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(B) Given that the mean of the first four observations is $11$,so $\sum_{i=1}^{4} x_i = 4 	imes 11 = 44$.The mean of the remaining six observations i

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$2$

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(B) Given that the mean of the first four observations is $11$,so $\sum_{i=1}^{4} x_i = 4 	imes 11 = 44$.The mean of the remaining six observations is $16$,so $\sum_{i=5}^{10} x_i = 6 	imes 16 = 96$.The total sum of the observations is $\sum_{i=1}^{10} x_i = 44 + 96 = 140$.The mean of the data is $\bar{x} = \frac{140}{10} = 14$.The variance is given by $\sigma^2 = \frac{\sum_{i=1}^{10} x_i^2}{n} - (\bar{x})^2$.Substituting the given values,$\sigma^2 = \frac{2000}{10} - (14)^2 = 200 - 196 = 4$.Therefore,the standard deviation is $\sigma = \sqrt{4} = 2$.

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

If the data $x_1, x_2, ..., x_{10}$ is such that the mean of the first four of these is $11$,the mean of the remaining six is $16$,and the sum of squares of all of these is $2,000$; then the standard deviation of this data is

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