Normal Approximation of a Binomial Distribution (Cambridge International AS Mathematics)

Normal Approximation of Binomial

When can I use a normal distribution to approximate a binomial distribution?

• A binomial distribution $X \sim \mathrm{B} ( n , p )$ can be approximated by a normal distribution $X_{N} \sim \mathrm{N} ( \\mu , \\sigma ^{2} )$ provided

  • n is large **

  • p is close to 0.5**

    • $np>5$
    • $nq>5whereq=1-p$

• The mean and variance of a binomial distribution can be calculated by:

  • $μ=np$
  • $σ^{2}=np(1-p)$

Why do we use approximations?

• 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

What are continuity corrections?

• 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 $P(3.5\leq X_{N}<4.5)=P(3.5<X_{N}<4.5)$

How do I apply continuity corrections?

• 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
  • $P(a+0.5<X_{N}<b+0.5)$

How do I approximate a probability?

• STEP 1: Find the mean and variance of the approximating distribution

  • $μ=np$
  • $σ^{2}=np(1-p)$

• 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:

    • $np>5$
    • $np>5$where$q=1-p$

Worked Example The random variable $X \sim B ( 1250 , 0.4 ).$

Use a suitable approximating distribution to approximate $P(485\leq X\leq530)$.

Answer:

Exam Hint

• 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.