+0

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

0
467
1

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

Guest Sep 10, 2014

#1
+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
Sort:

#1
+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

26 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details