3.4.2 Normal Approximation of Binomial

Normal Approximation of Binomial

A binomial distribution $X ~ B (n, p)$ can be approximated by a normal distribution $X_{N} ~ N (μ, σ^{2})$ provided
- n is large
- p is close to 0.5
  - $n p > 5$
  - $n q > 5 w h e r e q = 1 - p$
The mean and variance of a binomial distribution can be calculated by:
- $μ = n p$
- $σ^{2} = n p (1 - p)$

4-4-2-normal-approximation-of-binomial-diagram-1

If there are a large number of values for a binomial distribution there could be a lot of calculations involved and it is inefficient to work with the binomial distribution
- These days calculators can calculate binomial probabilities so approximations are no longer necessary
- However it is easier to work with a normal distribution
  - You can calculate the probability of a range of values quickly
  - You can use the inverse normal distribution function (most calculators don't have an inverse binomial distribution function)
In your exam you must use the formula and not a calculator to find binomial probabilities so you are limited to small values of n

The binomial distribution is discrete and the normal distribution is continuous
A continuity correction takes this into account when using a normal approximation
The probability being found will need to be changed from a discrete variable, X, to a continuous variable, X_N
- For example, X = 4 for binomial can be thought of as $3.5 \leq X_{N} < 4.5$ for normal as every number within this interval rounds to 4
- Remember that for a normal distribution the probability of a single value is zero so $P (3.5 \leq X_{N} < 4.5) = P (3.5 < X_{N} < 4.5)$

Think about what is largest/smallest integer that can be included in the inequality for the discrete distribution and then find its upper/lower bound
$P (X = k) \approx P (k - 0.5 < X_{N} < k + 0.5)$
$P (X \leq k) \approx P (X_{N} < k + 0.5)$
- You add 0.5 as you want to include k in the inequality
$P (X < k) \approx P (X_{N} < k - 0.5)$
- You subtract 0.5 as you don't want to include k in the inequality
$P (X \geq k) \approx P (X_{N} > k - 0.5)$
- You subtract 0.5 as you want to include k in the inequality
$P (X > k) \approx P (X_{N} > k + 0.5)$
- You add 0.5 as you don't want to include k in the inequality
For a closed inequality such as $P (a < X \leq b)$
- Think about each inequality separately and use above
- $P (X > a) \approx P (X_{N} > a + 0.5)$
- $P (X \leq b) \approx P (X_{N} < b + 0.5)$
- Combine to give
- $P (a + 0.5 < X_{N} < b + 0.5)$

STEP 1: Find the mean and variance of the approximating distribution
- $μ = n p$
- $σ^{2} = n p (1 - p)$
STEP 2: Apply continuity corrections to the inequality
STEP 3: Find the probability of the new corrected inequality
- Find the standard normal probability and use the table of the normal distribution
The probability will not be exact as it is an approximate but provided n is large and p is close to 0.5 then it will be a close approximation
- To decide if n is large enough and if p is close enough to 0.5 check that:
  - $n p > 5$
  - $n p > 5$ where $q = 1 - p$

The random variable $X ~ B (1250, 0.4) .$

Use a suitable approximating distribution to approximate $P (485 \leq X \leq 530)$ .

3-4-2-normal-approximation-of-binomial-we-solution-with-addition

In the exam, the question will often tell you to use a normal approximation but sometimes you will have to recognise that you should do so for yourself. Look for the conditions mentioned in this revision note, n is large, p is close to 0.5, np > 5 and nq > 5.