# What is the normal approximation of binomial distribution?

I can't find a specific formula for this problem where I have to use the normal approximation of the binomial distribution.

Suppose we have a Bernoulli experiment with $p=0.8$ and we want to know how many trials we need for $\mathbb{P}(X\ge 30)$ to get the probability $0.975$

So what we have is

$n$ = unknown

$p$ = 0.8

$q$ = 0.2

$\mathbb{P}(X\ge 30)$

$0.975$

I have to use the normal approximation of the binomial distribution to solve this problem but I can't find any formula for this.

The only thing I found is $$Z=\frac{Y-np}{\sqrt{np(1-p)}}\rightarrow N(0,1)$$

Will be this the approximation formula?

For large $n$ and/or $p$ close to $1/2$, the $\text{Bin}(n,p)$ distribution looks roughly like a normal distribution with the same mean and standard deviation. Since the mean of a $\text{Bin}(n,p)$ random variable is $np$ and the standard deviation is $\sqrt{np(1-p)}$, this means we can approximate a $\text{Bin}(n,p)$ random variable by a $N(np,np(1-p))$ random variable.
To actually do that approximation, we have to be a little careful because binomial random variables take on whole number quantities, but normal random variables take on real values. If $X \sim \text{Bin}(n,p)$ and $Y\sim N(np, np(1-p))$, then for any whole number $k$, $$P(X=k) \approx P\left(k-\frac{1}{2} \leq Y \leq k+\frac{1}{2}\right).$$ That is, to account for the fact that $Y$ does not always take on whole number values, we consider every value of $Y$ in the interval $\left[k-\frac{1}{2},k+\frac{1}{2}\right]$ to be roughly $k$. Similarly, if we have integers $k<\ell$, then \begin{align*} P(k \leq X \leq \ell) &\approx P\left(k-\frac{1}{2} \leq Y \leq \ell+\frac{1}{2}\right)\\ P(X \leq \ell) &\approx P\left(Y \leq \ell+\frac{1}{2}\right)\\ P(k \leq X) &\approx P\left(k-\frac{1}{2} \leq Y\right). \end{align*} All of the probabilities on the right-hand side can be found by computing the $z$-scores (with $\mu=np$ and $\sigma=\sqrt{np(1-p)}$) and using a $z$-chart.
For your problem, $n$ is unknown, so we have to work backward. We want \begin{align*} 0.975 &= P(X \geq 30) \\ &\approx P(Y \geq 29.5) \\ &= P\left( Z \geq \frac{29.5-np}{\sqrt{np(1-p)}}\right) \\ &= 1-P\left( Z \leq \frac{29.5-np}{\sqrt{np(1-p)}}\right) \end{align*} where $Y \sim N(np,np(1-p))$ and $Z\sim N(0,1)$. Now you can use a $z$-chart to find the value of $\frac{29.5-np}{\sqrt{np(1-p)}}$, plug in the value of $p$, and solve for $n$. If $n$ turns out not to be in integer, then you need to figure out whether it should be rounded up or down. (Hint: you want $P(X \geq 30)$ to be at least $0.975$. Does this probability get bigger or smaller as $n$ increases?)
• Thank you this helped me a lot. That means I have to solve $0.975=1-P\left(Z \le \frac{29.5-0.8n}{\sqrt{(0.16n))}}\right)$ – Anil Dec 7 '17 at 15:38
• but what will be my $Z$ ? – Anil Dec 7 '17 at 15:41
• As I said in my answer, $Z$ is a $N(0,1)$ random variable, i.e., a normal random variable with mean $0$ and variance $1$. Are you familiar with $z$-charts like this: sixsigmastudyguide.com/wp-content/uploads/2014/04/z-table.jpg? You should find the entry in the chart that is closest to $0.975$ and use that to obtain the $z$-value. This will be the value of $\frac{29.5-np}{\sqrt{np(1-p)}}$. – kccu Dec 7 '17 at 15:52
• It's been a while since I used z charts. Last time was in school and now in University we have probability theory and we recently started normal distribution. I looked at the table which you provided me in the link and the closest z value to $0.975$ is in line $1.9$ and row $0.06$ with the z value $0.9750$ – Anil Dec 7 '17 at 16:00
• $0.975$ is not the $z$-score, rather it is the probability that a $N(0,1)$ random variable is less than or equal to the $z$-score. You said it is in row $1.9$ and column $0.06$, which means the $z$-score is $1.96$. – kccu Dec 7 '17 at 16:03