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The average of  n numbers x_{1}, x_{2}, x_{3},.......x_{n} is M. 

If x_{n}\: is\: replaced by x^{1} then new average is 

  • Option 1)

    M-X_{n}+x^{1}

  • Option 2)

    \frac{nM-x_{n}+x^{1}}{n}

  • Option 3)

    \frac{(n-1)M+x^{1}}{n}

  • Option 4)

    \frac{M-x_{n+x^{1}}}{n}

 

Answers (1)

best_answer

As we discussed in concept

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by 

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.

-

 

 M\:=\:\frac{x_{1}+x_{2}+x_{3}...x_{n}}{n}

=> x_{4}+x_{2}+x_{3}+\:-----\:+x_{n}\:=\:Mn

Now x_{4}+x_{2}+x_{3}+-------x^{1}\:=\:Mn-x_{n}+x^{1}

New Average:

\frac{x_{1}+x_{2}+x_{3}+-------x^{1}}{n}\:=\:\frac{Mn-x_{n}+x^{1}}{n}

 


Option 1)

M-X_{n}+x^{1}

This option is incorrect.

Option 2)

\frac{nM-x_{n}+x^{1}}{n}

This option is correct.

Option 3)

\frac{(n-1)M+x^{1}}{n}

This option is incorrect.

Option 4)

\frac{M-x_{n+x^{1}}}{n}

This option is incorrect.

Posted by

prateek

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