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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≥5np \geq 5 and nq≥5nq \geq 5.
Aproksimasi normal terhadap binomial sesuai ketika np≥5np \geq 5 dan nq≥5nq \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.

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The Normal Approximation to the Binomial Distribution - Example

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Question 1

Prevention magazine reported that 6% of American drivers read the newspaper while driving.
If 300 drivers are selected at random,find the probability that exactly 25 say they read the newspaper while driving.
Solution :
  • Here p = 0.06, q = 0.94, and n = 300.
  • Check for normal approximation:
    •      np=300(0.06)=18~~~~~np = 300(0.06) = 18 and
    •      nq=300(0.94)=282~~~~~nq = 300(0.94) = 282. Since both values are at least 5, the normal approximation can be used.
  •  ~
  • μ=np=300(0.06)=18\mu = np = 300(0.06) = 18 and
  • σ=npq=(300)(0.06)(0.94)=4.11\sigma = \sqrt{npq} = \sqrt{(300)(0.06)(0.94)} = 4.11.
  •  ~
  • So P(X=25)=P(24.5≤X≤25.5)P(X = 25) = P(24.5 \leq X \leq 25.5).
  • Find the z-values for 24.5 and 25.5:
    •  ~
    •      z1=24.5−184.11=1.58~~~~~z_1 = \dfrac{24.5 - 18}{4.11} = 1.58 and
    •  ~
    •      z2=25.5−184.11=1.82~~~~~z_2 = \dfrac{25.5 - 18}{4.11} = 1.82 .
    •  ~
  • Thus, P(24.5≤X≤25.5)=P(1.58≤z≤1.82)=0.4656−0.4429=0.0227P(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%.