Find the minimum percentage of values within two bounds subject to moment constraints I am facing the following problem:

Variable Z has a mean of 15 and a standard deviation of 2. What is the minimum percentage of Z values that lie between 8 and 17?

I have tried the following: Here on right side the value is 17; that is, 1 sd.  On the left side it is (8-15)/2 =3.5 sd.
Since it has asked for a minimum percentage I tried Chebyshev's rule. That states the minimum is $1-1/k^2,$ where $k$ is number of standard deviations from mean. For 1 sd the minimum percentage of Z values is 0. For 3.5 sd the minimum percentage of Z values is 91.8.
But here, the number of sds is unequal on both sides of the mean. On the left side of the mean it is 3.5 sd and on the right side of the mean it is 1 sd.
Am I missing something? Please help in solving this>
 A: Suppose $Z$ consists of $n$ values.  Consider the dual problem:

Minimize the standard deviation,
given that at most $c$ of the values lie between $8$ and $17$ and the mean is $15.$

Because greater values of $c$ will lead to smaller possible SDs, if the minimum SD for a given value of $c$ exceeds $2,$ then $c$ must be too small.  In this fashion we can determine an upper bound for $c.$
As a matter of notation, let $a$ of those values be less than $8$ and $b$ of those values exceed $17.$  Thus,
$$n = a + b + c.\tag{1}$$
It is intuitive -- and relatively straightforward to demonstrate mathematically -- that the SD (or, equivalently, the variance) is minimized by (1) moving the $a$ smaller values as close as possible to $8;$ (2) moving the $b$ larger values as close as possible to $17;$ and (3) setting the remaining $c$ values to the average of $15.$  In such a configuration the mean is arbitrarily close to
$$15 = 8a + 17b + 15c\tag{2}$$
and the variance exceeds $2^2:$
$$2^2 \lt (8-15)^2a + (17-15)^2b + (15-15)^2 c = 49a + 4b.\tag{3}$$
At the extreme, this will be an equality.  Given $n,$ the system of three linear equations $\{(1),(2),(3)\}$ has the unique solution
$$(a,b,c) = \frac{n}{63}(4,14,45).$$
Indeed, with $n=63$ and $(a,b,c)=(4,14,45)$ we find the mean is $15$ and the standard deviation is exactly $2.$
It is also clear that if $c$ is any less than $45n/63,$ the standard deviation must exceed $2.$
The solution to the original problem is now apparent.  I leave it to you to turn this sketch into a demonstration that meets your standards of rigor.
