Measures of Dispersion Test

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Measures of Dispersion Test - 8

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Question 1
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$$5$$ students of a class have an average height $$150\ cm$$ and variance $$18\ cm^{2}$$. A new student, whose height is $$156\ cm$$, joined them. The variance (in $$cm^{2})$$ of the height of these six students is

A
$$22$$

B
$$20$$

C
$$16$$

D
$$18$$

Solution

Given $$\bar {x} = \dfrac {\sum x_{i}}{5} = 150$$
$$\Rightarrow \displaystyle \sum_{i = 1}^{5} x_{i} = 750 .... (i)$$
$$\sigma^2=18$$
$$\dfrac {\sum x_{i}^{2}}{5} - (\bar {x})^{2} = 18$$
$$\dfrac {\sum x_{i}^{2}}{5} - (150)^{2} = 18$$
$$\sum x_{i}^{2} = 112590 ..... (ii)$$
Given height of new student
$$x_{6} = 156$$
Now, $$\bar {x}_{new} = \dfrac {\displaystyle \sum_{i = 1}^{5}x_{i}}{6} = \dfrac {750 + 156}{6} = 151$$
Also, $$\sigma_{new}^2= \dfrac {\displaystyle \sum_{i = 1}^{5}x_{i}^{2}}{6} - (\overline {x}_{new})^{2}$$
$$= \dfrac {112590 + (156)^{2}}{6} - (151)^{2}$$
$$= 22821 - 22801 = 20$$.
Question 2
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A student scores the following marks in five test: $$45, 54, 41, 57, 43$$. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six test is

A
$$\dfrac{10}{\sqrt{3}}$$

B
$$\dfrac{100}{\sqrt{3}}$$

C
$$\dfrac{130}{3}$$

D
$$\dfrac{10}{3}$$

Solution

Let $$x$$ be the $$6^{th}$$ observation
$$\Rightarrow 45 + 54 + 41 + 57 + 43 + x = 48 \times 6 = 288$$
$$\Rightarrow x = 48$$
Variance $$= \left(\dfrac{\sum x_i^2}{6} - (\bar{X})^2\right)$$
$$\Rightarrow$$ variance $$=\dfrac{14024}{6} - (48)^2$$
$$=\dfrac{100}{3}$$
$$\Rightarrow$$ standard deviation $$=\dfrac{10}{\sqrt{3}}$$
Question 3
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If the data $$x_1, x_2, .., x_{10}$$ is such that the mean of first four of these is $$11$$, the mean of the remaining six is $$16$$ and the sum of square of all of these is $$2,000$$; then the standard deviation of this data is?

A
$$4$$

B
$$2$$

C
$$\sqrt{2}$$

D
$$2\sqrt{2}$$

Solution

$$x_1+...+x_4=44$$
$$x_5+.....+x_{10}=96$$
$$\vec{x}=14, \displaystyle\sum x_i=140$$
Variance $$=\dfrac{\displaystyle\sum x^2_i}{n}=\vec{x^2}=4$$
Standard deviation $$=2$$.
Question 4
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For two data sets, each of size $$ 5$$, the variances are given to be $$4$$ and $$5$$ and the corresponding means are given to be $$2$$ and $$4,$$ respectively. The variance of the combined data set is

A
$$\displaystyle \frac{11}{2}$$

B
$$6$$

C
$$\displaystyle \frac{13}{2}$$

D
$$\displaystyle \frac{5}{2}$$

Solution

$$\sigma_{\mathrm{x}}^{2}=4$$
$$\sigma_{\mathrm{y}}^{2}=5$$
$$\overline{\mathrm{x}}=2$$
$$\overline{\mathrm{y}}=4$$
$$\displaystyle \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{5}=2, \Sigma \mathrm{x}_{\mathrm{i}}=10;\Sigma \mathrm{y}_{\mathrm{i}}=20$$
$$
\sigma_{\mathrm{x}}^{2}=(\frac{1}{5}\Sigma \mathrm{x}_{\mathrm{i}}^{2})-(\overline{x})^{2}=\frac{1}{5}(\Sigma \mathrm{x}_{1}^{2})-4
$$
$$\sigma_{\mathrm{y}}^{2}=(\frac{1}{5}\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\overline{y})^{2}=\frac{1}{5}(\Sigma \mathrm{y}_{1}^{2})-4$$
$$
\Sigma \mathrm{x}_{\mathrm{i}}^{2}=40
$$
$$
\Sigma \mathrm{y}_{\mathrm{i}}^{2}=105
$$
$$\sigma_{\mathrm{z}}^{2}=\displaystyle \frac{1}{10}(\Sigma \mathrm{x}_{\mathrm{i}}^{2}+\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\frac{\overline{\mathrm{x}}+\overline{\mathrm{y}}}{2})=\frac{1}{10}(40+105)-9=\frac{145-90}{10}=\frac{55}{10}=\frac{11}{2}$$
Question 5
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The variance of first $$ 50$$ even natural numbers is

A
$$ \displaystyle \frac{833}{4}$$

B
$$ 833$$

C
$$ 437$$

D
$$ \displaystyle \frac{437}{4}$$

Solution

$${\textbf{Step-1: Use mean formula and find the mean.}}$$
$$\Rightarrow \sigma^2 = \dfrac{1}{n} \sum x_i^2 - \bar{(x)}^2$$

$$\Rightarrow n = 50, \sum x_i = 2 +4+6+8+...+100$$
$${\text{We know that,}}$$
$$\Rightarrow \bar {x} = \dfrac{\sum x_i}{n}$$

$$\Rightarrow \bar {x} = \dfrac{2 +4+6+8+...+100}{50}$$

$$\Rightarrow \bar {x} = \dfrac{50 \times 51}{50}$$ $$[\because \sum 2n = n(n+1)]$$

$$\Rightarrow \bar {x} = 51$$
$${\textbf{Step-2: Put the values in variance formula.}}$$
$$\Rightarrow \sigma^2 = \dfrac{1}{n} \sum x_i^2 - \bar{(x)}^2$$
$$= \dfrac{1}{50} (2^2 +4^2+6^2+8^2+...+100^2) - {(51)}^2$$
$$= \dfrac{1}{50} [({2.1})^2 +({2.2})^2+({2.3})^2+({2.4})^2+...+({2.50})^2] - {(51)}^2$$
$$= \dfrac{1}{50} 2^2[({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2] - {(51)}^2 ...(1)$$
$${\text{But,}}$$
$$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({n})^2 = \dfrac{n(n+1)(2n+1)}{6}$$
$$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(50+1)(2 \times 50+1)}{6}$$
$$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(51)(101)}{6}$$
$${\text{Equation (1) become,}}$$
$$= \dfrac{1}{50} 2^2[\dfrac{50(51)(101)}{6}] - {(51)}^2$$
$$= 34 \times 101 -2601$$
$$ = 3434-2601$$
$$=833$$
$${\textbf{Thus, option B is correct.}}$$
Question 6
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Find the Standard Deviation of of first $$10$$ multiples of $$3$$.

A
$$8.61$$

B
$$8.33$$

C
$$9.09$$

D
$$9.69$$

Solution

First $$10$$ multiples of $$3$$ are $$3,6,9...30$$.
This is an A.P.
$$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$$
$$\therefore sum=165$$.
Mean, $$u=\dfrac {sum}{n}=\dfrac{165}{10}$$.
Variance, $$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$$.
$$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$$.
$$\therefore \sigma ^2=346.5-272.25$$
$$\therefore \sigma ^2=74.25$$
Standard Deviation, $$S.D=\sqrt{ \sigma ^2}$$
$$\therefore S.D=\sqrt{74.25}$$
Thus, $$S.D=8.61$$
Ans-Option $$A$$.
Question 7
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If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, then its mean is

A
40

B
30

C
20

D
none of these

Solution

Given $$\sigma = 20$$, coefficient of variation $$=50$$ %
We know coefficient of variation $$=\cfrac{\sigma }{\bar{x}}\times 100=50$$
$$\Rightarrow \bar{x} = 2\times \sigma = 40$$
Question 8
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The variance of first n natural numbers is

A
$$\displaystyle \frac{n^2 + 1}{12}$$

B
$$\displaystyle \frac{n^2 - 1}{12}$$

C
$$\displaystyle \frac{n^2 - 1}{6}$$

D
$$\displaystyle \frac{n^2 + 1}{2}$$

Solution

$$\Rightarrow$$ First $$n$$ natural numbers are $$1,2,3,4,....n$$
$$\Rightarrow$$ $$Mean=\dfrac{1+2+3+4+....+n}{n}$$
$$=\dfrac{\dfrac{n(n+1)}{2}}{n}$$
$$=\dfrac{n+1}{2}$$
$$\Rightarrow$$ $$Variance (\sigma)^2=\dfrac{\sum(x_i)^2}{n}-(mean)^2$$
$$=\dfrac{(1)^2+(2)^2+....n^2}{n}-\left(\dfrac{n+1}{2}\right)^2$$
Since, $$(1)^2+(2)^2+(3)^2+....+(n)^2=\dfrac{n(n+1)(2n+1)}{6}$$
$$\therefore$$ $$Variance(\sigma)^2=\dfrac{n(n+1)(2n+1)}{6n}-\dfrac{(n+1)}{4}$$

$$=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)}{4}$$

$$=\dfrac{(n+1)}{2}\left(\dfrac{2n+1}{3}-\dfrac{n+1}{2}\right)$$

$$=\dfrac{(n+1)}{2}\left(\dfrac{4n+2-3n-3}{6}\right)$$

$$=\dfrac{n+1}{2}\times \dfrac{n-1}{6}$$

$$=\dfrac{n^2-1}{12}$$
Question 9
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Coefficient of deviation is calculated by the formula:

A
$$\cfrac { \bar { X } }{ \sigma } \times 100$$

B
$$\cfrac { \bar { X } }{ \sigma }$$

C
$$\cfrac { \sigma } {\bar { X }} \times 100$$

D
$$\cfrac{ \sigma } { \bar { X }}$$

Solution

It is a fundamental concept.
coefficient of deviation $$=\cfrac{\sigma}{\bar{x}}\times 100$$
where $$\sigma$$ and $$\bar{x}$$ are standard deviation and mean respectively.
Question 10
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The mean deviation about median from the data
$$340,150,210,240,300,310,320$$ is

A
$$50$$

B
$$52.8$$

C
$$55$$

D
$$45$$

Solution

Arrange in increasing order
150,210,240,300,310,320,340 N=number of data = 7
median (M) =300
The mean deviation about median $$= \frac{150+90+60+0+10+20+40}{7 }= 52.857$$

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Measures of Dispersion Test - 8

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Measures of Dispersion Test - 8

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Question 1

$$22$$

$$20$$

$$16$$

$$18$$

Solution

Question 2

$$\dfrac{10}{\sqrt{3}}$$

$$\dfrac{100}{\sqrt{3}}$$

$$\dfrac{130}{3}$$

$$\dfrac{10}{3}$$

Solution

Question 3

$$4$$

$$2$$

$$\sqrt{2}$$

$$2\sqrt{2}$$

Solution

Question 4

$$\displaystyle \frac{11}{2}$$

$$6$$

$$\displaystyle \frac{13}{2}$$

$$\displaystyle \frac{5}{2}$$

Solution

Question 5

$$ \displaystyle \frac{833}{4}$$

$$ 833$$

$$ 437$$

$$ \displaystyle \frac{437}{4}$$

Solution

Question 6

$$8.61$$

$$8.33$$

$$9.09$$

$$9.69$$

Solution

Question 7

40

30

20

none of these

Solution

Question 8

$$\displaystyle \frac{n^2 + 1}{12}$$

$$\displaystyle \frac{n^2 - 1}{12}$$

$$\displaystyle \frac{n^2 - 1}{6}$$

$$\displaystyle \frac{n^2 + 1}{2}$$

Solution

Question 9

$$\cfrac { \bar { X } }{ \sigma } \times 100$$

$$\cfrac { \bar { X } }{ \sigma }$$

$$\cfrac { \sigma } {\bar { X }} \times 100$$

$$\cfrac{ \sigma } { \bar { X }}$$

Solution

Question 10

$$50$$

$$52.8$$

$$55$$

$$45$$

Solution

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