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WHAT IS THE STANDARD DEVIATION OF 1.25,1,1.5,1.25,1

Guest Sep 10, 2014

Best Answer 

 #1
avatar+169 
+5

First you need to calculate the mean:

 

$${\frac{\left({\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{6}}}{{\mathtt{5}}}} = {\mathtt{1.2}}$$

 

No you need to calculate the square of the deviation of each value to the mean:

 

$$\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2\\(1.2-1.5)^2=(-0.3)^2\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2$$

 

Now we need the mean of the squared deviations:

 

$${\frac{\left({\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.3}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{7}}}{{\mathtt{200}}}} = {\mathtt{0.035}}$$

 

The square root of this mean is equal to the standard deviation:

 

$${\sqrt{{\mathtt{0.035}}}} = {\mathtt{0.187\: \!082\: \!869\: \!338\: \!697\: \!1}}$$

 

Hope this helps!

Honga  Sep 10, 2014
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1+0 Answers

 #1
avatar+169 
+5
Best Answer

First you need to calculate the mean:

 

$${\frac{\left({\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.25}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{6}}}{{\mathtt{5}}}} = {\mathtt{1.2}}$$

 

No you need to calculate the square of the deviation of each value to the mean:

 

$$\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2\\(1.2-1.5)^2=(-0.3)^2\\(1.2-1.25)^2=(-0.05)^2\\(1.2-1)^2=(0.2)^2$$

 

Now we need the mean of the squared deviations:

 

$${\frac{\left({\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.3}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left(-{\mathtt{0.05}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\mathtt{0.2}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{5}}}} = {\frac{{\mathtt{7}}}{{\mathtt{200}}}} = {\mathtt{0.035}}$$

 

The square root of this mean is equal to the standard deviation:

 

$${\sqrt{{\mathtt{0.035}}}} = {\mathtt{0.187\: \!082\: \!869\: \!338\: \!697\: \!1}}$$

 

Hope this helps!

Honga  Sep 10, 2014

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