Using the Tables for the standard normal distribution

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Presentation transcript:

Using the Tables for the standard normal distribution

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

If X has a normal distribution with mean m and standard deviation s then has a standard normal distribution. Hence

Example: Suppose X has a normal distribution with mean m =160 and standard deviation s =15 then find:

This also can be explained by making a change of variable Make the substitution when and Thus

The Normal Approximation to the Binomial

The Central Limit theorem If x1, x2, …, xn is a sample from a distribution with mean m, and standard deviations s, Let Then the distribution of approaches the standard normal distribution as

Hence the distribution of approaches the Normal distribution with or the distribution of approaches the normal distribution with

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

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

The binomial distribution

The normal distribution m = np, s2 = npq

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

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

Approximating Normal distribution P[X = a] Binomial distribution

P[X = a]

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

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

Hence = 0.4052 - 0.2327 = 0.1725 Compare with 0.1643

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

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

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

Hence = 0.5948 - 0.0436 = 0.5512 Compare with 0.5357

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

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

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

= 0.9004 - 0.0436 = 0.8008

Solution – part 2. = 1.000