Tag: karl's pearson method

Questions Related to karl's pearson method

 Given  $\sum xy=120, \sigma _x=8  ,\sum x^2=90,\sigma _ y=7 ,n=10$ ; where x and y are deviations from mean, the  correlation coefficient is 

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. $0.21$

  2. $0.9$

  3. $0.36$

  4. $0.6$

Show answer
Correct Option: A

Karl Pearson's coefficient of skewness of a distribution is 0.32.Its s.d.is 6.5 and mean is 29.6.The mode and median of the distribution are

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. 27.52,28.91

  2. 26.92,27.23

  3. 25.67,26.34

  4. none of these

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Correct Option: A
Explanation:

Karl Pearson's coefficient of skewness $\displaystyle =\frac{Mean-Mode}{S.D.}$ $\displaystyle \therefore 0.32=\frac{29.6-Mode}{6.5}\Rightarrow Mode=27.52$ Also Karl Pearson's coeff.of skewness $\displaystyle =\frac{3\left ( Mean-Median \right )}{S.D}$ $\displaystyle \because 0.32=\frac{3\left ( 29.6-Median \right )}{6.5}$ $\displaystyle \Rightarrow Median=28.91$

The sum of the deviations of the variates 6,8,10,16,20,24 

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. -1

  2. 1

  3. 0

  4. 4

Show answer
Correct Option: C
Explanation:

$\Rightarrow$   $Mean = \dfrac{6+8+10+16+20+24}{6}=14$

$\Rightarrow$   Sum of the deviations = $(6-14)+(8-14)+(10-14)+(16-14)+(20-14)+(24-14)$
$\Rightarrow$   Sum of the deviation = $-8-6-4+2+6+10$
$\therefore$    Sum of the deviation = $-18+18$
$\therefore$    Sum of the deviation = $0$

Choose the statement which consists of two correlated variables.

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. Increase in the intensity of cold results in greater sale of woollen clothes

  2. Increase in temperature of delhi has led to congestion

  3. Increase in weight of children s accompanied by increase in weight of their mother

  4. None of the above

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Correct Option: A
Explanation:

Since increase in intensity of cold result in greater scale of woolen clothes,

Calculate Pearson's coefficient of correlation between the values of $X$ and $Y$.

X 1 2 3 4 5
Y 7 6 5 4 3
statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. $-0.3$

  2. $0.3$

  3. $1$

  4. $-1$

Show answer
Correct Option: D
Explanation:
$X\\ 1\\ 2\\ 3\\ 4\\ 5\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { x } \cfrac { 15 }{ 5 } =3$                $Y\\ 7\\ 6\\ 5\\ 4\\ 3\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \overline { y } =\cfrac { 25 }{ 5 } =5$                  $Y=y-\overline { y } \\ \quad 02\\ \quad 01\\ \quad 00\\ -1\\ -1\\ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$                    $XY\\ -4\\ -1\\ \quad 0\\ -1\\ -4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { XY=-10 } $                 ${ X }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { X }^{ 2 }=10 } $        

${ Y }^{ 2 }\\ 4\\ 1\\ 0\\ 1\\ 4\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { Y }^{ 2 }=10 } $

Therefore, $r=\cfrac { \sum { XY }  }{ \sqrt { \sum { { X }^{ 2 }\sum { { Y }^{ 2 } }  }  }  } \\ =\cfrac { -10 }{ \sqrt { 10*10 }  } \\ =\cfrac { -10 }{ 10 } \\ =-1.$

For $ n=25,\sum x=125,\sum x^2=650,\sum y=100,\sum y^2=460,\sum xy=508$, correlation coefficient is 

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. $0.99$

  2. $0.207$

  3. $0.66$

  4. $0.89$

Show answer
Correct Option: B
Explanation:

$n=25$

$\sum x=125,\ \sum{{x}^{2}}=650,\ \sum{xy}=334$  $\sum y=100,\ \sum{{y}^{2}}=508$
$r=\cfrac { n\sum { xy } -(\sum { x } \times \sum { y } ) }{ \sqrt { (n\sum { { x }^{ 2 } } -\sum { { x }^{ 2 } } )(n\sum { { y }^{ 2 } } -\sum { { y }^{ 2 } } ) }  } =\cfrac { 25\times 508 -100\times 125 }{ \sqrt { (25\times 650 -{ 125 }^{ 2 }  )(25\times 460 -{100}^{2}) }  }=0.207 $

$n=25,\sum x=125,\sum x^2=650, \sum y=100,\sum y^2=460,\sum xy=508$. It was observed that two pair of values of $(x,y)$ were copied as $ (6,14)$ and $(8,6) $ instead of $(8,12),(6,8).$ The correct correlation coefficient is 

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. $0.667$

  2. $0.87$

  3. $-0.25$

  4. $0.356$

Show answer
Correct Option: A
Explanation:

Corrected $\sum x=125-6-8+8+6=125$

Corrected $\sum y=100-14-6+12+8=100$
Corrected $\sum x^2=650-(6)^2-(8)^2+(6)^2+(8)^2=650$
Corrected $\sum y^2=460-(14)^2-(6)^2+(12)^2+(8)^2=436$
Corrected $\sum xy=508-6\times 18-8\times 6+8\times 12+6\times 8=520$
The formula for Pearson product moment correlation is
$r=\dfrac{n\sum xy-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}$

    $=\dfrac{25\times 520-125\times 100}{\sqrt{[25\times 650-(125)^2][25\times 436-(100)^2]}}$

   $=\dfrac{13000-12500}{\sqrt{[16250-15625][10900-10000]}}$

   $=\dfrac{500}{\sqrt{[625][900]}}$

   $=\dfrac{500}{(25)(30)}$

   $=\dfrac{2}{3}$

   $=0.667$

The coefficient of correlation when coefficients of regression are $0.2$ and $1.8$ is

statistics skewness of frequency distribution karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. $0.36$

  2. $0.2$

  3. $0.6$

  4. $0.9$

Show answer
Correct Option: C
Explanation:

$Given\quad { b } _{ xy }=0.2\\ { b } _{ yx }=0.8\\ r=\sqrt { { b } _{ xy }\times { b } _{ yx } } \\ r=\sqrt { 0.36 } \\ r=0.6$

Where r is coefficient of correlation.

Standard error of regression analysis is classified as

business maths correlation and regression analysis karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. average of coefficient

  2. variance of residual

  3. mean of residual

  4. average of residual

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Correct Option: B
Explanation:

Standard error of regression analysis is classified as variance of residuals.

Variance of residuals also known as error variance.

The independent variable is also called: 

business maths correlation and regression analysis karl pearson coefficient of correlation karl pearson's product moment method karl's pearson method

  1. Regressor

  2. Regressand

  3. Predictand

  4. None of these

Show answer
Correct Option: A
Explanation:

A linear regression line has an equation of the form $Y=a+bX$


where $Y$ is called dependent variable or response or regressand
$X$ is called independent variable or predictors or explanatory variable or regressor
$a$ is the y-intercept and
$b$ is the slope of the line.