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Statistik topic8 special probability distribution

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Statistik topic8 special probability distribution

1. 1. Topic 8 Special Probability Distribution LEARNING OUTCOMES By the end of this topic, you should be able to: 1. use binomial distribution to solve binomial problem; 2. apply table of binomial distribution to calculate exact probability in binomial experiment; 3. formulate normal distribution to standard z-score; and 4. apply table of standard normal distribution to calculate probability. INTRODUCTIONIn the last topic, we have learnt about probability function p(x) for discretedistribution, and f(x) for continuous distribution. We mentioned there that p(x)and f(x) is of any algebraic expressions which comply with the respectiveprobability rules mentioned in last topic. However, in this topic we will discuss aspecial expression of p(x) for binomial distribution and a special expression off(x) for normal distribution. Binomial distribution is an example of discretedistribution and normal distribution is a continuous distribution.8.1 BINOMIAL DISTRIBUTION The binomial distribution is one of the most commonly used discrete probability distribution. It is used to obtain the exact probability of X successes in n repeated trials of a binomial experiment
2. 2. 124 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONThe trial in this experiment has only two outcomes which complementing eachother. As an example, a trial of tossing of a Malaysian coin might land picture (G)or number (N). A student taking a final examination will either pass or fail.Another feature of trial is that its results can be reduced to two outcomes whichcomplement each other. For example, we are interested of getting even number inthe trial of throwing a fair dice. The results can be reduced to getting event E (2 or4 or 6) as one outcome, and event O (1, or, 3 or 5) as another outcome. Theseevents E and O complement to each other. A trial which possess ONLY two outcomes one complementing each other is categorised as Bernoulli trial. (i) Give two examples of Bernoulli trial which has ONLY two outcomes. For each trial, explain its outcomes and state how they are complementing each other. (ii) Give one example of Bernoulli trial which has two reduced outcomes. For each trial, explain its outcomes and state how they are complementing each other.8.1.1 Binomial ExperimentBinomial experiment is a probability experiment consisting of repetition ofBernoulli trial. Conducting a urine test on 50 students is an example of binomialexperiment. In this experiment, the Bernoulli trial is the process of giving urinetest on each student whose outcome is “positive” or “negative” result. Thebinomial experiment here is the repetition of the urine test 50 times.The experiment of throwing fair dice 30 times where the interest of researcher isof getting even numbers on each throw is another example of binomialexperiment. The Bernoulli trial here is “throwing the dice” whose outcomes canbe reduced to getting event E i.e. {2,4,6} as one outcome and event O i.e. {1,3,5}as another outcome. This trial is repeated 30 times.
3. 3. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 125Binomial RequirementsThe binomial experiment should comply with the following requirements: Each trial should have only two outcomes or outcomes that can be reduced to two outcomes. These outcomes which complement to each other can be considered as either success or failure. This trial is categorised as Bernoulli trial. There must be a fixed number of repetitions, n of such trial. The outcome of each trial must be independent of each other.Sometimes, researcher tend to represent event success by using code “1” andevent failure by code “0” The Pr(Success) = p, and Pr(Failure) = q = 1 – p. Forexample in one of the above examples, let event E, getting even number becomethe success, then p = Pr(E) = 0.5, and q = 0.5. Let digit ‘1’ represent success in aBernoulli trial which has been repeated 10 times. For each trial, when eventsuccess occurs, a digit 1 is recorded, otherwise digit 0 is recorded. Then thefollowing are example outcome of binomial experiment. It consists of strings ofoutcomes of repeated Bernoulli trial.(a) 1010011101, this outcome of binomial experiment produces 6 successes, and 4 failures; with probability Pr(1010011101) = Pr(6 successes and 4 failures) = Pr(6 successes ) Pr(4 failures) = p6 q4(b) 0111001010, this outcome of binomial experiment produces 5 successes, and 5 failures, with probability Pr(1010011101) = Pr(5 successes and 5 failures) = Pr(5 successes ) Pr(5 failures) = p5 q5However, for any x number of successes in n repetition Bernoulli trial, there are n , or n choose x possible outcomes of binomial experiment each has a xcombination of success and failures as given in (a) and (b) above.It is not impossible to have all non-successful in a binomial experiment, which isrepresented by the string 0000000000 that consist of 10 failures. Likewise, it also
4. 4. 126 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONpossible to get full string of 10 successes i.e. 1111111111. Thus, the number ofsuccess in a binomial experiment can range from zero to a maximum of nsuccesses. The outcomes of binomial experiment and their correspondingprobabilities generate binomial distribution. ACTIVITY 8.1 1. Give two examples of binomial experiment; one for trial having two outcomes and one example for trial with reduced outcomes. 2. For each binomial trial, give samples of outcomes and its possible probability in term of p and q.8.1.2 Binomial Probability FunctionLet Y be a discrete random variable representing the total number of success in abinomial experiment with n repetition of Bernoulli trial. Then the probability of ysuccesses in n repetitions of Bernoulli trial is given by: n p y (1–p)n-y, y = 0,1, ..., n; P (y) y 0 others Formula 8.1Where n n! , and n! = n (n–1) (n–2) … 2 1 y y! (n-y)! Formula 8.1(a)Then we say that Y follow binomial distribution written as Y b(y; n, p).The probability of getting y successes in n repetition of Bernoulli trial can beobtained via Formula 8.1 or using table of binomial cumulative probabilitydistribution.
5. 5. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 127Example 8.1Consider an experiment of throwing unbiased dice 10 times. Find the probabilityof obtaining even numbers six times.SolutionThe interested event in this experiment is: E = even numbers = {2,4,6},The trial is: throwing an unbiased diceTrial outcomes: {E, O}, a Bernoulli trial with Pr(Success) = p = Pr(E) = 0.5. HereO stands for odd numbers.This trial is repeated 10 times to generate binomial experiment.Total umber of repetition = n = 10;Let Y be the number of successes in the experiment = Y b(y; 10, 0.5).Pr(getting 6 successes) is given by Using Formula 8.1 as 10 10! Pr(Y = 6) = p(6) = (0.5)6 (0.5) 4 (0.5)6 (0.5) 4 = 0.20508 0.2 6 6! 4!About 20% of ten throws will result in even numbers.The Mean and Variance of Binomial DistributionThe mean of binomial distribution is given by: n p Formula 8.2The variance is given by: 2 np (1 p) Formula 8.38.1.3 Probability Table of Binomial DistributionWe are referring to the Table of Binomial Distribution published by OUM. In thistable Y is referred as binomial discrete random variable. The number of success isrepresented by r, and Pr(success) = . For each n, the column of r is ranging
6. 6. 128 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONfrom 0 to n. The content of the table is accumulating probabilities of getting 0until r successes written in probability language as Pr(Y r). It is given by thefollowing formula: r P (r) = Pr (Y r) = p (y) y=0 Formula 8.4For example, Let Y is from binomial distribution with n = 10, p = = 0.2. Findthe probability of the following events:(a) Obtain 3 or less successes.(b) Obtain more than 3 successes.(c) Obtain exactly 4 successes.Refer to page 5 of the table published by OUM, look at the block on column n =10 (See Table 8.1) below.(a) Pr(Y 3) = 0.8791. Look at row r = 3, and column = 0.2.(b) By complement, Pr( Y > 3) = 1 - Pr(Y 3) = 1 – 0.8791 = 0.1209(c) Pr(r = 4) = Pr(Y 4) - Pr(Y 3) = 0.9672 – 0.8791 = 0.0881(The answer obtained by Formula 8.1 is 0.08808). Table 8.1: Binomial Distribution n r 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 10 0 0.5987 0.3487 0.1969 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010 1 0.9139 0.7361 0.5443 0.3758 0.2440 0.1493 0.0860 0.0464 0.0233 0.0107 2 0.9885 0.9298 0.8202 0.6778 0.5256 0.3828 0.2616 0.1673 0.0996 0.0547 3 0.9990 0.9872 0.9500 0.8791 0.7759 0.6496 0.5138 0.3823 0.2660 0.1719 4 0.9999 0.9984 0.9901 0.9672 0.9219 0.8497 0.7515 0.6331 0.5044 0.3770 5 1. 0.9999 0.9986 0.9936 0.9803 0.9527 0.9051 0.8338 0.7384 0.6230 6 1. 1. 0.9999 0.9991 0.9965 0.9894 0.9740 0.9452 0.8980 0.8281 7 1. 1. 1. 0.9999 0.9996 0.9984 0.9952 0.9877 0.9726 0.9453 8 1. 1. 1. 1. 1. 0.9999 0.9995 0.9983 0.9955 0.9893 9 1. 1. 1. 1. 1. 1. 1. 0.9999 0.9997 0.9990 11 0 0.5688 0.3138 0.1673 0.0859 0.0422 0.0198 0.0088 0.0036 0.0014 0.0005 1 0.8981 0.6974 0.4922 0.3221 0.1971 0.1130 0.0606 0.0302 0.0139 0.0059 2 0.9848 0.9104 0.7788 0.6174 0.4552 0.3127 0.2001 0.1189 0.0652 0.0327
7. 7. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 129Probabilities RULES using Table of Discrete DistributionIn general, we have the following properties in getting probabilities using table:Rule 5 Pr (Y < r) = Pr (Y r –1) = P (r –1)Rule 6 P(r) = Pr (Y = r) = Pr (Y r) = P (r –1) – Pr (Y r–1) = P(r) – P (r –1)Rule 7 Pr (Y > r) = 1 – Pr (Y r) = 1 – P(r)Rule 8 Pr (Y r) = 1 –Pr (Y < r) = 1 –Pr (Y r –1) = 1 –P(r –1)In this table Greek letter (red theta) has been used for the probability of successwhich ranges from 0 to 0.5.Example 8.2Let Y is from binomial distribution with n = 6, p = = 0.2. Find the probabilityof the following events:(a) Obtain 2 or less successes(b) Obtain less than 2 successes(c) Obtain exactly 2 successes.
8. 8. 130 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONSolutionRefer to page 5 of the binomial table published by OUM, with p = = 0.35, n = 6.(a) Pr (Y 2) = 0.6471, Refer at block n = 6, row r = 2; column = 0.35.(b) By Rule 5, Pr( r< 2) = Pr(r 1), since r discrete; Pr (r< 2) = Pr(r 1) = 0.3191. Refer at block n = 6, row r = 1, column = 0.35.(c) By Rule 6; Pr( r = 2) = Pr ( r 2) – Pr ( r 1) = 0.6471 – 0.3191 = 0.3280.Using Table by Complementary Method when > 0.5The table in the book is prepared for probability of success 0.5 or less. In practice,we may have probability of success greater than 0.5. For > 0.5, acomplementary method has to be used as follows:When r n, and > 0.5, then the followings are true:(a) Event {Y = r successes} is equivalent to Event {X = (n – r) failures} Where Y b(y; n, ), and X b(x; n, 1- ). Thus, Pr(Y = r) Pr(X = n - r), Y success, X failures Formula 8.5 For example, let Y b(y; 6, = 0.8), and find Pr(Y = 4). Define X b(x; 6, 1- = 0.2), and X is number of failures. Then from table page 5, and Rule 6, we have: Pr(Y = 4) Pr(X = 6 - 4) = Pr (X = 2) = Pr(r 2) – Pr(r 1) = 0.9011 – 0.6554 = 0.2457(b) Event {Y r successes} is equivalent to Event {X (n – r ) failures} where Y b(y; 6, ), and X b(x; 6, 1- ). Thus, Pr(Y r) Pr{X (n – r )}= 1- Pr( X < n – r) = 1- Pr( X n – r -1) for discrete X. Formula 8.6
9. 9. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 131For example, let Y b(y; 6, = 0.8), and find Pr(Y 2 ).Define X b(x; 6, 1- = 0.2). Pr(Y 2) Pr{X (6 – 2)} = 1- Pr( X < 6 – 2) = 1- Pr( X 6 – 2 -1) = 1 - Pr(X 3) = 1 - 0.9830 = 0.0170 (By the exact Formula 8.1, the answer is 0.01696).Example 8.3Let Y be a binomial distribution with n = 6, p = = 0.6. Find the probability ofobtaining 3 or less successes.SolutionDefine X b(x; 6, 1- = 0.4).By using Formula 8.6, we have Pr(Y 3) Pr{X (6 – 3)} = 1- Pr(X < 6 – 3) = 1- Pr(X 6 – 3 - 1) = 1- 0.5443 = 0.4557.
10. 10. 132 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION ACTIVITY 8.2 1. Compute the probability of Y successes using exact probability Formula 8.1. (a) n = 5, y = 3; p = 0.1 (b) n = 8, y = 5; p = 0.4 (c) n = 7, y = 4; p = 0.7 2. Compute the probability of Y successes in Question 1 by using table of binomial distribution published by OUM. Make a comparison of their values. 3. It is found that 40% of the first year students are using learner study system in one semester. Find the probability in a sample of 10 students, exactly 5 of which use learner study system. 4. Given that Y~ b(y; 4, 0.4) by using table of binomial distribution, find the probabilities of the following events: (a) (Y< 2), (c) (Y > 2) (b) (Y = 2), (d) (Y 2) 5. Given that Y~ b(y; 4, 0.65) by using table of binomial distribution, find the probabilities of the following events: (a) (Y< 2), (b) (Y = 2) (c) (Y 2) 6. A student answers 8 questions of MCQ type. Each question has 5 answers with only 1 correct answer. Compute the probability the student obtaining 4 correct questions. 7. It is known that only 60% of a defected computer can be repaired. A sample of 8 computers is selected randomly, find the probability of (a) At most 3 computers can be repaired. (b) 5 or less can be repaired (c) None can be repaired
11. 11. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 133 8.2 NORMAL DISTRIBUTIONIt is a continuous distribution having the following properties:(a) Its probability curve is bell-shaped.(b) Its mean, mode, and median are equal and located at the centre of the distribution.(c) It is a unimodal distribution.(d) The curve is symmetrical about the mean. Thus it has same shape on both sides of a vertical line drawn through the centre.(e) The curve never touches the horizontal axis or mathematically it touches at infinity.(f) The total area under the curve is equal to 1.0.(g) The area under the normal curve that lies within one standard deviation of the mean is 0.68 (or 68%); within two standard deviations of the mean is 0.95 (or 95%); within three standard deviations of the mean is 0.997 (or 99.7%). See Figure 8.1(h) The normal distribution has two parameters known as its mean and 2 variance .(i) Continuous random variable X from normal distribution with mean and 2 variance is denoted by X N( , 2 ). Figure 8.1: Areas under the normal distribution curve
12. 12. 134 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 2(j) Figure 8.2, shows some other properties. Let X N ( 1, 1 ), and X 2 N( 2 , 2 ). If 2 greater than 1 , then normal curve 2 is located on the 2 2 right of the normal curve 1. If 1 less than 2 , then normal curve 2 is more spread than normal curve 1. Figure 8.2: Some other properties, comparison of parameters
13. 13. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 135 ACTIVITY 8.3 Using the same X-Y axes, sketch the following normal curves: N(10, 2), N(10,4), N(10, 16), N(20, 2), N(20, 9)8.2.1 Standard Normal DistributionLet X be a continuous random variable from a normal distribution N( , 2), thenX can be transformed to Z score of standard normal distribution by the followingFormula: X Z Formula 8.7The random score Z is said to have standard normal distribution with mean 0 anda known variance 1. Standard normal curve preserves the same normal properties.Figure 8.3 depicts the above transformation. Figure 8.3: Transformation of X to Z score
14. 14. 136 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONTable of Standard Normal Curve N(0, 1)With the above transformation, almost all probability problems can be solvedeasily by using standard normal table. In the table published by OUM, standardnormal is given on page 23. It is developed based on cumulative distribution givenin Formula 8.8. z ( z) Pr( Z z) Pr( Z z) (u )du Formula 8.8A portion of this table is shown in Table 8.2 where (u ) is standard normaldensity function and (z ) is the area on the left of z as shown in Figure 8.4. Thisformula is for positive z. Figure 8.4: The area (z)For example let random variable X come from normal distribution with mean 20and variance 16. Find the Pr(X < 22). It is the area to the left of Z = (22-20)/4 =0.5. So from the table on page 23 (see Table 8.1), look at row 0.50 and undercolumn 0.00, the answer is 0.69146.
15. 15. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 137 Table 8.2 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.00 0.50000 0.50399 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53188 0.53586 0.10 0.53983 0.54380 0.54776 0.55172 0.55567 0.55962 0.56356 0.56749 0.57142 0.57535 0.20 0.57926 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61026 0.61409 0.30 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173 0.40 0.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.68793 0.50 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240 0.60 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490 0.70 0.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524 0.80 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327 0.90 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891 1.00 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 0.86214 1.10 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.88298 1.20 0.88493 0.88686 0.88877 0.89065 0.89251 0.89435 0.89617 0.89796 0.88730 0.90147 1.30 0.90320 0.90490 0.90658 0.90824 0.90988 0.91149 0.91308 0.91466 0.91621 0.91774 1.40 0.91924 0.92073 0.92220 0.92364 0.92507 0.92647 0.92785 0.92922 0.93056 0.93189 1.50 0.93319 0.93448 0.93574 0.93699 0.93822 0.93943 0.94062 0.94179 0.94295 0.94408 1.60 0.94520 0.94630 0.94738 0.94845 0.94950 0.95053 0.95154 0.95352 0.95352 0.95449 1.70 0.95543 0.95637 0.95728 0.95818 0.95907 0.95994 0.96080 0.96246 0.96246 0.96327 1.80 0.96407 0.96485 0.96562 0.96637 0.96712 0.96784 0.96856 0.96995 0.96995 0.97062 1.90 0.97128 0.97193 0.97257 0.97320 0.97381 0.97441 0.97500 0.97558 0.97615 0.97670Area on the Left of Negative zThe area on the left of negative z can be obtained by symmetry property and usingcomplement method as follows. Pr (Z < – z) = Pr (Z > z) = 1 – (z) Formula 8.9For example let random variable X is coming from normal distribution with mean20, and variance 16. Find the Pr(X < 18). It is the area to the left of Z = (18-20)/4= -0.5. So from Formula 8.9, it is equivalent to the area on the right of Z = 0.5,and by complement the area is 1 – 0.69146 = 0.30854.Area under the Normal Curve Lying between Two Values of z Pr( a < Z < b) = (b) - (a) Formula 8.10
16. 16. 138 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONIt is the shaded area given in Figure 8.5. Figure 8.5: Area between two values of zExample 8.4Using standard normal table, find the Pr(a<Z<b) for the following values of a andb.(i) a =1.5, b = 2.55(ii) a = –2.0, b = –1.5(iii) a = –1.5, b = 1.5SolutionFigure 8.6 depicts the necessary shaded area and from table we have: (1.50) = 0.93319, (2.00) = 0.97725, (2.55) = 0.99461From Formula 8.10, we have:(i) Pr(1.50< Z<2.55) = (2.55) – (1.50) = 0.06142
17. 17. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 139(ii) Pr (–2.00 < Z < –1.50) = (–1.50) - (–2.00), = {1– (1.50)} – {1- (2.00)} = 0.06681 – 0.02275 = 0.04406(iii) Pr (–1.50 < Z < 1.50) = (1.50) – (–1.50), = (1.50) – (1- 1.50) = 0.93319 – 0.06681 = 0.86638Example 8.5Let a continuous random variable X follows normal distribution N(4.0, 16). Find(a) Pr(X < 2).(b) Pr(X > 4.6).SolutionUse Formula 8.7 to get the corresponding z score. Then use Formula 8.8, 8.9,8.10. X 4 2 4(a) Pr( X 2) Pr = Pr( Z 0.5) =1 - (0.5) 16 4 = 1 – 0.69146 = 0.380854 X 4 4.6 4(b) Pr( X 4.6) Pr = Pr( Z 0.15) =1 - (0.15) 16 4 = 1 – 0.55962 = 0.44038
18. 18. 140 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTIONExample 8.6The long-distance calls made by executives of a private university are normallydistributed with a mean 10 minutes and a standard deviation of 2.0 minutes. Findthe probability that a call:(a) lasts between 8 and 13 minutes(b) lasts more than 9.0 minutes(c) lasts less than 7 minutes.SolutionLet random variable X represents duration of long-distance call and X N(10, 4). 8 10 X 10 13 10(a) Pr(8 X 13) Pr Pr 1.0 Z 1 .5 2 2 2 = (1.5) – (-1.0) = (1.5) – {1 - (1.0)} = (1.5) + (1.0) – 1 = 0.93319 + 0.84134 – 1 = 0.77453 9 10(b) Pr( X 9.0) Pr Z Pr Z 0.5 2 = (0.5) = 0.69146 7 10(c) Pr( X 7.0) Pr Z Pr Z 1.5 2 =1– (1.5) = 1 – 0.93319 = 0.06681
19. 19. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 141 ACTIVITY 8.4 1. Let a continuous random variable X follows normal distribution N(4.0, 16). Find Pr(2 < X < 3) 2. Let a continuous random variable X follow normal distribution N(50.0, 4). Find (a) Pr( X < 45) (b) Pr(X>55) (c) Pr(45.0 < X < 55.0) 3. A continuous random variable X is normally distributed with mean 50 and standard deviation 5. What value of K is such that Pr(X < K) = 0.08.8.2.2 Application to Real ProblemsBinomial and normal distributions are widely used in solving various day to dayproblems. If sample size is large, normal distribution can be used to replacebinomial distribution in solving binomial problems. However, some numericalcorrections have to be done.Correction for Conversion Binomial Discrete Variable to Normal ContinuousVariableLet Y be discrete random variable from a binomial distribution b(y: n, p) withmean np and variance np(1-p). Then, the corresponding continuous X followsnormal distribution with mean np and variance np(1-p). The following arenecessary corrections when shifting discrete Y to continuous X. The normaldistribution will give a good approximation to binomial, if n is large and np > 5;also np(1-p)> 5. The approximation is recommended because when n is large thenumerical calculation using binomial function become tedious.
20. 20. 142 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION Table 8.3 Corrections in Continuous Variable X, y is theDiscrete Variable Y; y is Observed Value Same Observed Value for Discrete Y (a) Y=y y - 0.5 < X < y + 0.5 (b) Y y X y + 0.5 (c) Y<y X < y – 0.5 (d) Y y X y – 0.5 (e) Y>y X > y + 0.5Correspondingly, we have the equivalent probability statement in the followingTable 8.4. Table 8.4 X N(np, np(1-p)); y is the Same Y b(y;n, p); y is Observed Value Observation Value for Discrete Y (a) Pr(Y = y) Pr(y - 0.5 < X < y + 0.5) (b) Pr(Y y) Pr(X y + 0.5) (c) Pr(Y < y) Pr(X < y – 0.5) (d) Pr(Y y) Pr(X y – 0.5) (e) Pr(Y > y) Pr(X > y + 0.5)Example 8.7Let Y from binomial distribution b(y; 100, 0.4). Find(a) Pr(Y > 50).(b) Pr(50< Y < 60)SolutionY is binomial with n = 100; mean = np = 40, variance = np(1-p) = 24.Define continuous variable X from normal N(40, 24). The observation value is y= 50.
21. 21. TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION 143(a) From conversion Table 8.4(e), we have Pr(Y > 50) Pr(X > 50 + 0.5) = Pr(X > 50.5) = 1 – Pr(X 50.5) 50.5 40 = 1 – Pr(X )=1– (2.143) 24 =1- (2.14) 1 – 0.98382 = 0.01618.(b) In this problem, Y is strictly greater than 50 and strictly less than 60, so use conversion Table 8.3 (c) & (e). Thus we have, Pr( 50 < Y < 60) Pr( 50 + 0.5 < X < 60 – 0.5) 59.5 40 50.5 40 = - = (3.98) – (2.14) 24 24 = 0.99997 – 0.98382 = 0.01615Correction Due to Rounding Error in ObservationSometimes the observations are recorded after being round up or round down tothe nearest integer, or figure. Thus correction is recommended as in the followingexample.Example 8.8Let the lifetime of electric bulb follows normal distribution with mean 1000 hoursand standard deviation 100 hours. The observation of the lifetime is recorded tothe nearest hours. An electric bulb is selected at random, find the probability thatit will have lifetime between 850 hours and 1050 hours.SolutionLet X be the lifetime of the electric bulb, then X N(1000, 1002 ). 849.5 1,000 1,050.5 1,000 Pr(849.5 X 1,050.5) Pr Z 100 100 (0.505) ( 1.505) = 0.69497 + 0.93448 – 1 = 0.62945
22. 22. 144 TOPIC 8 SPECIAL PROBABILITY DISTRIBUTION ACTIVITY 8.5 1. The length of yard stick is said to follow normal distribution with mean 150mm, and standard deviation 10mm. The length is recorded to the nearest integer. A yard stick is taken at random: (a) find the probability its length falls in the following intervals: (i) between 125mm and 155mm; (ii) more than 180mm. (b) If 500 such yard stick has been selected randomly, find the number of them who has lengths in the respective intervals given in (a).Two types of distributions have been introduced. A binomial distribution is ofdiscrete type and normal distribution of continuous type. It has been clearlydescribed the difference between these two types of distributions. The binomialdistribution is concerning exact probability whose values can be obtained throughFormula 8.1 or by using table of binomial distribution. For the normaldistribution, probability value can best be obtained by using table of standardnormal. As such, all normal problems can be transformed to standard normal viaFormula 8.7 and then the standard normal table can be used to determine theprobability. At the end of the topic, approximation to binomial by normaldistribution has been introduced. However, some correction has to be done whenconverting discrete variable to continuous variable as given in Table 8.3 & 8.4.