Question

Find the standard deviation by actual mean and assumed mean method of the following data :

6,7,10,12,13,4,8,12
Please tell all the steps so that I can understand

Answer 1

☢ Find the Standard deviation by Actual mean method and Assumed mean method of the following data :

Values(X) : 6, 7, 10, 12, 13, 4, 8, 12

⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

☯ By Actual Mean Method ;

$\begin{array}{|c|c|c|} \dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{} \\\bf Values(X) & \bf {x = X - \overline{X}}& \bf X^{2}\\\dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{} \\\sf 6 & \sf 6 - 9 = -3 & \sf 9 \\ &\\\sf 7 & 7 - 9 = -2 & 4 \\ & \\ \sf 10 & 10 - 9 = 1 & 1 \\&\\ \sf 12 & 12 - 9 = 3 & 9 \\&\\ \sf 13 & 13 - 9 = 4 & 16 \\&\\ \sf 4 & 4 - 9 = -5 & 25 \\&\\ \sf 8 & 8 - 9 = -1 & 1 \\&\\ \sf 12 & 12 - 9 = 3 & 9 \\&\\ \dfrac{\qquad\qquad\qquad\qquad}{\sum\limits X = 72} & \dfrac{}{} & \dfrac{\qquad\qquad\qquad\qquad}{\sum\limits X^{2}} \\ \dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{}\end{array}$

$\underline{\dag \: {\mathfrak{ Now, \: as \: we \: know \: that}}}$

$\bigstar{\boxed{\orange{\sf\: \overline{X} = \frac{\sum\limits\:X}{N} }}}$

$\bigstar{\boxed{\green{\sf\: \sigma = \sqrt{\frac{\sum\limits\:X^{2}}{N}} }}}$

$\ : \implies\sf\: \overline{X} \: = \frac{\sum\limits\:X}{N}$

$\ : \implies\sf\: \overline{X} \: = \frac{\cancel{72}}{\cancel{8}}$

$\ : \implies{\boxed{\boxed{\red{\sf\: \overline{X} \: = 9 }}}}$

$\ : \implies\sf\: \sigma \: = \sqrt{\frac{\sum\limits X^{2}}{N}}$

$\ : \implies\sf\: \sigma \: = \sqrt{\frac{74}{8} }$

$\ : \implies\sf\: \sigma \: = \sqrt{9.25}$

$\ : \implies{\boxed{\boxed{\red{\sf\: \sigma\: = 3.04}}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

☯ By Assumed Mean Method ;

$\begin{array}{|c|c|c|} \dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{} \\\bf Values(X) & \bf {d = X - A}& \bf d^{2}\\\dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{} \\\sf 6 & \sf 6 - 12 = -6 & \sf 36 \\ &\\\sf 7 & 7 - 12 = -5 & 25 \\ & \\ \sf 10 & 10 - 12 = -2 & 4 \\&\\ \sf 12(A) & 12 - 12 = 0 & 0 \\&\\ \sf 13 & 13 - 12 = 1 & 1 \\&\\ \sf 4 & 4 - 12 = -8 & 64 \\&\\ \sf 8 & 8 - 12 = -4 & 16 \\&\\ \sf 12 & 12 - 12 = 0 & 0 \\&\\ \dfrac{}{} & \dfrac{\qquad\qquad\qquad\qquad}{\sum\limits d = -24} & \dfrac{\qquad\qquad\qquad\qquad}{\sum\limits d^{2} = 146} \\ \dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} & \dfrac{\qquad\qquad\qquad\qquad}{}\end{array}$

$\underline{\dag \: {\mathfrak{ Now, \: as \: we \: know \: that}}}$

$\bigstar{\boxed{\blue{\sf\: \sigma = \sqrt{ \frac{\sum\limits\:d^{2}}{N} - \left( \frac{\sum\limits\: d}{N}\right) ^{2} } }}}$

$\ : \implies\sf\: \sigma = \sqrt{ \frac{\sum\limits\:d^{2}}{N} - \left( \frac{\sum\limits\: d}{N} \right)^{2} }$

$\ : \implies\sf\: \sigma = \sqrt{ \frac{146}{8} - \frac{576}{64} }$

$\ : \implies\sf\: \sigma = \sqrt{ 18.25 - 9}$

$\ : \implies\sf\: \sigma = \sqrt{ 9.25}$

$\ : \implies{\boxed{\boxed{\red{\sf\: \sigma\: = 3.04}}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\therefore\:\underline{\textsf{Hence, \: the \: standard \: deviation \: is \: \textbf{3.04}}}$