Question

What is the difference between the Range and the Standard Deviation of all the natural numbers between 81
and 90, both inclusive?​

Answer 1

Answer:

$we \: know \: standard \: deviation \: of \: first \: n \: natural \: numbers \: = \sqrt{ \frac{ {n}^{2} - 1 }{12} } \\ so \: standard \: deviation \: = \sqrt{ \frac{ {90}^{2} - 1 }{12} } - \sqrt{ \frac{ {80}^{2} - 1 }{12} } \\ = \sqrt{ \frac{8099}{12} } - \sqrt{ \frac{6399}{12} } \\ = 25.98 - 23.09 \\ = 2.89 \\ now \: range \: = 90 - 81 = 9 \\ so \: require \: difference = 9 - 2.89 = 6.11$