Normal Approximation to the Binomial

• Poisson distribution can be used to approximate binomial probabilities when $n$ is quite large and $p$ is very close to 0 or 1.
• Normal distribution not only provide a very accurate approximation to binomial distribution when $n$ is large and $p$ is not extremely close to 0 or 1,
• But also provides a fairly good approximation even when $n$ is small and $p$ is reasonably close to $\frac{1}{2}$.

  Figure 1: Normal approximation of $b(x; 15,0.4)$.

  

• Theorem 6.2:
  
• $P(7\leq X \leq 9)$
  $$\sum_{x=7}^{9} b(x;15,0.4)=\sum_{x=0}^{9} b(x;15,0.4)-\sum_{x=0}^{6} b(x;15,0.4)$$
  $$=0.9662-0.6098=0.3564$$
  $$\mu=np=15*0.4=6,~\sigma^2=15*0.4*0.6=3.6,~\sigma=1.897$$
  $$z_1=\frac{6.5-6}{1.897}=0.26,~and~z_2=\frac{9.5-6}{1.897}=1.85$$
  $$P(7\leq X \leq 9) \approx P(0.26 <Z < 1.85)=P(Z<1.85)-P(Z<0.26)$$
  $$=0.9687-0.6026=0.3652$$

$P(X=4)=b(4;15,0.4)=0.1268$

$$z_1=\frac{3.5-6}{1.897}=-1.32,~and~z_2=\frac{4.5-6}{1.897}=-0.79$$

$$P(X=4) \approx P(3.5 < X <4.5)=P(-1.32 <Z < -0.79)$$

$$=P(Z<-0.79)-P(Z<-1.32)$$

$$=0.2148-0.0934=0.1214$$

Figure 2: Normal approximation of $b(x; 15,0.4)$ and $\sum_{x=7}^9 b(x; 15,0.4)$.

• The degree of accuracy, which depends on how well the curve fits the histogram, will increase as n increases.
• If both $np$ and $nq$ are greater than or equal to 5, the normal approximation will be good.

  Figure 3: Histogram for $b(x; 6, 0.2)$.

  

  Figure 4: Histogram for $b(x; 15, 0.2)$.

  

• Let $X$ be a binomial random variable with parameters $n$ and $p$.
• Then $X$ has approximately a normal distribution with mean $\mu = np$ and variance $\sigma^2 = npq$ and
  $$P(X \leq x)=\sum_{k=0}^x b(k;n,p)$$
  $$\approx area~under~ normal~curve~ to~ the~ left~ of~x+0.5$$
  $$=P\left( Z \leq \frac{x+0.5-np}{\sqrt{npq}}\right)$$
  and the approximation will be good if $np$ and $nq$ are greater than or equal to 5.

• Example 6.15: The probability that a patient recovers from a rare blood disease is 0.4.
• If 100 people are known to have contracted this disease, what is the probability that less than 30 survive?
• Solution:
  $$\mu=np=100*0.4=40$$
  $$\sigma=\sqrt{100*0.4*0.6}=4.899$$
  $$z_1=\frac{29.5-40}{4.899}=-2.14$$
  $$P(X<30)\approx P(Z<-2.14)$$
  $$=0.0162$$

  Figure 5: Area for Example 6.15.

  

• Example 6.16: A multiple-choice quiz has 200 questions each with 4 possible answers of which only 1 is correct answer.
• What is the probability that sheer guess-work yields from 25 to 30 correct answers for 80 of the 200 problems about which the student has no knowledge?
• Solution:
  $$\mu=np=80*\frac{1}{4}=20$$
  $$\sigma=\sqrt{80*\frac{1}{4}*\frac{3}{4}}=3.873$$
  $$z_1=\frac{24.5-20}{3.873}=1.16,$$
  $$~z_2=\frac{30.5-20}{3.873}=2.71$$
  $$P(25 \leq X \leq 30)=\sum_{x=25}^{30} b(x;80,\frac{1}{4})$$
  $$\approx P(1.16 < Z <2.71)$$
  $$=0.9966-0.8770=0.1196$$

  Figure 6: Area for Example 6.16.