Question 198809
First sort the numbers into ascending order (from least to greatest):

0.2,0.2,0.3,0.4,0.6,0.6,0.9,1.1,1.2,1.3,1.4



Range:


To find the range, subtract the smallest number (in this case 0.2) from the largest number (in this case 1.4) like this:



{{{Range=1.4-0.2=1.2}}}


So the range is 1.2




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Interquartile Range:



First, find the median for the first half of the data set (ignore the median):



Median of 0.2, 0.2, 0.3, 0.4, 0.6 is 0.3 (since this is the middle most value)



Now find the median of the last half of the data set  (ignore the median):



Median of 0.9, 1.1, 1.2, 1.3, 1.4 is 1.2 (since this is the middle most value)



So the 1st and 3rd quartiles are 0.3 and 1.2 respectively.



This means that the interquartile range is 


1.2 - 0.3 = 0.9



Note: the interquartile range is the difference between the 1st and 3rd quartiles



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Note: I'm assuming that you want the sample standard deviation.



Standard Deviation:



Step 1) Find the mean:



*[Tex \LARGE Mean=\frac{0.2+0.2+0.3+0.4+0.6+0.6+0.9+1.1+1.2+1.3+1.4}{11}=\frac{8.2}{11}=0.745454545454546]




So the mean is roughly 0.75 (rounded to the nearest hundredth).



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Step 2) Find the differences between the mean and every value:



Now subtract EVERY value from the mean (0.75) to get:

0.2-0.75 = -0.55, 0.2-0.75 = -0.55, 0.3-0.75 = -0.45, 0.4-0.75 = -0.35, 0.6-0.75 = -0.15, 0.6-0.75 = -0.15, 0.9-0.75 = 0.15, 1.1-0.75 = 0.35, 1.2-0.75 = 0.45, 1.3-0.75 = 0.55, 1.4-0.75 = 0.65, 



So you should have a new column of values:   

-0.55, -0.55, -0.45, -0.35, -0.15, -0.15, 0.15, 0.35, 0.45, 0.55, 0.65, 

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Step 3) Squaring the differences:



Now square EVERY value in the last column to get: 

(-0.55)^2 = 0.3, (-0.55)^2 = 0.3, (-0.45)^2 = 0.2, (-0.35)^2 = 0.12, (-0.15)^2 = 0.02, (-0.15)^2 = 0.02, (0.15)^2 = 0.02, (0.35)^2 = 0.12, (0.45)^2 = 0.2, (0.55)^2 = 0.3, (0.65)^2 = 0.42, 



So you should have a new column of values:   

0.3, 0.3, 0.2, 0.12, 0.02, 0.02, 0.02, 0.12, 0.2, 0.3, 0.42, 

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Step 4) Adding up the squares:



Now add up the values of the last column:   

0.3 + 0.3 + 0.2 + 0.12 + 0.02 + 0.02 + 0.02 + 0.12 + 0.2 + 0.3 + 0.42 = 2.02


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Step 5) Divide the sum by n-1:



Divide that sum by one less than the number of values in the list (in this case, it is n-1=11-1=10).



{{{2.02/(11-1)=0.202}}}



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Step 6) Take the square root of the quotient:



Take the square root of {{{0.202}}} to get {{{sqrt(0.202)=0.449444101084885}}}



Now round to the nearest hundredth to get {{{0.45}}}.



So the standard deviation is roughly {{{0.45}}}.



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If you don't like that method, then here's another way:



First sort the numbers into ascending order (from least to greatest):

0.2,0.2,0.3,0.4,0.6,0.6,0.9,1.1,1.2,1.3,1.4



Standard Deviation: *[Tex \LARGE   \sigma=\sqrt{ \frac{1}{N-1}\displaystyle\sum_{i=0}^N (x_i-\bar{x})^2}] where *[Tex \LARGE \bar{x}] is the average, *[Tex \LARGE x_i] is the ith number, and *[Tex \LARGE N] is the number of numbers


So we can replace N with 11


*[Tex \LARGE\sqrt{ \frac{1}{11-1}\displaystyle\sum_{i=0}^11 (x_i-\bar{x})^2}]


Subtract {{{11-1}}} to get 10


*[Tex \LARGE\sqrt{ \frac{1}{10}\displaystyle\sum_{i=0}^11 (x_i-\bar{x})^2}]


Replace  *[Tex \LARGE \bar{x}] with 0.75


*[Tex \LARGE\sqrt{ \frac{1}{10}\displaystyle\sum_{i=0}^11 (x_i-0.75)^2}]


Expand the summation (replace each {{{x[i]}}} with the respective number)



{{{sqrt((1/10)((0.2-0.75)^2+(0.2-0.75)^2+(0.3-0.75)^2+(0.4-0.75)^2+(0.6-0.75)^2+(0.6-0.75)^2+(0.9-0.75)^2+(1.1-0.75)^2+(1.2-0.75)^2+(1.3-0.75)^2+(1.4-0.75)^2))}}}


Subtract the terms in the parenthesis


{{{sqrt((1/10)((-0.55)^2+(-0.55)^2+(-0.45)^2+(-0.35)^2+(-0.15)^2+(-0.15)^2+(0.15)^2+(0.35)^2+(0.45)^2+(0.55)^2+(0.65)^2))}}}


Square each term


{{{sqrt((1/10)(0.3025+0.3025+0.2025+0.1225+0.0225+0.0225+0.0225+0.1225+0.2025+0.3025+0.4225))}}}


Add up all of the terms


{{{sqrt((1/10)2.0475)}}}


Multiply


{{{sqrt(0.20475)}}}


Take the square root


{{{0.452493093869951}}}



Round to the nearest hundredth


{{{0.45}}}



So the standard deviation is roughly {{{0.45}}}


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25th and 75th percentiles:


The 25th and 75th percentiles are simply the first and third quartiles. We found these earlier to be 0.3 and 1.2. 




So the 25th and 75th percentiles are 0.3 and 1.2