> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.

# The Normal Approximation to the Binomial Distribution

The normal distribution is often used to solve problems that involve the binomial distribution since when n is large (say, 100), the calculations are too difficult to do by hand using the binomial distribution.

> Distribusi normal sering digunakan untuk menyelesaikan masalah yang melibatkan distribusi binomial karena ketika n besar (misalnya, 100), perhitungan terlalu sulit dilakukan secara manual menggunakan distribusi binomial.

The normal approximation to the
binomial is appropriate when $np \geq 5$
and $nq \geq 5$.

> Aproksimasi normal terhadap binomial sesuai ketika $np \geq 5$ dan $nq \geq 5$.

In addition, a correction for
continuity may be used in the normal
approximation.

> Selain itu, koreksi untuk kontinuitas dapat digunakan dalam aproksimasi normal.

A **correction for continuity** is a correction
employed when a continuous distribution
is used to approximate a discrete
distribution.

> **Koreksi untuk kontinuitas** adalah koreksi yang digunakan ketika distribusi kontinu digunakan untuk mendekati distribusi diskrit.

The continuity correction means that for
any specific value of X, say 8, the
boundaries of X in the binomial
distribution (in this case 7.5 and 8.5) must
be used.

> Koreksi kontinuitas berarti bahwa untuk nilai X tertentu, katakan 8, batas X dalam distribusi binomial (dalam hal ini 7,5 dan 8,5) harus digunakan.

### The Normal Approximation to the Binomial Distribution - Example

#### Question 1

<Card>
  Prevention magazine reported that 6%
  of American drivers read the newspaper
  while driving.

  <br />

  If 300 drivers are selected at random,find the probability
  that exactly 25 say they read the
  newspaper while driving.

  <br />

  Solution :

  * Here p = 0.06, q = 0.94, and n = 300.

  * Check for normal approximation:
    * $~~~~~np = 300(0.06) = 18$ and
    * $~~~~~nq = 300(0.94) = 282$. Since both values are at least 5, the normal
      approximation can be used.

  * $~$

  * $\mu = np = 300(0.06) = 18$ and

  * $\sigma = \sqrt{npq} = \sqrt{(300)(0.06)(0.94)} = 4.11$.

  * $~$

  * So $P(X = 25) = P(24.5 \leq X \leq 25.5)$.

  * Find the z-values for 24.5 and 25.5:
    * $~$
    * $~~~~~z_1 = \dfrac{24.5 - 18}{4.11} = 1.58 $ and
    * $~$
    * $~~~~~z_2 = \dfrac{25.5 - 18}{4.11} = 1.82 $.
    * $~$

  * Thus, $P(24.5 \leq X \leq 25.5) = P(1.58 \leq z \leq 1.82) = 0.4656 - 0.4429 = 0.0227$.

  * Hence, the probability that exactly 25
    people read the newspaper while
    driving is 2.27%.
</Card>