Statistics Test - 54

Mean deviation about mean MD \(=\left(\frac{1}{N}\right)\left(\sum\left|x_{i}-\bar{x}\right|\right) \quad \quad \ldots\) (1)

also, \(\overline{ x }= a +\sum \frac{fd _{ i }}{N}\)

Put \(\bar{x}\) in equation (i) we get

\(MD =\left(\frac{1}{N}\right)\left(\sum x _{ i }-\left( a +\sum \frac{fd _{ i }}{N} \right)\right.\)

After shifting or changing the origin \(x _{ i }= d _{ i }+ a\)

\(MD =\left(\frac{1}{N}\right)\left(\sum \mid d _{ i }+ a - a -\sum \frac{fd_{i}}{N} \right)\)

\(\Rightarrow MD =\left(\frac{1}{N}\right)\left(\sum \mid d _{ i }-\sum \frac{fd _{i}}{N} \right)\)

\(\Rightarrow MD =\left(\frac{1}{N}\right)\left(\sum \mid d _{ i }-\overline{ d }\right)\)

Which is independent of \(a\).

N is the number of observations, a is the origin\(\ fd _{ i }\) is the frequency distribution \(\sum\) is the summation.

\(\therefore\) Absolute mean deviation about mean is independent of change of the origin.

Mean Deviation from Mean \(=\frac{\sum|x-\bar{x}|}{n}\)

Mean Deviation from Median \(=\frac{\sum|x-M|}{n}\)

Here,

\(\sum\) represents the summation

\(x=\) Observations given

\(\bar{x}=\) Mean

\(n =\) The number of observations

\(M =\) Median

Mean \(=\frac{(1+0+2+3+1+1+15+1+3) }{9}=\frac{27}{9}=3\)

Mean Deviation (M.D) about mean will be = Subtract Mean from each given data set values

Example, \(0-3=3\) (don't consider the sign here, only focus on the values)

\(1-3=2,2-3=1,3-3=0,15-3=12\)

M.D (Mean) \(=\frac{(3+2+2+2+2+1+0+12)}{9}=\frac{24}{9}=\frac{8}{3}\)

As we have calculated above the value of Mean Deviation(M.D) \(=3,2,2,2,2,1,12\)

The value of Variance \(=\frac{\left(3^{2}+2^{2}+2^{2}+2^{2}+2^{2}+1^{2}+12^{2}\right)}{9}=\frac{\sqrt{170}}{\sqrt{9}}={\frac{\sqrt{170}}{3}}\)

\(\therefore\) Standard Deviation (S.D) \(=\sqrt{\left(\frac{170}{9}\right)^{2}}=\frac{170}{9}\)

Given scores 10, 12, 13, 15, 15, 13, 12, 10, x and mode = 15

The mode of the n observation is the number that has the highest frequency.

Frequency of score '12' = 2

Frequency of score '15' = 2

But for mode to be 15, x should be '15'.

We know that:

Coefficient of variation \(=\frac{\text {standard deviation}}{\text{mean}}\)

Coefficient of variation for Group A \(=\frac{10}{22}=0.4545\)

Coefficient of variation for Group \(B=\frac{12}{23}=0.522\)

Group A is less variable than Group B because Group A's coefficient of variation is smaller.

Let, the missing number be \(x\)

Given Mean \(= 25\), Total elements \(= 8\)

Mean of n elements \(= \frac{\text{Sum of all n elements}}{\text{Total number of elements (n)}}\)

\(\frac{30+24+27+22+18+26+32+x830+24+27+22+18+26+32+x}{8} = 25\)

\(179 + x = 200\)

\(x = 21\)

Case 1:

It is a fundamental property.

Example

Consider a set of 5 data points

{1, 2, 3, 4, 5}

median = 3

Deviation from mean:

1 - 3 = -2

2 - 3 = -1

3 - 3 = 0

4 - 3 = 1

5 - 3 = 2

Mean of deviations (about mean) = \(\frac{0}{5}\) = 0

Case 2:

Consider data point 2

Deviations of data about 2

1 - 2 = - 1

2 - 2 = 0

3 - 2 = 1

4 - 2 = 2

5 - 2 = 3

Mean of deviation of data about 2 = \(\frac{(-1 + 0 + 1 + 2 + 3)}{5}\) = 1

Mean of deviation of data about median is minimum and this is true for all data.

From the given data it is cleared data the highest frequency is 14 and it belongs to 20 - 30 class interval so 20 - 30 is the modal class.

L = 20, \(f_1=14, f_2=12, f_0=10,\)  I = 10

Mode \(=L+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h\)

⇒ 20 + [\(\frac{(14 - 10)}{(2 × 14 - 10 - 12)}\)] × 10 = 26.66

Given data is \(170,430,300,600\) and 470

Mean \(\bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\)

\(=\frac{170+430+300+600+470}{5}\)

\(=394\)

For '\(n\)' observation \(x_{1}, x_{2} \ldots \ldots \ldots \ldots . . x_{n}\), the mean deviation about their mean \(\bar{x}\) is given by,

\(M . D=\frac{\sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|}{N}\) where, \(N\) is the number of observations.

Mean deviation \(=\frac{|170-394|+|430-394|+|300-394|+|600-394|+|470-394|}{5}\)

\(=\frac{636}{5}\)

\(=127.2\)

Mean Deviation for ungrouped data:

For 'N' observation \(x_{1}, x_{2} \ldots \ldots \ldots \ldots . x_{n}\), the mean deviation about their mean \(\bar{x}\) is given by \(M . D=\frac{\sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|}{N}\)

where, \(N\) is the number of observations

Given data of numbers are \(p, 6,6,7,8,11,15\), and 16.

Mean \(\bar{x}=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\)

\(=\frac{p+6+6+7+8+11+15+16}{8}=3 p\)

\(\therefore 23 p=69\)

\(\Rightarrow p =3\)

So, given data is \(3,6,6,7,8,11,15,16\) and mean is \(3 p\), i.e., 9.

\(\therefore\) Mean deviation \(=\frac{|3-9|+|6-9|+|6-9|+|7-9|+|8-9|+|11-9|+|15-9|+|16-9|}{8}\)

Mean deviation \(=\frac{\left|x_{i}-\bar{x}\right|}{n}=\frac{30}{8}=3.75\)

Given: Number of observation \((N) =20\)

\(\sum x_{i}=1000\) and \(\sum x_{i}^{2}=84000\)

Variance \(=\frac{\sum x_{ i }^{2}}{ N }-\frac{\left(\sum x _{ i }\right)^{2}}{ N ^{2}}\)

\(=\frac{84000}{20}-\frac{1000^{2}}{20^{2}}\)

\(=4200-2500\)

\(=1700\)