1. Using the Tables for the standard normal distribution

2. Tables have been posted for the standard normal distribution. Namely The values of z ranging from -3.5 to 3.5

3. If X has a normal distribution with mean  and standard deviation  then has a standard normal distribution. Hence

4. Example: Suppose X has a normal distribution with mean  =160 and standard deviation  =15 then find:

6. This also can be explained by making a change of variable Make the substitution whenand Thus

7. The Normal Approximation to the Binomial

8. Let The Central Limit theorem Then the distribution of approaches the standard normal distribution as If x 1, x 2, …, x n is a sample from a distribution with mean , and standard deviations 

9. the Normal distribution with or the distribution of approaches the normal distribution with Hence the distribution of  approaches

10. Thus The Central Limit theorem states Suppose that X has a binomial distribution with parameters n and p. Then That sums and averages of independent R.Vs tend to have approximately a normal distribution for large n. whereare independent Bernoulli R.V.’s

11. Thus for large n the Central limit Theorem states that has approximately a normal distribution with Thus for large n where X has a binomial (n,p) distribution and Y has a normal distribution with

12. The binomial distribution

13. The normal distribution  = np,  2 = npq

14. Binomial distribution Approximating Normal distribution Binomial distribution n = 20, p = 0.70

15. Normal Approximation to the Binomial distribution X has a Binomial distribution with parameters n and p Y has a Normal distribution

16. Binomial distribution Approximating Normal distribution P[X = a]

19. Example X has a Binomial distribution with parameters n = 20 and p = 0.70

20. Using the Normal approximation to the Binomial distribution Where Y has a Normal distribution with:

21. Hence = 0.4052 - 0.2327 = 0.1725 Compare with 0.1643

22. Normal Approximation to the Binomial distribution X has a Binomial distribution with parameters n and p Y has a Normal distribution

25. Example X has a Binomial distribution with parameters n = 20 and p = 0.70

26. Using the Normal approximation to the Binomial distribution Where Y has a Normal distribution with:

27. Hence = 0.5948 - 0.0436 = 0.5512 Compare with 0.5357

28. Comment: The accuracy of the normal appoximation to the binomial increases with increasing values of n

29. Example The success rate for an Eye operation is 85% The operation is performed n = 2000 times Find 1.The number of successful operations is between 1650 and 1750. 2. The number of successful operations is at most 1800.

30. Solution X has a Binomial distribution with parameters n = 2000 and p = 0.85 where Y has a Normal distribution with:

31. = 0.9004 - 0.0436 = 0.8008

32. Solution – part 2. = 1.000