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Lecture 9

Lecture 9. Normal Distribution, Central Limit Theorem, Gamma Distribution. The Reproductive Property of Normal Distribution. Examples

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Lecture 9

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  1. Lecture 9 Normal Distribution, Central Limit Theorem, Gamma Distribution

  2. The Reproductive Property of Normal Distribution

  3. Examples Three steel pipes are selected randomly from N(12.00, 0.0016) population and assembled end-to-end. Compute the probability that the total length of the assembly exceeds 36.12. n=3, each ci = 1, and each Xi ~ N(12.00, 0.0016) Suppose 4 steel pipes are selected at random from the N(12, 0.0016) population and placed end-to-end. (a) Compute the pr that the length of the assembly does not exceed 47.90". (b) Compute the pr that the average length of the 4 pipes exceeds 12.04. (a) n=4, each ci = 1, and each X1 ~ (12.00, 0.0016) (b) n=4, each ci = 1/4, and each X1 ~ (12.00, 0.0016)

  4. The Central Limit Theorem

  5. Central limit theorem Refer to the “The Moments of a Random Variable.doc” handout for the values of 3 and 4.

  6. The Normal Approximation to the Binomial

  7. Example • Consider a binomial distribution with n=50 trials and p=0.30. (So that =np=15). Note that we would like to have np>15, but n=50>25(3)2 is easily satisfied. Then the Range space of the discrete rv XD={0,1,2,,50}. The exact probability of attaining exactly 12 successes in 50 trials is given by b(12; 50, 0.30) = P(XD=12) = 0.08383. • To apply the normal distribution approximation (with =15 and 2=npq=10.50), we first need to select an interval on a continuous scale to represent the discrete XD=12. Clearly, this interval needs to be (11.5,12.5)C . Thus, P(XD=12) = P(11.5<XC<12.5)

  8. 0.08383 Gray Area = 0.080157 XC=11.5 XD=12 XC=12.5

  9. Green Area = 0.77980 0.08383 Gray Area = 0.080157 XC=11.5 XD=12 XC=12.5

  10. The Gamma Distribution and its Relatives The Gamma Function: Properties of the Gamma function: • For any >1, () = (1)· (1) • For any positive integer, n, (n) = (n1)! A continuous rv X has a Gamma Distribution if the pdf of X is: Where >0 and >0. The standard Gamma distribution has =1

  11. =0.6, =1 =2, =1/3 =1, =1 =1, =1 =2, =1 =2, =1 =5, =1 =2, =2 Why do we need gamma distribution? Any normal distribution is bell-shaped and symmetric. There are many practical situations that do not fit to symmetrical distribution. The Gamma family pdfs can yield a wide variety of skewed distributions.  is called the scale parameter because values other than 1 either stretch or compress the pdf in the x-direction.

  12. The mean and the variance of a gamma distribution • The cdf of the standard gamma rv X: is called the incomplete gamma function. There are extensive tabulations available for F(x;). We will refer to page 742 on Devore, for  = 1,2,…,10 and x = 1,2,..,15.

  13. Example • The reaction time X of a randomly selected individual to a certain stimulus has a standard gamma distribution with =2. Compute the probability that the reaction time is between 3 and 5 time units. P(3X5) = F(5;2)  F(3;2) = 0.960  0.801 = 0.159 Compute the probability that the reaction time is longer than 4 time units. P(X>4) = 1  F(4;2) = 1  0.908 = 0.092

  14. Let X have a gamma distribution with parameters  and . For any x>0, the cdf of X is given by: where F(;) is the incomplete gamma function. Example: X = the survival time in weeks. =8, =15. Expected survival time E(X)=(8)(15)=120 weeks. V(X) = (8)(15)2 = 1800, x= 42.43 P(60X120) = P(X120) - P(X60) = F(120/15;8) – F(60/15;8) = F(8;8) – F(4;8) = 0.547 – 0.051 = 0.496 P(X30) = 1 – P(X<30) = 1 – P(X30) = 1 – F(30/15;8) = 0.999

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