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The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by p and then reduced by q, where p\neq 0 and q\neq 0. If the new mean and new s.d. become half of their original values, then q is equal to :
Option: 1 -20
Option: 2 -5
Option: 3 10  
Option: 4 -10

Answers (1)




Some Important Point Regarding Statistics -

Some Important Point Regarding Statistics

  1. The sum of the deviation of an observation from their mean is equal to zero. i.e. \sum^{n}_{i=1}\left ( x_i-\bar x \right )=0.
  2. The sum of the square of the deviation from the mean is minimum, i.e. \sum^{n}_{i=1}\left ( x_i-\bar x \right )^2\;\text{is least.}
  3. The mean is affected accordingly if the observations are given a mathematical treatment i.e. addition, subtraction multiplication by a constant term.
  4. If set of n1 observations has mean \bar x_1 and set of n2 observations has mean \bar x_2, then their combined mean is \frac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2} and the .
  5. If set of  n1 observations has mean \bar x_1 and set of n2 observations has mean \bar x_2  then their combined variance is given by

            \sigma ^{2}= \frac{n_{1}\left ( \sigma _{1}^{2}+d_{1}^{2} \right )+n_{2}\left ( \sigma _{2}^{2}+d_{2}^{2} \right )}{n_{1}+n_{2}}

          where, \\d_1=\bar x_2-\bar x\;\;\;\text{and}\;\;\;d_2=\bar x_ 1-\bar x 

          \\ {\sigma_{1}^{2}=\frac{1}{n_{1}} \sum_{i=1}^{n_{1}}\left(x_{1 i}-\overline{x_{1}}\right)^{2}} \\\\ {\sigma_{2}^{2}=\frac{1}{n_{2}} \sum_{j=1}^{n_{2}}\left(x_{2 j}-\overline{x_{2}}\right)^{2}}

          \bar x_1,\;\;\bar x_2are the means and \sigma_1,\;\;\sigma_2  are the standard deviations of two series.


\text{old series} \Rightarrow x_{old}=x_1,x_2,x_3\ldots\ldots x_n

Old Mean

\overline x_{old}=\frac{\left (x_1+x_2+x_3\ldots\ldots x_n \right ) }{n}=20

Old Standard Deviation

\sigma^{2}_{old}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

If each observation is multiplied with p & then q is subtracted

\text{new series} \Rightarrow x_{new}=px_1-q,\;px_2-q,\;px_3-q\ldots\ldots \;px_n-q

New mean is the half of old mean

\\ \overline x_{new}=\frac{p\left (x_1+x_2+x_3\ldots\ldots x_n \right )-nq}{n} =10\\\\ \overline x_{new}=p\;\overline x_{old}-q=10\\\\ \overline x_{new}=20p-q=10\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i) 

and new standard deviations is the half of old mean

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{new}-\bar{x}_{new}\right)^{2}

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left[(px_{i}-q)-(p\bar{x}_{old}-q)\right]^{2}

\sigma^{2}_{new}=\frac{1}{n} \sum_{i=1}^{n}\left(px_{i}-p\bar{x}_{old}\right)^{2}

\sigma^{2}_{new}=\frac{1}{n}p^2 \sum_{i=1}^{n}\left(x_{i}-\bar{x}_{old}\right)^{2}




Case 1 p=0.5

Using equation 1

we get


Case 1 p=-0.5

Using equation 1

we get


Correct Option (1)

Posted by

Kuldeep Maurya

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