How would I go about calculating a specific standard deviations range? I know we have the 68-95-99.7 rule, but that's only for 1 standard deviation, 2 standard deviation and 3 standard deviation. What if I wanted .5 or 1.2 or 2.1 standard deviation etc... Is there a formula for this? I can't seem to find any answers anywhere online or in my book. How would I go about calculating this if the measurements are normal distributed and I'm wanting percentage? Thanks!
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1No, there isn't a simple formula. In elementary statistics you generally get these other values from a normal distribution table. They are also available from software. – Ian Nov 27 '16 at 02:44
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Yeah I guess my answer doesn't suffice lol. I sort of skimmed through the question – Brandon Nov 27 '16 at 02:51
2 Answers
Yes. Have you heard of a Z score??
a Z score is the measure of how far away a value is from the mean (in deviations)..
For example, say you have a dataset like
$\{1,2,3,4,5,6,7,8,9,10\}$
Now just bear with me for this example..We can compute the mean which is just $\frac{11}{2}$ or $5.5$...You can also compute the sample standard deviation which is just $\sqrt{\frac{55}{6}}$ or $3.02765$
To compute a Z score we use $\frac{X-\mu }{s }$
X is just a data value you want to test, so try 1 or 10
Now you can find how far a particular value is from the mean with these 3 pieces of data
Maybe this isn't what your looking for though.
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Yes, I'm familiar with the Z score. I'm just confused on how to calculate something like the range within 1.2 standard deviation and getting a percentage. – Nack Nov 27 '16 at 02:51
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You mean like, "what is the percentage of values within 1.2 standard deviations"? As Ian commented above, its no simple formula. – Brandon Nov 27 '16 at 02:53
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Yea, basically that's what I'm looking for. If there's no simple formula, would a calculator suffice in producing the answer? – Nack Nov 27 '16 at 02:55
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There is a theorem called "Chebyshevs Theorem" and it is used to find a probability or percent that certain values fall within a specific range..
Given something like:
$\mu =124$
and
$\sigma =7$
One could find the minimum probability that the number of values falls between 110 and 138. Note that this is 2 deviations away. For the sake of the problem lets say that 124 represents the amount of crimes per week in some neighborhood
the formula to this theorem looks like this:
$P(\mu-k \sigma <x<k \sigma +\mu)\geq 1-\frac{1}{k^2}$
where k is the number of deviations, so since above I noted that the values between 110 and 138 are 2 deviations away then we will use k = 2.
We can plug in the values we have above:
$P(124-2 \sigma <x<2 \sigma +124)\geq 1-\frac{1}{2^2}$
=
$P(124-2 \sigma <x<2 \sigma +124)\geq 0.75$
=
$P(110 <x< 138)\geq 0.75$
We would conclude that there is at least a 75% chance that the amount of crimes per week is between 110 and 138..
I'm sorry if this wasn't what you were looking for, but its a useful theorem to know..I don't want to confuse you though... But you can use this for any deviation you like, such as 1.5 or 2.5 and so on.
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