The standard deviation and mean of five obser | vedclass

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(B) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given that the mean is $9$,we have $\frac{x_1+x_2+x_3+x_4+x_5}{5} = 9$,which implies $x_1+

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$2$

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(B) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given that the mean is $9$,we have $\frac{x_1+x_2+x_3+x_4+x_5}{5} = 9$,which implies $x_1+x_2+x_3+x_4+x_5 = 45$. Since the standard deviation is $0$,all observations must be equal to the mean. Thus,$x_1 = x_2 = x_3 = x_4 = x_5 = 9$. Now,one observation $x_5$ is changed to $y$ such that the new mean is $10$. So,$\frac{x_1+x_2+x_3+x_4+y}{5} = 10$,which implies $x_1+x_2+x_3+x_4+y = 50$. Substituting $x_1+x_2+x_3+x_4 = 36$ (since $x_1=x_2=x_3=x_4=9$),we get $36+y = 50$,so $y = 14$. The new set of observations is ${9, 9, 9, 9, 14}$. The new mean is $10$. The new standard deviation is $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{(9-10)^2 + (9-10)^2 + (9-10)^2 + (9-10)^2 + (14-10)^2}{5}}$. $= \sqrt{\frac{(-1)^2 + (-1)^2 + (-1)^2 + (-1)^2 + (4)^2}{5}} = \sqrt{\frac{1+1+1+1+16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$.

Height in cms	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

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

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