1. THE NORMAL APPROXIMATION TO THE BINOMIAL

2. Under certain conditions the Normal distribution can be used as an approximation to the Binomial, thus reducing the large number of calculations needed. If X ~ B ( n, p ) For large n, and not too small or too large p approximately X ~ N ( np, npq ) The approximation is better for larger values of n and when p is close to 0.5, however as a rough guideline, np and nq should both be greater than 5.

3. Example 1: A coin is biased with the probability of a head being 0.4. The coin is tossed 20 times. Using the Normal approximation estimate the probability of obtaining at least 7 heads. Let X be the r.v. “ The number of Heads”.X ~ B ( 20, 0.4) So ≈ X ~ N ( 8, 4.8 ) np > 5 ( and also nq > 5 ) 6.5 8 x z 0 z1z1 z 1 = 6.5 – 8 [ From the tables, z = 0.68, p = 0.7517 ] = – 0.68 By symmetry, the probability of at least 7 heads = 0.7517 ( np =20 × 0.4 = 8 ) ( npq =20 × 0.4 × 0.6 = 4.8 ) * * A continuity correction has been used. See next slide. Note, the exact probability here can be found using the Binomial tables.

4. Continuity Correction In the previous example, the required probability is the sum of the areas of the rectangles. This is approximated by the area under the normal curve. We have to take into account the fact that we are using a continuous distribution for a discrete variable. The bar representing 7 heads on a continuous scale starts at 6.5 and ends at 7.5. 6.57.5 Hence 6.5 rather than 7 was used to calculate the z score. 7 8910 1112

5. Below is shown the Binomial distribution B ( 20, 0.4 ) in Red, and the Normal distribution N ( 8, 4.8 ) in Blue. Note that the condition for the approximation is only just satisfied however there is still a good match between the distributions

6. Example 2: In a large college 55% of students are female and 45% are male. A random sample of 80 of the students is chosen. Using a suitable approximation find the probability that less than half of the sample are female. Let X be the r.v. “ The number of female students”.X ~ B ( 80, 0.55) np > 5 ( and also nq > 5 )( np =80 × 0.55 = 44 ) ( npq =80 × 0.55 × 0.45 = 19.8 ) So ≈ X ~ N ( 44, 19.8 ) 44 x 0 z z1z1 We want to know the probability of less than half are female i.e. 39 must be in the tail so use 39.5 to find the z score. 39.5 z 1 = 39.5 – 44 [ From the tables, z = 1.01, p = 0.8438 ] = – 1.01 0.1562 The probability of at most 39 females = 1 – 0.8438 =

7. Below is shown the Binomial distribution B ( 80, 0.55 ) and the Normal distribution N ( 44, 19.8 ). Note that the condition for the approximation is easily satisfied and hence there is a very good match between the distributions.

8. Summary of key points: This PowerPoint produced by R.Collins ; Updated Nov. 2010 If X ~ B ( n, p ) For large n, and not too small or too large p approximately X ~ N ( np, npq ) The approximation is better for larger values of n and when p is close to 0.5, however as a rough guideline, np and nq should both be greater than 5. Since we are using a continuous distribution to approximate a discrete variable, a continuity correction must be used.