The mean and the standard deviation (s.d.)&nb | vedclass

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Question

Here mean $ = \bar x = 9$

$ \Rightarrow \bar x = \frac{{\sum {{x_i}} }}{n} = 9$

$ \Rightarrow \sum {{x_i}}  = 9 	imes 5 = 45$

Now,

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Here mean $ = \bar x = 9$

$ \Rightarrow \bar x = \frac{{\sum {{x_i}} }}{n} = 9$

$ \Rightarrow \sum {{x_i}}  = 9 	imes 5 = 45$

Now, standard deviation $=0$

$	herefore $ all the five terms are same i.e.;$9$.

Now for changed observation 

${{\bar x}_{new}} = \frac{{36 + {x_5}}}{5} = 10$

$ \Rightarrow {x_5} = 14$

$	herefore {\sigma _{new}} = \sqrt {\frac{{\sum {{{\left( {{x_i} - {{\bar x}_{new}}} ight)}^2}} }}{n}} $

$ = \sqrt {\frac{{4{{\left( {9 - 10} ight)}^2} + {{\left( {14 - 10} ight)}^2}}}{5}} $

$ = 2$

${x_i}$	$3$	$8$	$13$	$18$	$25$
${f_i}$	$7$	$10$	$15$	$10$	$6$

${x_i}$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
${f_i}$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

The mean and the standard deviation $(s.d.)$ of five observations are $9$ and $0,$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10,$ then their $s.d.$ is?

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