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?
Best Answer
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 $0.17103699497659\ldots$.