Upcoming SlideShare
×

# Probabilitydistributionlecture web

308 views

Published on

it is well a good

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
308
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
2
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Probabilitydistributionlecture web

1. 1. PROBABILITY DISTRIBUTIONSThere are two types of random variables: A Discrete random variable can take on only specified, distinct values. A Continuous random variable can take on any value within an interval.A probability distribution for a discrete random variable is a mutuallyexclusive listing of all possible numerical outcomes for that randomvariable, such that a particular probability of occurrence is associated witheach outcome.Probability Distribution for the Toss of a Die: Xi P(Xi)1 1/62 1/63 1/64 1/65 1/66 1/6This is an example of a uniform distribution.Discrete Probability Distributions have 3 major properties:1) ∑ P(X) = 12) P(X) ≥ 03) When you substitute the random variable into the function, you find outthe probability that the particular value will occur. Three major probability distributions: Binomial distribution, Hypergeometric distribution, Poisson distribution. p. 1
2. 2. MATHEMATICAL EXPECTATIONA random variable is a variable whose value is determined by chance.Expected value is a single average value that summarizes a probabilitydistribution. E(X) = ∑ XiP(Xi)If X is a discrete random variable that takes on the value Xi with probabilityP(Xi), then the expected value of X – E(X) – is obtained by multiplying eachvalue that random variable X can assume by its probability P(Xi) and summingthese products. (In other words, it is a weighted average over all possibleoutcomes.)The expected value is normally used as a measure of central tendency forprobability distributions (where ∑P(Xi) = 1). Hence, E(X) = μ. μ = E(X) = ∑ XiP(Xi)Example, the probability distribution for the random variable D, the number onthe face of a die after a single toss:D P(D) D·P(D)1 1/6 1/62 1/6 2/63 1/6 3/64 1/6 4/65 1/6 5/66 1/6 6/6 21/6μ = E(X) = 21/6 = 3.5The expected value is a single average value that summarizes a probabilitydistribution. On average, the value you expect from a toss of a die is 3.5. Thisis the population mean. p. 2
3. 3. Variance of a random variable:σ2 = Var (D) = E[(Di– μ)2] = ∑( Di – μ)2 P(DXi) σ D = ∑( Di − µ) 2 P ( Di ) 2 = (1 − 3.5) 2 ⋅1 / 6 + ( 2 − 3.5) 2 ⋅1 / 6 + (3 − 3.5) 2 ⋅1 / 6 + ( 4 − 3.5) 2 ⋅1 / 6 + (5 − 3.5) 2 ⋅1 / 6 + (6 − 3.5) 2 ⋅1 / 6 = 2.9166 σ D =1.71 p. 3
4. 4. Expected Monetary ValueExample:In the following game, there is an equally likely chance of making \$300, \$120,and \$0. How much would you be willing to pay to play?V (Dollar Value) P(V)\$300 1/3\$120 1/30 1/3What is the expected value of this lottery?E(V) = \$140. [\$300 (1/3) + \$120 (1/3) + \$0 (1/3)]. On average, you will make\$140 per game if you play the game for a long, long time. p. 4
5. 5. Example:In the particular game, a coin is tossed. If the coin comes up heads, the playerwins \$100. If the coin comes up tails, the player loses \$50. What is theexpected value of the game? X (Dollar Value) P(X) X·P(X) \$100 1/2 \$50 -\$50 1/2 -\$25 \$25The expected value of this game is \$25. Over the long term, this game is worth\$25 per toss. If you play this game many, many times (say, 1,000 times) on theaverage you can expect to make \$25 per toss.Out of, say, 100 tosses, you would expect to win \$100 50 times and to lose \$50fifty times. Thus, you will make \$2500. This works out to an average winningof \$2500 / 100 = \$25.Don’t pay more than \$25 to play. p. 5
6. 6. Example:In the following game, there is a one in 4 chance of winning \$80; a one in 4chance of losing \$100; and a one-half chance of coming out even. How muchwould you be willing to pay to play?Vi (Dollar Value) P(Vi)-\$100 1/40 1/2+\$80 1/4E(V) = -\$5 [-\$100 (1/4) + \$0 (1/2) + \$80 (1/4)] p. 6
7. 7. Example: a lottery ticketHow much would you be willing to pay for a lottery ticket with a one in 5,000chance of winning \$1 million dollars, and a 4 in 5,000,000 chance of winning\$100,000?X P(X) X·P(X)\$1,000,00 1 .20 5,000,0000\$100,000 4 .08 5,000,000\$0 4,999,995 0 5,000,000 \$0.28Answer: Don’t pay more than 28 cents! p. 7
8. 8. Example:Would you be willing to pay \$9 for a lottery that gives you one chance in amillion of making \$5,000,000?Vi (Dollar Value) P(Vi)\$5,000,000 .000001\$0 .999999The expected value of the above lottery is \$5.00. Mathematically, it does notmake sense to spend \$9 for something that has an expected value of only \$5.00.Of course, people do not think this way. Many will spend the \$9 or even morefor a chance to make \$5 million. Utility theory is used to explain why peopleact in this seemingly irrational manner. This is beyond the scope of this course. p. 8
9. 9. Continuous Probability DistributionsCalled a Probability density function. The probability is interpreted as"area under the curve."1) The random variable takes on an infinite # of values within a giveninterval2) the probability that X = any particular value is 0. Consequently, we talkabout intervals. The probability is = to the area under the curve.3) The area under the whole curve = 1.Some continuous probability distributions: Normal distribution, StandardNormal (Z) distribution, Students t distribution, Chi-square ( χ2 )distribution, F distribution. p. 9
10. 10. THE NORMAL DISTRIBUTIONThe probability density function for the normal distribution:f(X), the height of the curve, represents the relative frequency at which thecorresponding values occur.There are 2 parameters: μ for location and σ for shape.Probabilities are obtained by getting the area under the curve inside of aparticular interval. The area under the curve = the proportion of times underidentical (repeated) conditions that a particular range of values will occur.The total area under the curve = 1.Characteristics of the Normal distribution:1. it is symmetric about the mean μ.2. mean = median = mode. [“bell-shaped” curve]3. f(X) decreases as X gets farther and farther away from the mean. Itapproaches horizontal axis asymptotically: -∞<X<+∞This means that there is always some probability (area) for extreme values. p. 10
11. 11. Since there are 2 parameters – μ for location and σ for shape. – This means thatthere are an infinite number of normal curves – even with the same mean.Curves A and B are both normal distributions. They have the same mean butdifferent standard deviations. μ=30 μ=80 μ=120 σ=5 σ = 10 σ=2 p. 11
12. 12. However, there is only ONE Standard Normal Distribution. This distributionhas a mean of 0 and a standard deviation of 1. μ=0 σ=1Any normal distribution can be converted into a standard normal distribution bytransforming the normal random variable into the standard normal r.v.: X −µ Z= σThis is called standardizing the data. It will result in (transformed) data with μ= 0 and σ = 1.The Standard Normal Distribution (Z) is tabled:Please note that you may find different tables for the Z-distribution. The tablewe prefer (below), gives you the area from 0 to Z. Some books provide aslightly different table, one that gives you the area in the tail. If you check thediagram that is usually shown above the table, you can determine which tableyou have. In the table below, the area from 0 to Z is shaded so you know thatyou are getting the area from 0 to Z. Also, note that table value can never bemore than .5000. The area from 0 to infinity is .5000.The Normal Distribution is also referred to as the Gaussian Distribution,especially in the field of physics. In the social sciences, it is sometimes calledthe bell curve because of the way it looks (lucky for us it does not look like achicken). p. 12
13. 13. THE STANDARDIZED NORMAL (Z) DISTRIBUTIONEntry represents area under the standardized normal distribution from the mean to ZZ .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .03590.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .07530.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .11410.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .15170.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .18790.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .22240.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .25490.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .28520.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .31330.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .33891.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .36211.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .38301.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .40151.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .41771.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .43191.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .44411.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .45451.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .46331.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .47061.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .47672.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .48172.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .48572.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .48902.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .49162.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .49362.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .49522.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .49642.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .49742.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .49812.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .49863.0 .49865 .49869 .49874 .49878 .49882 .49886 .49889 .49893 .49897 .499003.1 .49903 .49906 .49910 .49913 .49916 .49918 .49921 .49924 .49926 .499293.2 .49931 .49934 .49936 .49938 .49940 .49942 .49944 .49946 .49948 .499503.3 .49952 .49953 .49955 .49957 .49958 .49960 .49961 .49962 .49964 .499653.4 .49966 .49968 .49969 .49970 .49971 .49972 .49973 .49974 .49975 .499763.5 .49977 .49978 .49978 .49979 .49980 .49981 .49981 .49982 .49983 .499833.6 .49984 .49985 .49985 .49986 .49986 .49987 .49987 .49988 .49988 .499893.7 .49989 .49990 .49990 .49990 .49991 .49991 .49992 .49992 .49992 .499923.8 .49993 .49993 .49993 .49994 .49994 .49994 .49994 .49995 .49995 .499953.9 .49995 .49995 .49996 .49996 .49996 .49996 .49996 .49996 .49997 .49997 p. 13
14. 14. REMEMBER THESE PROBABILITIES (percentages): # s.d. from the mean approx area under the normal curve ±1 .68 ±1.645 .90 ±1.96 .95 ±2 .955 ±2.575 .99 ±3 .997 p. 14
15. 15. USING THE NORMAL DISTRIBUTION TABLEExample:If the weight of males is N.D. with μ=150 and σ=10, what is the probability thata male will weight between 140 lbs and 155 lbs? [Important Note: The probability that X is equal to any one particular value is zero – P(X=value) = 0 since the N.D. is continuous.] 140 −150 − 10Z= 10 = 10 = -1 s.d. from mean Area under the curve = .3413 (from Z table) 155 −150 5Z= 10 = 10 = .5 s.d. from mean Area under the curve = .1915 (from Z table)Answer: .3413 .1915 .5328  p. 15
16. 16. Example:If IQ is ND with a mean of 100 and a s.d. of 10, what percentage of thepopulation will have(a) IQs ranging from 90 to 110?(b) IQs ranging from 80 to 120?(a) Z = (90 – 100) / 10 = -1 area = .3413 Z = (110 – 100) / 10 = +1 area = .3413 .6826 Answer: 68.26% of the population(b) Z = (80 – 100) / 10 = -2 area = .4772 Z = (120 – 100) / 10 = +2 area = .4772 .9544 Answer: 95.44% of the populationExample:Suppose that the average salary of college graduates is N.D. with μ=\$40,000and σ=\$10,000.(a) What proportion of college graduates will earn less than \$24,800? Z = (\$24,800 – \$40,000) / \$10,000 = −1.52 area = .06436.43% of college graduates will earn less than \$24,800(b) What proportion of college graduates will earn more than \$53,500? Z = (\$53,500 – \$40,000) / \$10,000 = +1.35 area = .08858.85% of college graduates will earn more than \$53,500 p. 16
17. 17. (c) What proportion of college graduates will earn between \$45,000 and\$57,000? Z = (\$57,000 – \$40,000) / \$10,000 = +1.70 area = .4554 Z = (\$45,000 – \$40,000) / \$10,000 = +0.50 area = .1915 Answer: .4554 − .1915 = .2639(d) Calculate the 80th percentile. Find the area that corresponds to an area of .3000 from 0 to Z (this meansthat there will be .2000 in the tail). A Z value of + 0.84 corresponds to the 80thpercentile. +.84 = (X − \$40,000) / \$10,000 X = \$40,000 + \$8,400 = \$48,400.  [Incidentally, the 20th percentile would be \$40,000 − \$8,400 = \$31,600](e) Calculate the 27th percentile. Find the area that corresponds to an area of .2300 from 0 to Z (this meansthat there will be .2700 in the tail). A Z value of − 0.61 corresponds to the 27thpercentile. − 0.61 = (X − \$40,000)/ \$10,000 X = \$40,000 − \$6,100 = \$33,900 p. 17
18. 18. Exercise:The GPA of college students is ND with μ=2.70 and σ=0.25.(a) What proportion of students have a GPA between 2.40 and 2.50?(b) Calculate the 97.5th percentile.[97.5% of college students have a GPA below _______?](c) Calculate the 10th Percentile.[90% of students will have higher GPAs.]Answers:The Z-value for the 2.40 GPA converts to -1.20 [ (2.40 – 2.70) / .25 ];The Z-value for the 2.50 GPA converts to -.80 [ (2.50 – 2.70) / .25 ];The area from 0 to -1.20 is .3849The area from 0 to - .80 is .2881Answer is .3849 - .2881 = .0968 or 9.68% of college students(b) A z-score of 1.96 is equal to the 97.5th percentile (.5000 + .4750).Thus, 1.96 = (X – 2.70) / .25Solve for X. X = .49 + 2.70 = 3.19 Answer = A GPA of 3.19 is the 97.5thpercentile.(c) A Z score of -1.28 is approximately the 10th percentile. Find the area that corresponds to an area on the left side (negative) of theZ-distribution of .4000 from 0 to Z (this means that there will be .1000 in thetail). A Z value of -1.28 corresponds to the 10th percentile. A Z-score of + 1.28is approximately the 90th percentile (actually it is .50 + .3997). Thus, - 1.28 = (X – 2.70) / .25 Solve for X. X = 2.70 - .32 = 2.38 Answer = GPA of 2.38 is the 10thpercentile. p. 18
19. 19. Exercise:Chains have a mean breaking strength of 200 lbs, σ=20 lbs.(a) What proportion of chains will have a breaking strength below 180 lbs?(b) 99% of chains have breaking points below _________? [99th percentile]Hint: 50% have breaking points below 200 lbs which is equal to the populationmean. The answer has to be more than 200 lbs. We are on the right side of theZ distribution.Answers:Z = (180 – 200) / 20 = -1.00 The area that is between -1.000 and 0 in the Zdistribution is .3413. We want the left tail below the -1.000. The entire area tothe left of the 0 in the Z-distribution is .5000. Thus, (a) .5000 - .3413 = .1587 Answer is 15.87% (b) The value of + 2.33 corresponds to the 99th percentile .5000 + .4901 = . 9901. That is close enough for our purposes.2.33 = ( X – 200) / 20X = 246.60 pounds. p. 19