Minimum percentage of a random variable within two bounds 
Random 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 the 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 minimum percentage I tried Chebyshev rule, which states that the probability is no less than $1-1/k^2$, 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 number of sd is unequal on both side of mean.
Am I missing something? Please help in solving this.
 A: 
$\def\F{\mathscr{F}}\def\R{\mathbb{R}}\def\d{\mathrm{d}}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$Proposition: For any $σ > 0$, define$$
\F_σ = \brace{ F: \R → [0, 1] \,\middle|\, F\ \text{is c.d.f.}, \int_\R x \,\d F(x) = 0, \int_\R x^2 \,\d F(x) = σ^2 }.
$$
If $a, b > 0$, then$$
\min_{F \in \F_σ} \int_{(-a, b)} \d F(x) = \begin{cases}
0; & σ^2 > ab\\
\dfrac{4(ab - σ^2)}{(a + b)^2}; & \dfrac{1}{2} (ab - \min(a^2, b^2)) < σ^2 \leqslant ab\\
\dfrac{\min(a^2, b^2)}{\min(a^2, b^2) + σ^2}; & σ^2 \leqslant \dfrac{1}{2} (ab - \min(a^2, b^2))
\end{cases}.
$$

Proof: For $F \in \F_σ$, suppose $F$ is the c.d.f. of random variable $X_F$.
Case 1: $σ^2 > ab$. Obviously $P(X_F \in (a, b)) \geqslant 0$ for any $F \in \F_σ$, and the equality is attained if$$
P(X_F = -a) = \frac{σ^2}{a^2 + σ^2}, \quad P\paren{ X_F = \frac{σ^2}{a}} = \frac{a^2}{a^2 + σ^2}.
$$
Case 2: $\dfrac{1}{2} (ab - \min(a^2, b^2)) < σ^2 \leqslant ab$. Note that for $F \in \F_σ$,$$
E((a + X_F)(b - X_F)) = ab + (b - a) E(X_F) - E(X_F^2) = ab - σ^2,
$$
and $(a + x)(b - x) \leqslant 0$ for $x \in (-a, b)^c$, thus\begin{gather*}
E((a + X_F)(b - X_F) I_{(-a, b)}(X_F)) \geqslant E((a + X_F)(b - X_F)) = ab - σ^2. \tag{1}
\end{gather*}
By AM-GM inequality,$$
(a + x)(b - x) \leqslant \paren{ \frac{1}{2} ((a + x) + (b - x)) }^2 = \frac{1}{4} (a + b)^2, \quad \forall x \in (a, b)
$$
therefore$$
\frac{1}{4} (a + b)^2 P(X_F \in (-a, b)) \geqslant E((a + X_F)(b - X_F) I_{(-a, b)}(X_F)) \geqslant ab - σ^2,
$$
which implies that$$
P(X_F \in (-a, b)) \geqslant \frac{4(ab - σ^2)}{(a + b)^2}.
$$
The equality is attained if\begin{align*}
\small P(X_F = -a) = \frac{b^2 - ab + 2σ^2}{(a + b)^2}, \quad P\paren{ X_F = \frac{1}{2}(b - a) } = \frac{4(ab - σ^2)}{(a + b)^2}, \quad P(X_F = b) = \frac{a^2 - ab + 2σ^2}{(a + b)^2}.
\end{align*}
Case 3: $σ^2 \leqslant \dfrac{1}{2} (ab - \min(a^2, b^2))$. Without loss of generality, suppose $a \leqslant b$. Note that for $F \in \F_σ$,$$
E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } } = (a^2 + 2σ^2) + \frac{2σ^2}{a} E(X_F) - E(X_F^2) = a^2 + σ^2,
$$
and $a + \dfrac{2σ^2}{a} \leqslant b$ implies that $(a + x) \paren{ a + \dfrac{2σ^2}{a} - x } \leqslant 0$ for $x \in (-a, b)^c$, thus\begin{gather*}
{\small E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } I_{(-a, b)}(X_F) } \geqslant E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } } = a^2 + σ^2.} \tag{2}
\end{gather*}
By AM-GM inequality,$$
\small (a + x) \paren{ a + \frac{2σ^2}{a} - x } \leqslant \paren{ \frac{1}{2} \paren{ (a + x) + \paren{ a + \frac{2σ^2}{a} - x } } }^2 = \paren{ a + \frac{σ^2}{a} }^2, \quad \forall x \in (-a, b)
$$
therefore$$
\small \paren{ a + \frac{σ^2}{a} }^2 P(X_F \in (-a, b)) \geqslant E\paren{ (a + X_F) \paren{ a + \frac{2σ^2}{a} - X_F } I_{(-a, b)}(X_F) } \geqslant a^2 + σ^2,
$$
which implies that$$
P(X_F \in (-a, b)) \geqslant \frac{a^2}{a^2 + σ^2}.
$$
The equality is attained if\begin{gather*}
P(X_F = -a) = \frac{σ^2}{a^2 + σ^2}, \quad P\paren{ X_F = \frac{σ^2}{a} } = \frac{a^2}{a^2 + σ^2}. \tag*{$\Box$}
\end{gather*}
Now return to the question. Since $E(Z) = 15$ and $D(Z) = 2^2 = 4$, define $X = Z - 15$, then $E(X) = 0$ and $E(X^2) = 4$. Because $8 - 15 = -7$, $17 - 15 = 2$, and $\dfrac{1}{2} (7 · 2 - 2^2) > 4$, so the proposition shows that$$
P(Z \in (8, 17)) = P(X \in (-7, 2)) \geqslant \frac{2^2}{2^2 + 4} = \frac{1}{2},
$$
and the equality is attained if $P(X = -2) = P(X = 2) = \dfrac{1}{2}$, i.e. $P(Z = 13) = P(Z = 17) = \dfrac{1}{2}$.
A: You have correctly used Chebyshev’s Rule to determine that 91.8% of values are within 3.5 standard deviations of the mean. That is the band between 8 and 22. You may also know that about 68.3% of values are within 1 standard deviation of the mean. That’s the band of values between 13 and 17. The difference between those gives us 91.8% - 68.3% = 23.5% of values in the two bands between 8 and 13 and between 17 and 22. Assuming symmetry, that gives you half of 23.5% = 11.75% between 8 and 13. So between 8 and 17 is 11.75+ 68.3% = 80.05%
