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Chapter 7 The Normal Distribution and Its Applications 7.1 Continuous Probability Distribution From Frequency Distribution to Probability Distribution height = Probability Density Relative Frequency Density approximation fi / N Class Width areas = 1 area under curve = 1 pdf of continuous Distribution 2. f(x) 0 for all value of x conditions for f(x) to be a pd. f (x)dx 1 3. P(X = c) = 0 1. k 4. P(hXk)= P(h<Xk)= P(hX<k)= P(h<X<k)= f (x)dx h = x f ( x ) dx and 2 = var(x) = ( x )2 f ( x)dx = x 2 f ( x)dx 2 7.2 Skipped (Basic Knowledge of A Continuous Probability Distribution - included in 7.1) 7.3 The Normal Distribution Denoted by X N(, 2) pdf. : 1 x 2 ) ( 1 2 e f(x) = 2 -<x< when = 0, 2 = 1 then Z N(0, 1) it is called standard normal Probabilities of normal distribution X N(, 2) X can be transformed to standard normal by Z X IIT_web_material P(h < X <k) = P( h k <Z< ) then table of standard normal can by used. 7.5 Normal Approximation To The Binomial (Optional) XB B(n, p) generally if np > 5 and npq > 5 then 1 1 < XN < x + ) 2 2 1 1 P(x1 XB x2) = P(x - < XN < x + ) 2 2 P(XB = x) = P(x - 7.6 Normal Approximation to The Poisson Generally if > 5, normal approximation can be used. e.g. Calls come in to a telephone exchange at random but at a rate of 400 per hour. Find the probability that in a given 1 hour period a) there are 350 - 450 calls, b) there are 360 or more calls Solution : The distribution is Poisson type with = 400. Since is large, so can approximate it by a Normal distribution with = 400 and = 400 = 20. a) P(350 X 450) = P(349.5 < XN < 450.5) 349.5 400 450.5 400 Z ) = P( 20 20 = P(-2.525<Z<2.525) = 9.884 b) P(360 X) 359.5 400 Z) = P( 20 = 0.9786 IIT_web_material