into the formula for u and we get Il I fL x n pXl p X xIl x xO Then making use

Into the formula for u and we get il i fl x n pxl p x

This preview shows page 15 - 18 out of 38 pages.

into the formula for u, and we get Il I fL = '" n. pX(l- p)"-X ~ x!(Il-x)! x=O Then, making use of the fact that x 1 = x! (x - I)!
Image of page 15
96 Chapter 4 ProbabilityDistributions EXAMPLE Solution Mean ofhypergeometric distribution EXAMPLE and n! = n(n - I)!, and factoring out nand p, we obtain /I (n-l)! px-I(l_p)n-x fJ. = np L (x _ 1)1(n - x)! x=l th ti starts with x = 1since the original summand iszerofor x := O. where esumma ton· . If we now let y = x-I and m = n- 1, we obtain m , ~ m. pY(1 _ p)m- y fJ.=npLJ y!(m-y)! y=o and this last sum can easily be recognized as tbat of all the terms of thebinomial distribution with the parameters m and p. Hence, this sum equals 1 and Itfollows that u. = np. Using JL = lip to findthe mean number of heads inthree tosses Find the mean of the probability distribution of the number of heads obtainedin 3 flips of a balanced coin. 113 For a binomial distribution with n = 3 and p = 2' we get fJ. = 3 . 2 = 2' andthis agrees with the result obtained on page 95. The formula fJ. = np applies, of course, only to binomial distributions, butfor other special distributions, we can, similarly, express the mean in termsoftheir parameters. For instance, for the mean of the hypergeometric distribution withthe parameters 11, a, and N, we can write a fJ.=n· - N In Exercise 4.43, the reader will be asked to derive this formula by a method similar to the one we used to derive the formula for the mean of a binomial distribution. Usingthe formula for the mean of a hypergeometric distribution With reference to the example on page 90 in which 5 of 20 car chargers were defective, find the mean of the probability distribution of the number of defectives in a sample of 10 randomly chosen for inspection. Solution Substituting n = 10, a = 5, and N = 20 into the above formula for u; we get 5 "=10'-=25 ~ 20' In other words, if we inspect 10 of the car chargers, we can expect 2.5 defectives, where expect is to be interpreted in the sense, it represents the long-run average number of defectives if 10 chargers are repeatedly selected from 20 chargersof which 5 are defective. To study the second of the two properties of probability distributions mentioned on page 94, their vanation, let us refer again to the two probability distributionsof Figure a.S. Forthe one where n = 4, there is a high probability of getting valuesclcs tothe mean, but for the one where II = 16, there is a high probability of getting valu~s scattered over considerable distances away from the mean. Using this property, It
Image of page 16
Varianceofprobability distribution Standarddeviation of probabilitydistribution EXAMPLE Solution Sec4.4 The Mean and the Variance of aProbability Distribution 97 may seem reasonable to measure thevariation ofaprobability distribution withthe quantity L (x - jh) . f(x) all x namely, theaverage amount bywhich thevalues oftherandom variable deviate from the mean. Unfortunately, .L (x - jh) .
Image of page 17
Image of page 18

You've reached the end of your free preview.

Want to read all 38 pages?

  • Fall '09
  • CHUNG
  • Probability, Probability theory, Binomial distribution, Discrete probability distribution, Hypergeometric Distribution

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture