I am struggling with the following textbook problem:
Let $X \sim B(16, 0.5)$ with mean $\mu$ and standard deviation $\sigma$. Calculate the probability that a realization of X is within $[\mu-\sigma, \mu+\sigma]$ exactly and approximately using Chebychev's inequality.
From that I can immediately calculate $\mu=np=8$ and $\sigma=\sqrt{np(1-p)}=2$.
Then I calculated the exact solution as follows:
$$P(X \in [6, 10])=\sum_{k=6}^{10}{16 \choose k}0.5^k0.5^{16-k}=2^{-16}\sum_{k=6}^{10}{16 \choose k}\approx 0.7899$$
Then I tried the approximation using Chebychev:
$$P(X \in [6, 10])=1-P(X \notin [6, 10])=1-P(|X-8|\ge 3)$$ $$ \le 1 - \frac{1}{3^2}\sigma^2=1-\frac{4}{9}=\frac{5}{9}\approx 0.5556$$
Now obviously this contradicts the first result. I think that I made a mistake assuming that $X$ is integral but I don't know why I would not be allowed to do that ($X$ is the number of successful trials after all). When writing the above without assuming $X$ to be integral I get the following:
$$P(X \in [6, 10])=1-P(X \notin [6, 10])=1-P(|X-8| > 2)$$
However, now I have no idea how to appy Chebychev since I have a $>$ instead of a $\ge$.
Question: What mistakes did I make and how would I correctly solve the problem?
$$P(X \in [6, 10])=1-P(X \notin [6, 10])=1-P(|X-8|\ge 3) \geq 1 - \frac{1}{3^2}\sigma^2=1-\frac{4}{9}=\frac{5}{9}\approx 0.5556$$ with no contradiction with exact value of probability. When you subtract inequality $$ P(|X-8|\ge 3)\leq \frac{1}{3^2}\sigma^2$$ from $1$, change the sign.