How would I find the approximate percentage of values within a standard deviation of less than one on the normal model?
Chebyshev's rule is only used when the standard deviation is greater than or equal to 1.
[Math] Standard deviation of less than one
normal distributionstandard deviationstatistics
Related Solutions
Chebyshev's Inequality is true for any $c > 0$, but you are right that it only provides useful information for $c > 1$.
This is actually surprisingly easy prove. Define $\mu = E(X)$ and $\sigma^2 = E((X-\mu)^2)$. Observe that for any $c \geq 0$ we have $\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 c^2}$ where $\mathbb{1}\{\cdot\}$ is an indicator function equal to 1 if the event inside the brackets occurs and 0 otherwise. (To see this, suppose $\left|\frac{X-\mu}{\sigma}\right| \geq c$. Then the left-hand side equals 1 but the right-hand side is $\geq$ 1. If $\left|\frac{X-\mu}{\sigma}\right| < c$, the left-hand side is 0 but the right-hand side is positive.)
Since the expectation operator is monotonic, we can take expectations of both sides to obtain $$ \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) \leq \frac{E(X-\mu)^2}{\sigma^2 c^2} = \frac{1}{c^2} $$ Since $\Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) = 1- \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\leq c\right)= 1 -\Pr(\mu- \sigma c \leq X \leq \mu + \sigma c)$, we can rearrange to obtain the desired inequality.
The proof illustrates how you could derive many other inequalities of this type. Any function $g$ that satisfies
$$\left|\frac{X-\mu}{c\sigma}\right|\geq 1 \Rightarrow g\left(\left|\frac{X-\mu}{c\sigma}\right|\right) \geq 1$$
will work. For example, you could use an identical proof to show $$ \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) \leq \frac{E\left(|X-\mu|^k\right)}{\sigma^k c^k} $$ for all $k > 0$.
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.
Best Answer
$$P(Z < 1) = \alpha \approx 0.8413$$ $$P(Z > 1) = 1-\alpha = P(Z < -1)$$ $$P(-1 < Z < 1) = 1 - P(Z>1)-P(Z<-1)$$ $$P(-1 < Z < 1) = 1 - 2\cdot P(Z>1) = 1 - 2\cdot (1-\alpha)$$ $$P(-1 < Z < 1) = 2\alpha - 1 \approx 0.6826$$