If, tossing a coin 400 times, we count the heads, what is the probability that the number of heads is [160,190]? I wanted to solve the problem with the Central Limit Theorem.
Analyzing the question, I modeled the situation with a random variable :
$$\begin{cases} 1 & \text{with probability } 1/2; \\
0 & \text{with probability } 1/2; \end{cases}
$$
Calculating the mean $\mu = \frac{1}{2}$ and the variance $\sigma^2 = \frac{1}{4}$.
Then, I thougt that since the number of repetitions is >> 30 I tried to fit the probability function to a Gaussian normal $N(400\mu,400\sigma) = N(200,100)$.
Calculating and "normalizing"...
$$\begin{align}
& P(160 < x < 190)\\
&= P\left( \frac{160-200}{20 \times 100} < z < \frac{190-200}{20 \times 100}\right) \\
&= P(-0.02 < z < -0.005).
\end{align}$$
And here I'm stuck because I wanted to use the relation
$$        P(160 < x < 190) = \Phi(-0.005) - \Phi(-0.02) $$
That result negative..
Do you see any error in my strategy? 
 A: With $n = 400$ trials, the exact probability distribution for the number of heads $X$ observed is given by $X \sim {\rm Binomial}(n = 400, p = 1/2)$, assuming the coin is fair.  Since calculating $\Pr[160 \le X \le 200]$ requires a computer, and $n$ is large, we can approximate the distribution of $X$ as ${\rm Normal}(\mu = np = 200, \sigma^2 = np(1-p) = 100)$.  Thus $$\begin{align*} \Pr[160 \le X \le 200] &\approx \Pr[159.5 \le X \le 200.5] \\ &= \Pr\left[\frac{159.5 - 200}{\sqrt{100}} \le \frac{X - \mu}{\sigma} \le \frac{200.5 - 200}{\sqrt{100}} \right] \\ &= \Pr[-4.05 \le Z \le 0.05] \\ &= \Phi(0.05) - \Phi(-4.05) \\ &\approx 0.519913. \end{align*}$$  Note that we employed continuity correction for this calculation.  The exact probability is $0.5199104479\ldots$.
A similar calculation applies for $\Pr[160 \le X \le 190]$.  Using the normal approximation to the binomial, you would get an approximate value of $0.171031$.  Using the exact distribution, the probability is $0.17103699497659\ldots$.
A: While I was typing, an answer was given by heropup. So  the stuff below Old, though correct, can be disregarded.
We address only the question about $\Phi(-0.005) - \Phi(-0.02)$, which in the OP is said to be negative. The expression is not the correct  answer to the question, but it is not negative.
The usual tables do not give $\Phi(z)$ for negative $z$. This information is felt to be unnecessary, since by symmetry if $z$ is negative, then $\Pr(Z\le z)=\Pr(Z\ge -z)=1-\Phi(-z)$. Thus
$$\Phi(-0.005) - \Phi(-0.02)=(1-\Phi(0.005))-(1-\Phi(0.02))=\Phi(0.02)-\Phi(0.005),$$
and the last expression is positive, and can be found from standard tables. 
Old:
We will use the continuity correction.  Our probability is the probability that the binomial is $\le 200$ minus the probability that the binomial is $\le 159$. 
Let $Y$ be a normal with mean $200$ and variance $100$. Our approximation is
$$\Pr(Y\le 200.5)-\Pr(Y\le 159.5).$$
This is 
$$\Pr\left(Z\le \frac{0.5}{\sqrt{100}}\right)-\Pr\left(Z\le \frac{-40.5}{\sqrt{100}}\right),$$
where $Z$ is standard normal.
The first number is directly available from the usual tables. The second usually is not, but by symmetry it is equal to $1-\Phi(4.05)$. For all practical purposes, this is $0$. So our required probability is approximately $\Phi(0.05)$. 
