 50
350 350 - 250 = 100 100
400 400 - 250 = 150 150
N = 7 \(\sum\)\(\begin{vmatrix} x - mdn\\ \end{vmatrix}\) = 600

\begin{align*} Mean\; Deviation\; from\; median\; &= \frac {\sum\begin{vmatrix} x - mdn\\ \end{vmatrix}}N\\ &= \frac {600}7\\ &= 85.71\\ \end{align*}

∴ Mean Deviation from median = 85.71Ans

Given data is:

40, 44, 54, 60, 62

N = 5

\begin{align*} Median\; (M_d) &= \frac {(N+1)^{th}}2\; term\\ &= \frac {(5+1)^{th}}2\;term\\ &= \frac {6^{th}}2\;term\\ &= 3^{rd}\; term\\ \end{align*}

Position of 3rd term = 54

∴ median (mdn) = 54

Calculating mean deviation from median

x x - median \(\begin{vmatrix}x - median\end{vmatrix}\)
40 -14 14
44 -10 10
54 0 0
60 6 6
62 8 8
\(\sum\)\(\begin{vmatrix}x - median\end{vmatrix}\) = 38

\begin{align*} Mean\; deviation\; from\; median &= \frac {\sum\begin{vmatrix}x - median\end{vmatrix}}N\\ &= \frac {38}5\\ &= 7.6_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Mean

Score

(x)

Frequency

(f)

fx \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}x - \overline{X}\\ \end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
1 2 2 2 4
2 5 10 1 5
3 6 18 0 0
4 5 20 1 5
5 2 10 2 4
N = 20 \(\sum\)fx = 60 \(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 18

\begin{align*} Mean\; (\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {60}{20}\\ &= 3\\ \end{align*}

\begin{align*} Mean\; Deviation\; from\; mean\;(M.D) &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {18}{20}\\ &= 0.9_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;M.D &= \frac {M.D.}{Mean}\\ &= \frac {0.9}3\\ &= 0.3_{Ans}\\ \end{align*}

Calculating Mean Deviation from Mean

x f fx \(\begin{vmatrix}x - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}x - \overline{X}\end{vmatrix}\)
10 2 20 12 24
15 4 60 7 28
20 6 120 2 12
25 8 200 3 24
30 5 150 8 40
N = 25 \(\sum\)fx = 550 \(\sum\)f\(\begin{vmatrix}x - \overline{X}\end{vmatrix}\) = 128

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {550}{25}\\ &= 22\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}x - \overline{X}\end{vmatrix}}N\\ &= \frac {128}{25}\\ &= 5.12_{Ans}\\ \end{align*}

Calculation of M.D. from Mean

X f fx \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}X - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
20 2 40 8 16
18 4 72 6 24
16 9 144 4 36
14 18 252 2 36
12 27 324 0 0
10 25 250 2 50
8 14 112 4 56
6 1 6 12 6
N = 100 \(\sum\)fX = 1200 \(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 224

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fX}}N\\ &= \frac {1200}{100}\\ &= 12\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {224}{100}\\ &= 2.24_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;the\;Mean\;Deviation\;from\;Mean &= \frac {M.D.}{Mean}\\ &= \frac {2.24}{12}\\ &= 0.186_{Ans}\\ \end{align*}

Here,

The maximum frequencyis 40 for value of variable 30 so mode (Mo) = 30

Marks (X) No. of students (f) \(\begin{vmatrix}X - M_o \end{vmatrix}\) f\(\begin{vmatrix}X - M_o \end{vmatrix}\)
20 12 10 120
24 28 6 168
30 40 0 0
32 26 2 52
40 14 10 140
N = 120 \(\sum\)f\(\begin{vmatrix}X - M_o \end{vmatrix}\) = 480

\begin{align*} M.D.\;from\;Mode &= \frac {\sum{f}\begin{vmatrix}X - M_o \end{vmatrix}}N\\ &= \frac {480}{120}\\ &= 4_{Ans} \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mode &= \frac {M.D.\;from\;Mode}{mode}\\ &= \frac 4{30}\\ &= 0.133_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Mean

Class interval Mid-value (m) f fm \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
0-10 5 2 10 23 46
10-20 15 3 45 13 39
20-30 25 6 150 3 18
30-40 35 5 175 7 35
40-50 45 4 180 17 68
N = 20 \(\sum fm\) = 560 \(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 206

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {560}{20}\\ &= 28\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &=\frac {\sum {f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {206}{20}\\ &= 10.3_{Ans}\\ \end{align*}

Calculation of M.D. from Mean

Class interval

(x)

Mid-value

(m)

frequency

(f)

fm \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
10-20 15 3 45 29.28 87.84
20-30 25 5 125 19.28 96.4
30-40 35 4 140 9.28 37.12
40-50 45 5 225 0.72 3.6
50-60 55 4 220 10.72 42.88
60-70 65 4 260 20.72 82.88
70-80 75 3 225 30.72 92.16
N = 28 \(\sum\) N = 28 \(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 442.88

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {1240}{28}\\ &= 44.28\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum {f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {442.88}{28}\\ &= 15.82_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;M.D.\;from\;Mean &= \frac {M.D.}{Mean}\\ &= \frac {15.82}{44.28}\\ &= 0.36_{Ans}\\ \end{align*}

Let: Assumed mean (A) = 25

Calculation of Mean Deviation from Mean

Class interval mid-value f d = \(\frac {m-25}{10}\) fd \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m = \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
0-10 5 2 -2 -4 23 46
10-20 15 3 -1 -3 13 39
20-30 25 6 0 0 3 18
30-40 35 5 1 5 7 35
40-50 45 4 0 8 17 68
N = 20 \(\sum {fd}\) = 6 \(\sum\)f\(\begin{vmatrix}D\end{vmatrix}\) = 206

\begin{align*} Mean\;(\overline{X}) &= A + \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N× I\\ &= 25 + \frac 6{20}× 10\\ &= 28\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {206}{20}\\ &= 10.3_{Ans}\\ \end{align*}

Calculating Mean Deviation from Mean

Weight in kg. Mid-value (m) No. of men (f) fm \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
20-30 25 2 50 20 40
30-40 35 5 175 10 50
40-50 45 6 270 0 0
50-60 55 5 275 10 50
60-70 65 2 130 20 40
N = 20 \(\sum {fm}\) = 900 \(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 180

\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {900}{20}\\ &= 45\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean\;(M.D.) &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {180}{20}\\ &= 9\;kg_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Mean

Wages (in Rs.) Mid-value (m) f fm \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - \overline{X}\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
0-4 2 5 10 1.09 5.45
4-8 6 7 42 30.91 216.37
8-12 10 10 100 88.91 889.1
12-16 14 15 210 198.91 2983.65
16-20 18 7 126 114.91 804.37
N =44 \(\sum {fm}\) =488 \(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 4898.94

\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {488}{44}\\ &= 11.09\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mean &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {4898.94}{44}\\ &= 113.3_{Ans}\\ \end{align*}

Calculating Mean Deviation from Median

Marks (X) No. of students (f) c.f. \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}X - mdn\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
20 4 4 420 80
30 7 11 10 70
40 12 23 0 0
50 2 25 10 20
60 4 29 20 80
70 6 35 30 180
N = 35 \(\sum f\)\(\begin{vmatrix}D\end{vmatrix}\) = 430

\begin{align*} Position\;of\;Median &= (\frac {N+1}2)^{th}\;item\\ &= (\frac {35+1}2)^{th}\;item\\ &= 18^{th}\;item\\ \end{align*}

18thitem = 40

∴ Mdn = 40

\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {430}{35}\\ &= 12.28_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;Mean\;Deviation\;from\;Median &= \frac {M.D.\;from\;Median}{Median}\\ &= \frac {12.28}{40}\\ &= 0.307_{Ans}\\ \end{align*}

Calculation of M.D. from Median

Marks f c.f mid-value (m) \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - md.\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
0-10 4 4 5 30 120
10-20 6 10 15 20 120
20-30 10 20 25 10 100
30-40 10 30 35 0 0
40-50 15 45 45 10 150
50-60 5 50 55 20 100
N = 50 \(\sum {f}\)\(\begin{vmatrix}D\end{vmatrix}\) = 590

\begin{align*} Median &= \frac {N^{th}}2\;item\\ &= \frac {50^{th}}2\;item\\ &= 25^{th}\;term\\ \end{align*}

Median Class = (30-40)

Here,

L = 30

\(\frac N2\) = 25

cf = 20

f = 10

I = 10

\begin{align*} Median\;(mdn) &= L\;+\;\frac{\frac N2\;-\;cf}f\; ×\;I\\ &= 30\;+\;\frac {25-20}{10}\;×\;10\\ &= 30\;+\;5\\ &= 35\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {590}{50}\\ &= 11.8_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Median

Marks f c.f mid-value (m) \(\begin{vmatrix}D\end{vmatrix}\) = \(\begin{vmatrix}m - md.\end{vmatrix}\) f\(\begin{vmatrix}D\end{vmatrix}\)
10-20 6 6 15 25 150
20-30 8 14 25 15 120
30-40 11 25 35 5 55
40-50 14 39 45 5 70
50-60 8 47 55 15 120
60-70 3 50 65 25 75
N = 50 \(\sum {f}\)\(\begin{vmatrix}D\end{vmatrix}\) = 590

\begin{align*} Position\;of\;Median &= \frac {N^{th}}2\;item\\ &= \frac {50^{th}}2\;item\\ &= 25^{th}\;term\\ \end{align*}

Median Class = (30-40)

Here,

L = 30

\(\frac N2\) = 25

cf =14

f =11

I = 10

\begin{align*} Median\;(mdn) &= L\;+\;\frac{\frac N2\;-\;cf}f\; ×\;I\\ &= 30\;+\;\frac {25-14}{11}\;×\;10\\ &= 30\;+\;\frac {11}{11}\;×\;10\\ &= 40\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Median &= \frac {\sum{f}\begin{vmatrix}D\end{vmatrix}}N\\ &= \frac {590}{50}\\ &= 11.8\\ \end{align*}

\begin{align*} Coefficient\;of\;M.D.\;from\;Median &= \frac {M.D.}{Median}\\ &= \frac {11.8}{40}\\ &= 0.295_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Median

Age (in year)

X

No. of people

(f)

Cumulative Frequency

(c.f.)

\(\begin{vmatrix}X - M_d\end{vmatrix}\) f\(\begin{vmatrix}X - M_d\end{vmatrix}\)
10 6 6 5 30
12 14 20 3 42
15 20 40 0 0
16 13 53 1 13
20 7 60 5 35
N = 60 \(\sum f\)\(\begin{vmatrix}X - M_d\end{vmatrix}\) = 120

\begin{align*} Median\;(M_d) &= (\frac {N\;+\;1}2)^{th}\;item\\ &= (\frac {60\;+\;1}2)^{th}\;item\\ &= (30.5)^{th}\;item &= 15\\ \end{align*}

\begin{align*} M.D.\;from\;Median &= \frac {\sum {f}\begin{vmatrix}X - M_d\end{vmatrix}}N\\ &= \frac {120}{60}\\ &= 2_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Mode

Age (in years) f Mid-value (X) \(\begin{vmatrix}X - M_o\end{vmatrix}\) f\(\begin{vmatrix}X - M_o\end{vmatrix}\)
0-4 4 2 7.6 3.04
4-8 6 6 3.6 21.6
8-12 8 10 0.4 3.2
12-16 5 14 4.4 22
16-20 2 18 8.4 16.8
N = 25 \(\sum f\)\(\begin{vmatrix}X - M_o\end{vmatrix}\) = 34

Model Class = The class containing maximum frequency = (8-12)

Here,

L = 8

f0 = 6

f1 = 8

f2 = 5

\begin{align*} Mode\;(M_o) &= L\;+\;\frac {f_1\;-\;f_0}{2f_1\;-\;f_0\;-\;f_2}\\ &= 8 + \frac {8\;-\;6}{2\;×\;8\;-\;6\;-\;5}\\ &= 8\;+\;\frac 85\\ &= 9.6\;years\\ \end{align*}

\begin{align*} Mean\;Deviation\;from\;Mode &= \frac {\sum {f}\begin{vmatrix}X - M_o\end{vmatrix}}N\\ &= \frac {34}{25}\\ &= 1.36\;years_{Ans}\\ \end{align*}

Calculation of Mean Deviation from Median

Age (in years) Mid-value (X) Frequency (f) c.f. \(\begin{vmatrix}X - M_d\end{vmatrix}\) f\(\begin{vmatrix}X - M_d\end{vmatrix}\)
20-30 25 5 5 18.75 93.75
30-40 35 7 12 8.75 61.25
40-50 45 8 20 1.25 10
50-60 55 6 26 11.25 67.5
60-70 65 4 30 21.25 85
N = 30 \(\sum f\)\(\begin{vmatrix}X - M_d\end{vmatrix}\) = 317.5

\begin{align*} Class\;of\;Median &= (\frac N2)^{th}\;item\\ &= (\frac {30}2)^{th}\;item\\ &= 15^{th}\;item\\ \end{align*}

15th item represents (40-50) class

Here,

L = 40

\(\frac N2\) = 15

c.f. = 12

f = 8

I = 10

\begin{align*} Median\;(M_d) &= L\;+\;\frac {\frac N2 - c.f.}f\;×\;I\\ &= 40\;+\;\frac {15\;-\;12}8\;×\;10\\ &= 40\;+\;\frac {30}8\\ &= 43.75\\ \end{align*}

\begin{align*} M.D.\;from\;Median &= \frac {\sum {f}\begin{vmatrix}X - M_d\end{vmatrix}}N\\ &= \frac {317.5}{30}\\ &= 10.58years_{Ans}\\ \end{align*}

0%
  • Find the range and the coefficient of range from following data:

    Temp.o 0-10 10-20 20-30 30-40 40-50
    Days 5 12 25 18 2

    65, 1


    60,1


    90, 1


    50, 1


  • Compute mean deviation from mean:

    class 0-10 10-20 20-30 30-40
    f 2 4 5 1

    9.1, 0.27


    7.5,0.1


    M.D.(ar X) = 7.5, 0.39


    11.125, 0.23


  • x 10-19 20-29 30-39 40-49 50-59
    f 4 10 18 16 12

    9.91, 0.26


    8.4,0.23


    7.5, 0.37


    7.91, 0.19


  • comput

    class 0-10 10-20 20-30 30-40
    f 2 4 5 1

    e mean deviation from median. Also calculate coefficient of them.

     

    7.1,026


    MD = 7.5, 0.37


    MD = 6.5,0.24


    8.1, 0.32


  • class 10-20 10-30 10-40 10-50
    f 2 4 9 20

    8.2, 0.11


    6.5, 0.145


    7.91, 0.19


    7.62,0.23


  • Marks 0-10 0-20 0-30 0-40 0-50
    No. of students 4 12 24 44 62

    MD =9.7, 0.29


    7.8, 0.45


    7.9, 0.56


    8.5, 0.23


  • compute the mean deviation from the media and its coefficient from the data given below:

    x 1 2 3 4 5
    f 2 5 6 5 2

    0.5,0.8


    0.6, 0.21


    0.8, 0.3


    0.9, 0.3


  • Find the mean deviation from the mean  and its coeffiecent

    Marks  0-10 10-20 20-30 30-40 40-50
    No. of student 4 10 17 30 19

    8.5, 0.63


    7.2, 0.36


    9.1, 00.22


    10.09,0.32


  • Find the mean deviation and its coefficient from the following data:

    x 15 20 10 5 25 30
    f 4 6 5 8 4 4

    0.45


    0.31


    0.81


    0.22


  • Compute the mean deviation from media and its coefficient from the data given below:

    x 20 14 26 12 28
    f 18 12 12 4 4

    5, 0


    3, 1


    12, 5


    0, 0


  • Find the mean deviation from median and its coefficient from the given data:

    Marks obtained 20 30 50 40 60 70
    No. of Students 4 7 12 2 4 6

    21.75, 0.9


    4.0, 0.69


    13.72, 0.33


    12.29, 0.31


  • Find  mean, median  and its coefficient.

    x 10-19 20-29 30-39 40-49 50-59
    f 4 10 18 16 12

    M.D = 9.91, 0.27


    3.01,0.89,0.14


    1.02,2.2,0.33


    2.02,0.2,0.33


  • Compute mean deviation from mean. Also calculate the coefficient.

     

    x 15 20 10 5 25 30
    f 4 6 5 8 4 4

    3


    0.1


    1


    0.8


  • You scored /13


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