Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
2280
1
avatar

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

 Sep 10, 2014

Best Answer 

 #1
avatar+169 
+5

First you need to calculate the mean:

 

(1.25+1+1.5+1.25+1)5=65=1.2

 

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

 

(1.21.25)2=(0.05)2(1.21)2=(0.2)2(1.21.5)2=(0.3)2(1.21.25)2=(0.05)2(1.21)2=(0.2)2

 

Now we need the mean of the squared deviations:

 

((0.05)2+(0.2)2+(0.3)2+(0.05)2+(0.2)2)5=7200=0.035

 

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

 

0.035=0.1870828693386971

 

Hope this helps!

 Sep 10, 2014
 #1
avatar+169 
+5
Best Answer

First you need to calculate the mean:

 

(1.25+1+1.5+1.25+1)5=65=1.2

 

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

 

(1.21.25)2=(0.05)2(1.21)2=(0.2)2(1.21.5)2=(0.3)2(1.21.25)2=(0.05)2(1.21)2=(0.2)2

 

Now we need the mean of the squared deviations:

 

((0.05)2+(0.2)2+(0.3)2+(0.05)2+(0.2)2)5=7200=0.035

 

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

 

0.035=0.1870828693386971

 

Hope this helps!

Honga Sep 10, 2014

2 Online Users

avatar
avatar