When we hear about the statistical studies the first thing that comes to our mind is probability. We have already seen that **what is empirical rule** and how it is used for the symmetric normal distribution probability calculation. Today we will be discussing about the Normal Distribution to the Binomial Problems.

When the continuous normal distribution is used to compute the discrete distribution or the binomial distribution is known as the normal approximation to the binomial problems. We know that The Central Limit Theorem says, if the size of the sample is large, the sample distribution of the sample means will be approximately normal. There must be a question in your mind that **what is the meaning** of Normal Approximation To Binomial Problem, I will clear this in this article itself.

**Normal Approximation/Distribution To The Binomial**

The very first step for the normal approximation to the binomial is to determine that you have enough large samples for the calculations. How large is this sample? So you need to compute this with the equation n*p and n*q.

here, n is the sample size

p is probability

q is 1 – p, here you need to find the value of q by subtracting the probability as 1:q = 1 – p.

When, n*p and n*q are greater than 5, it is possible for you to use the normal approximation to the binomial.

**Example**

Now let us see the normal approximation problem and how to solve it.

In one of the districts, urban school around 62% of the students of 12^{th} grade attends the school. If 500 students of the 12^{th} grade are selected, then calculate that at least 290 students have active admission in the school.

**Solution: **

First, we need to compute the values of p, q, and n

Here, p = 62% = 0.62

q = 1:1 – 0.62 = 0.38

n = 500

now, calculate the normal approximation to the binomial, i.e. n*p and n*q and also check if these values are greater than 5, so that you can use the approximation

∴n*p = 500*0.62

∴n*p = 310

Similarly, n*q = 500*0.38

∴ n*q = 190

Here, both the values are greater than 5, hence we can compute the normal approximation now. Now, calculate the value of the mean, i.e. µ

∴µ = n * p = 310 ………………….. (which we have already calculated)

Now, we need to multiply the value of µ by q

∴310*0.38 = 117.8

Now, we have to compute the value of the standard deviation i.e., σ

∴σ = square root of 117.8

∴σ = √117.8

∴σ = 10.85

Now, using the Continuity Correction Factor

We have been asked to find out the probability that at least 290 students should have active admission in the school, hence re-write the continuity correction factor

∴P(X≥290)

If P(X=n) – P(n-0.5<X<n+0.5)

If P(X>n) – P(X>n+0.5)

If P(X≤n) – P(X<n+0.5)

If P(X<n) – P(X<n-0.5)

If P(X≥n) – P(X>n-0.5)

∴ P(X≥290-0.5) = P(X≥289.5)

Now, you have to draw the graph of the mean in the center as follows

Now, compute the value of z-score

∴z = ((new probability calculated) – µ) / σ

∴z = (289.5 – 310)/ 10.85

∴z = -1.89

From the z table, we can get the value for the area of -1.89 as 0.4706

So now, to find the probability we need to add 0.5 in the above value of z area

∴0.4706+0.5 = 0.9706

∴the probability that the 290 students can have active admissions in the school is 97.06%

**Conclusion: **This is all about the Normal Approximation to the Binomial Problems.