# standard deviation question

• August 5th 2009, 04:18 PM
fiurik
standard deviation question
hi all.

i have a problem. i am trying to figure out the SD for a data set where the largest value is 5 (it's a data set of customer service ratings from 1 - 5) My mean is 4 and my standard deviation is 2. How is this possible if my largest possible rating can only be 5?

Thanks for the help!(Hi)
• August 5th 2009, 05:21 PM
matheagle
I may try to construct a sample to see if I can recreate this, do you know the number of observations, i.e., sample size?

I played around with a sample of {2,5,5}.
I got the sample mean to be 4 and the st deviation to be $\sqrt 3$ which is close.
That's about the highest st deviation I can get with a mean of 4 and a small sample.
Maybe with 2 observations I might get a st deviation of 2, but 2 is high.
• August 5th 2009, 05:25 PM
fiurik
Thanks for your help, matheagle, i do appreciate it.

size is 85 data points, mean is 4, data points are:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
• August 5th 2009, 05:28 PM
matheagle
Quote:

Originally Posted by fiurik
Thanks for your help, matheagle, i do appreciate it.

size is 85 data points, mean is 4, data points are:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

how many
1's=
2's=
3's=
4's=
5's=
• August 5th 2009, 05:29 PM
fiurik
1 = 13
2 = 5
3 = 7
4 = 11
5 = 49
• August 5th 2009, 05:33 PM
matheagle
so $\sum x^2=(13)(1)+(5)(4)+(7)(9)+(11)(16)+(49)(25)=1497$

check my work

$\sum x=(13)(1)+(5)(2)+(7)(3)+(11)(4)+(49)(5)=333$

sample mean is 333/85=3.917647..............

so one thing is, we have a round off error, which is making things worse as I calculate the variance

sample VARIANCE is

${1497-{(333)^2\over 85}\over 84}=2.290756....$

some use 85 in the denomiator, but both are fine, then square root this to get the st deviation.

.... 1.5135244...
• August 5th 2009, 05:37 PM
fiurik
ok
• August 5th 2009, 05:45 PM
fiurik
sorry, the message truncated the first time and all I saw was this part of your first sentence. I refreshed the screen just now and got full explanation.

do, if i follow your steps, doesn't the SD still come to 2 (after rounding)?
• August 5th 2009, 05:46 PM
matheagle
Quote:

Originally Posted by fiurik
sorry, the message truncated the first time and all I saw was this part of your first sentence. I refreshed the screen just now and got full explanation.

do, if i follow your steps, doesn't the SD still come to 2 (after rounding)?

barely, 1.51
• August 5th 2009, 05:48 PM
fiurik
right, so if my maximum rating is a 5 and my mean is 4 (or 3.9...) how can my SD be 2

2 points away from four puts me beyond me max rating? is this ok?
• August 5th 2009, 06:40 PM
TKHunny
Let's see if I understand the question...

Sample Data Set
1
5
5
5

Mean = 4
Sample Standard Deviation = 2

4 + 2 = 6 > 5 and this gives you pause. How can a mere one standard deviation exceed the greatest possible value?!

This is a reasonable question.

Try to remember that the Mean and Standard Devaition are measures of Central Tendency and Variability, respectively. They are NOT population restrictions.

This does give cause for additional consideration when constructing confidence intervals. The task of using "truncated" distributions is a matter of much conversation.

Note: This is just the sort of thing that gives Tchebychev's Inequality its power. Tchebychev's Inequality Truly amazing stuff!
• August 5th 2009, 07:08 PM
fiurik
wow, thanks for your answer. it sounds great, but i wasn't kidding when I said I was a novice. that was way over my head! :)
• August 7th 2009, 08:07 PM
TKHunny
You don't need to worry about the awesome ineauality. For now, just get the hang of the arithmetic. Try to recall that some measures of Central Tendency are better than others. There are also various measures of variability. Some work better in some situations. Always keep your eye out for results that don't make sense - like your observation that motivated this question. Good work.