146 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Determine (a) the value α, (b) the probability density function, (c) the expected value, µ, of X, (d) the standard deviation, σ, of X, (e) P(| X – µ | . 1 3 ). 9. The continuous random variable X has probability density function 4k, 0 x 1, f(x) = k(x – 3)2 , 1 x 3, 0, otherwise. (a) Find the value of constant k. (b) Show that E(X) = 27 22 . (c) Find the value of c if E(X + c) = 3E(X – c). 10. The continuous random variable X has probability density function 4 3 x, 0 x 1, f(x) = 4 3 x–5, x 1. Show that E(X) = 8 9 , and find Var(X). Hence, find E(3 – X) and Var(3 – X). 11. The continuous random variable X takes values in the interval [0, 1] and has probability density function 3.75x + 0.1, 0 x 0.4, 1.6, 0.4 x 0.6, f(x) = 3.85 – 3.75x, 0.6 x 1, 0, otherwise. (a) Sketch the graph f and state the mean of X. (b) Find Var(X). The random variable Y is defined by Y = 2X + 1. Find E(Y) and Var(Y). 12. The continuous random variable X has probability density function 1 18 (x + 3), –3 x 3, f(x) = 0, otherwise. (a) Find E(X) and Var(X). (b) Assume that Y = aX + b, with a and b as constants and a 0. Find the values a and b if E(Y) = 0 and Var(Y) = 1. 13. X is a continuous random variable with p.d.f. 1 8 , –5 x 3, f(x) = 0, otherwise. Find (a) the mean, µ, (b) the variance, σ2 , (c) P(X 1).
147 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 14. The continuous random variable Y follows a uniform distribution on the interval [1, 8]. Find (a) the mean and variance of Y, (b) P(2 Y 5), (c) P(Y 3). Hence, sketch the graph of the probability density function of Y. 15. The continuous random variable X has probability density function f with its graph as shown on the right. Find (a) the value of b, (b) E(X), (c) Var(X), (d) P(2.7 X 5.1) 16. The random variable Z is uniformly distributed on the interval [m, 2]. If P(Z –1) = 1 7 , find (a) the value of m, (b) P(Z 0), (c) P(– 1 2 Z 1 2 ), (d) E(Z). 17. The continuous random variable X has probability density function 1 2π , –π x π, f(x) = 0, otherwise. (a) Find the mean and standard deviation of X. (b) Find the probability that X lies within one standard deviation from the mean. 18. The continuous random variable X has probability density function 1 d – c , c x d, f(x) = 0, other values of x. (a) Show that E(X) = c + d 2 and Var(X) = (d – c) 2 12 . (b) If the mean is 1 and variance is 4 3 , find (i) the values of c and d, (ii) P(X 0), (iii) the value of t such that P(X > t + σx ) = 1 4 with σx as the standard deviation of X. 19. A continuous random variable X has p.d.f as follows. 4e–4x , x 0, f(x) = 0, otherwise. Find (a) P(X 0.6), (b) P(0.1 X 0.3), (c) E(X), (d) P[X E(X)], (e) the median of X. 20. The lifespan, in thousand hours, of a ‘Super bright’ bulb has an exponential distribution with p.d.f. 0.2e–0.2x , x 0, f(x) = 0, otherwise. f(x) x 0 2 b 1 – 6
148 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (a) Find the mean lifespan of the ‘Super bright’ bulbs. (b) A bulb is chosen at random. Find the probability that this bulb will last (i) longer than 1 200 hours, (ii) shorter than 1 000 hours. (c) Two bulbs are chosen at random. Find the probability that one bulb lasts longer than the mean lifespan while the other bulb lasts shorter than the mean lifespan. (d) A sample of 5 bulbs is chosen at random. Find the probability that exactly 2 bulbs last longer than 1 200 hours. 21. A man shoots at a fixed target. The distance, X cm, measured from the centre of the target to where the bullet hits the target, has a probability density function 1 10 e–—x 10 , x 0, f(x) = 0, otherwise. The man shoots at the target once. If X 2, five points are given. If 2 X 4, three points are given. If 4 X 10, only one point is given and for other values of X, no points are given. Find, correct to two decimal places, the expected score obtained by the man. 3.3 Binomial Distribution In a coin throwing experiment, two outcomes are possible: a ‘head’ or a ‘tail’. If we consider the event ‘getting a head’ as a success, then the event ‘getting a tail’ is a failure. A fair dice is thrown and six outcomes are possible, that is {1, 2, 3, 4, 5, 6}. If we consider the event of obtaining an even number as a success, then the event of obtaining an odd number is a failure. A football match has three possible outcomes: win, draw or lose. If the event of a win is considered a success, then the event of a draw or lose is a failure. From the above examples, we observe that each experiment has two types of outcomes: success and failure. In general, success means that an event occurs and failure means that the event does not occur. These experiments are called Bernoulli experiments or Bernoulli trials. Usually, in Bernoulli experiments, the probability of success is represented by the symbol p whereas the probability of failure is represented by the symbol q and p + q = 1. Example 18 A biased coin, where the probability of obtaining a ‘tail’ is 1 3 , is thrown 4 times. Calculate the probability of getting ‘tail’ exactly 0 time, 1 time, 2 times, 3 times and 4 times. Solution: Let A be the event ‘a tail is obtained’, – A be the event ‘a head is obtained’, and X is a discrete random variable which represents the number of ‘tails’ obtained in 4 throws.
149 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Then P(A) = 1 3 and P(– A) = 2 3 P(X = 0) = P(– A – A – A – A) = 1 2 3 2 4 = 16 81 P(X = 1) = P(– A – A – AA) + P(– A – AA – A) + P(– AA – A – A) + P(A – A – A – A) = 41 2 3 2 3 1 1 3 2 = 32 81 P(X = 2) = P( – A – AAA) + P(– AA – AA) + P(A – A – AA) + P(A – AA – A) + P(AA– A – A)+ P(– AAA– A) = 6 1 2 3 2 2 1 1 3 2 2 = 24 81 P(X = 3) = P(– AAAA) + P(A – AAA) + P(AA– AA) + P(AAA– A) = 4 1 2 3 2 1 1 1 3 2 3 = 8 81 P(X = 4) = P(AAAA) = 1 1 3 2 4 = 1 81 The following table shows the result P(X = x) for each value of x. x P(X = x) 0 1 2 3 2 4 = 4 C0 1 1 3 2 0 1 2 3 2 4 1 41 1 3 21 2 3 2 3 = 4 C1 1 1 3 2 1 1 2 3 2 3 2 61 1 3 2 2 1 2 3 2 2 = 4 C2 1 1 3 2 2 1 2 3 2 2 3 41 1 3 2 3 1 2 3 2 = 4 C3 1 1 3 2 3 1 2 3 2 1 4 1 1 3 2 4 = 4 C4 1 1 3 2 4 1 2 3 2 0 If X is a discrete random variable which represents the number of ‘successes’ in n independent trials of an experiment, with p as the probability of ‘success’ and q = 1 – p as the probability of ‘failure’, then the probability distribution function (probability function) of X is given by P(X = x) = n Cx px qn – x, x = 0, 1, 2, ..., n.
150 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 19 In a ‘congkak’ competition between A and B, the probability of A winning each game is p and the probability of A losing is q. The competition requires A and B to play a total of 5 consecutive games. Find the probability of A winning exactly 0, 1, 2, 3, 4 and 5 games. Solution: Let the random variable X represent ‘the number of games won by A in the ‘congkak’ competition’. By using P(X = x) = n Cx px qn – x, x = 0, 1, 2, 3, 4, 5 and n = 5, P(X = 0) = 5 C0 p0 q5 = q5 P(X = 1) = 5 C1 p1 q4 = 5p1 q4 P(X = 2) = 5 C2 p2 q3 = 10p2 q3 P(X = 3) = 5 C3 p3 q2 = 10p3 q2 P(X = 4) = 5 C4 p4 q1 = 5p4 q1 P(X = 5) = 5 C5 p5 q0 = p5 Observe that (q + p) 5 = q5 + 5pq4 + 10p2 q3 + 10p3 q2 + 5p4 q1 + p5 ↓ 1 = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) 5 1 = ∑ P(X = x) x = 0 The terms of a binomial expansion are given as follows. (q + p) n = qn + nC1 pqn – 1 + nC2 p2 qn – 2 + ... + nCx px qn – x ↓ + ... + pn ↓ ↓ 1 = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = x) + ... + P(X = n) n 1 = ∑ P (X = x) x = 0 This means that P(X = x) for x = 0, 1, 2, ..., n can be obtained from the terms of a binomial expansion of (q + p) n with p as the probability of ‘success’ and q = 1 – p as the probability of ‘failure’ in n independent trials of an experiment. Since P(X = x) can be obtained from a binomial expansion, then the random variable X is called a binomial random variable and its probability distribution is called the binomial probability distribution or simply the binomial distribution. If the discrete random variable X is binomially distributed, we write X ~ B(n, p). We read X ~ B(n, p) as X is binomially distributed with parameters n and p. The probability distribution function of X is P(X = x) = n Cx px qn – x, x = 0, 1, 2, ..., n.
151 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 20 The probability that a resident of Pearl Island supports political party A is 0.7. A sample of 6 residents of Pearl Island is chosen at random. Find the probability that (a) exactly 4 residents support political party A, (b) less than 4 residents support political party A. Solution: Let X be a discrete random variable which represents the number of supporters of political party A. Then n = 6, p = 0.7 and q = 0.3. By using the formula P(X = x) = n Cx px qn – x, (a) P(X = 4) = 6 C4 (0.7)4 (0.3)2 = 15(0.7)4 (0.3)2 = 0.3241 (b) P(X 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 6 C0 (0.7)0 (0.3)6 + 6 C1 (0.7)1 (0.3)5 + 6 C2 (0.7)2 (0.3)4 + 6C3 (0.7)3 (0.3)3 = 0.000729 + 0.010206 + 0.059535 + 0.18522 = 0.2557 Alternative Method P(X 4) = 1 – P(X 4) = 1 – [P(X = 4) + P(X = 5) + P(X = 6)] = 1 – [6 C4 (0.7)4 (0.3)2 + 6 C5 (0.7)5 (0.3)1 + 6 C6 (0.7)6 ] = 1 – [0.324135 + 0.302526 + 0.117649] = 0.2557 Example 21 According to the Penang Education Department, 40% of the students go to school by bus. Students are chosen at random from a school in Penang. Find the minimum number of students that must be chosen so that the probability that at least one student goes to school by bus is more than 0.9. Solution: Let X be a discrete random variable which represents the number of students who go to school by bus. Then X ~ B(n, p) with p = 0.4 and n unknown. P(X = x) = n Cx px qn – x, x = 0, 1, 2, ..., n P(at least one student goes to school by bus) > 0.9 P(X 1) 0.9 1 – P(X 1) 0.9 P(X 1) 0.1 P(X = 0) 0.1 n C0 (0.4)0 (0.6)n 0.1 (0.6)n 0.1 n log10 0.6 log100.1 n log10 0.1 log10 0.6 n 4.508 The minimum value of n = 5. Therefore, the minimum number of students that must be chosen is 5.
152 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Exercise 3.8 1. A fair coin is tossed 5 times. Find the probability that (a) ‘head’ appears 3 times, (b) ‘head’ appears more than 3 times, (c) ‘head’ appears less than 3 times. 2. A fair dice is tossed 6 times. Find the probability that (a) the number 1 is obtained exactly 2 times, (b) the number 1 is obtained at least once, (c) the number 1 is obtained at most 2 times. 3. Four fair coins are tossed simultaneously. Find the probability of obtaining (a) exactly 1 ‘tail’, (b) at least 3 ‘tails’, (c) no ‘tails’ at all. 4. If X ~ B(8, 1 3 ), find (a) P(X = 3), (b) P(X 4). 5. If Y ~ B(6, 0.4), find (a) P(Y = 5), (b) P(Y = 0), (c) P(Y > 4). 6. Find the probability of obtaining not more than five ‘tails’ in seven tosses of a fair coin. 7. By assuming that a married couple has an equal opportunity of giving birth to a boy or girl, find the probability that a couple who has 5 children will have more boys than girls in the family. 8. The probability that a shooter strikes a target in one shot is 0.7. Find the probability that the shooter strikes a target 3 times out of 6 independent trials. 9. A man sends 4 application forms to apply for shares in Company ABC. If the probability of each form being successful is 4 5 , find the probability that he is successful in (a) 1 application, (b) less than 3 applications, (c) at least 3 applications. 10. 40% of the students in a school wear spectacles. From a sample of 10 students chosen at random, find the probability that (a) only 3 students wear spectacles, (b) more than 7 students wear spectacles. 11. A multiple choice test has 8 questions with 5 answer choices each. Each question has only one correct answer. If a student answers the questions by guessing, find the probability that he (a) will answer all questions wrongly, (b) will get more than 6 questions correct. 12. The probability that a housewife buys brand A soap is 0.72. Find the probability that in a sample of 10 housewives who have bought soap for a certain day, (a) exactly 4 of them bought brand A soap, (b) more than 6 of them bought brand A soap.
153 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 13. A box contains red and green marbles in the ratio 1 : 2. Five marbles are drawn from the box one by one with replacement before the next marble is drawn. Find the probability that (a) exactly 4 are red marbles, (b) more than 2 are green marbles. 14. 2% of the batteries produced by a factory are unsatisfactory. In a sample of 12 batteries chosen at random, find the probability that (a) exactly 3 batteries are unsatisfactory, (b) less than 2 batteries are unsatisfactory. 15. If X ~ B(n, 0.05) and P(X 1) 0.5, find the smallest value of n. 16. The probability that a football player scores a goal when a penalty is given is 0.6. Find the number of penalties which must be given so that the probability of scoring at least one goal is more than 0.99. Mean and variance If a random variable X has a binomial distribution with parameters n and p, then the mean of X is np and the variance of X is npq, where q = 1 – p. We can prove the above results. Consider the random variable X ~ B(n, p) and the random variable Yi with i = 1, 2, ..., n which represents the number of successes in the ith trial. E(Yi ) = 0 + 1 × p = p E(X) = E(Y1 ) + E(Y2 ) + ... + E(Yn ) = p + p + ... + p = np Var(Yi ) = 12 × p – p2 = p(1 – p) Var(X) = Var(Y1 ) + Var(Y2 ) + ... + Var(Yn ) = p(1 – p) + p(1 – p) + ... + p(1 – p) = np(1 – p) = npq, with q = 1 – p Example 22 The probability that a durian chosen at random from a basket of durians is bad is 0.08. Find the expectation and standard deviation of the number of bad durians in a sample of 20 durians chosen at random from the basket. Solution: Let the discrete random variable X represent the number of bad durians in a sample of 20. Then, X ~ B(20, 0.08) E(X) = np = 20(0.08) = 1.6 Var(X) = npq = 20(0.08)(0.92) = 1.472 Standard deviation = Var(X) = 1.472 = 1.213
154 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 23 A binomial random variable X, that is X ~ B(n, p), has mean 3 and variance 51 20 . Find the value of n and p. Hence, find P(X = 3). Solution: E(X) = 3 np = 3............................... a Var(X) = 51 20 npq = 51 20 ......................b b a : q = 51 60 = 17 20 p = 1 – q = 1 – 17 20 = 3 20 From a, n1 3 20 2 = 3 n = 20 P(X = x) = n Cx px qn – x where x = 0, 1, 2, ..., n P(X = 3) = 20C31 3 20 2 3 1 17 20 2 17 = 0.2428 Exercise 3.9 1. A fair dice is thrown 60 times. Find the number of times the number 6 is expected to appear and calculate the standard deviation. 2. 20% of the electric components produced by a factory are defective. If a sample of 40 electric components are taken, find the number of components expected to be defective and calculate the variance. 3. X is a random variable such that X ~ B(n, p). If the expectation of X = 3.2 and p = 0.4, find the value of n and the standard deviation of X. 4. X is a random variable such that X ~ B(10, p) with p 0.5 and Var(X) = 8 5 . Find (a) p, (b) E(X), (c) P(X = 5). 5. 20 fair coins are tossed simultaneously. X is a discrete random variable which represents the number of ‘heads’ obtained. Find the expectation and variance of X. 6. The probability of a mango chosen at random being infested by worms is 1 10. Find the expectation and standard deviation of the number of mangoes that are worm infested in a sample of 30 mangoes. 7. 1% of the torchlights sold at a shop are faulty, and 30 torchlights are bought from the shop. Find the expectation and the variance of the number of torchlights which are faulty in the sample. 8. In a group of teachers, the expected number of teachers who own Proton cars is 8 and variance is 1.6. Find the probability that (a) a teacher chosen at random owns a Proton car, (b) exactly 4 teachers from the group own Proton cars.
155 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Determining the mode of a binomial random variable The value of X which is most likely to occur is the value of X with the highest probability. This value of X is also called the mode of X. The following example shows two methods which can be used to determine the mode of X. The second method saves time if the sample size, n, is large. Example 24 A random variable X has a binomial distribution with n = 6 and p = 0.4. Calculate the value of X most likely to occur. Solution: Method I Calculate P(X = x) for x = 0, 1, 2, ..., 6 P(X = 0) = 6 C0 (0.4)0 (0.6)6 = 0.046656 P(X = 1) = 6 C1 (0.4)1 (0.6)5 = 0.186624 P(X = 2) = 6 C2 (0.4)2 (0.6)4 = 0.31104 Highest value P(X = 3) = 6 C3 (0.4)3 (0.6)3 = 0.27648 P(X = 4) = 6 C4 (0.4)4 (0.6)2 = 0.13824 P(X = 5) = 6 C5 (0.4)5 (0.6)1 = 0.036864 P(X = 6) = 6 C6 (0.4)6 (0.6)0 = 0.004096 The highest probability is 0.31104 when X = 2. Hence, the value of X most likely to occur is 2. Method II In this method, we compare P(X = x + 1) and P(X = x). P(X = x) = n Cx px qn – x = n! (n – x)! x! px qn – x …………………… a P(X = x + 1) = n Cx + 1 px + 1 qn – x – 1 = n! (n – x – 1)! (x + 1)! px + 1qn – x – 1 ……………… b b a : P(X = x + 1) P(X = x) = (n – x)! x! (n – x – 1)! (x + 1)! px + 1qn – x – 1 px qn – x = 1 n – x x + 1 21 p q 2 If P(X = x + 1) P(X = x) (n – x)p (x + 1)q (6 – x)(0.4) (x + 1)(0.6) 12 – 2x 3x + 3 9 5x x 1.8 Hence, P(X = 2) P(X = 1) P(X = 0). If P(X = x + 1) P(X = x), x 1.8. Hence, P(X = 2) P(X = 3) P(X = 4) P(X = 5) P(X = 6). P(X = 2) has the highest probability and the value of X most likely to occur is 2.
156 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 25 70% of the residents in a country are infected by a type of skin disease. If 15 residents from the country queue up to see a doctor, how many of them are most likely to be infected with skin disease? Solution: P(X = x + 1) P(X = x) = 1 n – x x + 1 21 p q 2 = 1 15 – x x + 1 21 0.7 0.3 2 = 1 15 – x x + 1 21 7 3 2 If P(X = x + 1) P(X = x), (15 – x)7 (x + 1)3 102 10x x 10.2 Hence, P(X = 11) P(X = 10) P(X = 9) … P(X = 0). If P(X = x + 1) P(X = x), x 10.2. Hence, P(X = 11) P(X = 12) P(X = 13) P(X = 14) P(X = 15). P(X = 11) has the highest probability. Hence, the most likely number of residents infected with skin disease is 11. Exercise 3.10 1. If the random variable X is binomially distributed with n = 8 and p = 1 4 , determine the mode of X. 2. If Y ~ B(10, 4 5 ), determine the value of Y most likely to occur. 3. If T ~ B(16, 0.3), determine the mode of T. Calculate the probability that the mode value of T occurs. 4. A random variable X is binomially distributed with mean 2 and variance 1.6. Find (a) the mode of X, (b) P(X 6). 5. In a bag, there are 6 white discs, 6 black discs and 8 red discs. A disc is drawn at random from the bag. Its colour is recorded and the disc is returned to the bag. This process is repeated 10 times. Find (a) the expected number of black discs, (b) the number of white discs most likely to be drawn, (c) the probability that less than 4 red discs are drawn. 3.4 Poisson Distribution If X is a discrete random variable which represents the number of times a random event occurs in an interval of time or space, then X is a Poisson random variable. Some examples of Poisson random variables are given below. (a) The number of accidents occuring on a highway in a day. (b) The number of telephone calls received by a school clerk from 8 a.m. to 8.30 a.m. (c) The number of typing mistakes on a page. (d) The number of insurance claims in a month. (e) The number of bacteria in 1 ml of lake water.
157 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 The probability distribution of a Poisson random variable is called the Poisson probability distribution or simply the Poisson distribution. The probability distribution function (probability function) of a Poisson random variable X is P(X = x) = e–λ λx x! for x = 0, 1, 2, … where λ is the mean number of times an event occurs in an interval of time or space. If X has a Poisson distribution, we write X ~ P0 (λ). We can prove ∑ P(X = x) = 1 as follows. all x ∞ ∑ P(X = x) = ∑ e–λ λx all x x = 0 x! = e–λ 1 λ0 0! + λ1 1! + λ2 2! + λ3 3! + …2 = e–λ 1 1 + λ + λ2 2! + λ3 3! + …2 = e–λ (eλ ) 1 + λ + λ2 —2! + λ3 —3! + … = eλ = 1 Mean and variance If a random variable X has a Poisson distribution, that is, X ~ P0 (λ), then the mean of X is λ and the variance of X is λ. Both the mean and variance can be proved as follows, E(X) = ∑ all x x P(X = x) = ∑ all x x1 e–λ λx x! 2 = 0(e–λ ) + 11 e–λ λ 1! 2 + 21 e–λ λ2 2! 2 + 31 e–λ λ3 3! 2 + 41 e–λ λ4 4! 2 + … = e–λ λ 1 1 + λ + λ2 2! + λ3 3! + …2 = e–λ λ(eλ ) 1 + λ + λ2 — 2! + λ3 — 3! + … = eλ = λ E(X2 ) = ∑ all x x2 P(X = x) = ∑ all x x2 1 e–λ λx x! 2 = 0(e–λ ) + 11 e–λ λ 1! 2 + 41 e–λ λ2 2! 2 + 91 e–λ λ3 3! 2 + 161 e–λ λ4 4! 2 + … = e–λ λ 1 1 + 2 λ + 3 λ2 2! + 4 λ3 3! + …2 = e–λ λ1 + λ + λ2 2! + λ3 3! + … + λ + 2 λ2 2! + 3 λ3 3! + …2
158 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 = e–λ λ3eλ + λ1 + λ + λ2 2! + …4 = e–λ λ[eλ + λeλ ] 1 + λ + λ2 —2! + λ3 —3! + … = eλ = e–λ λ eλ (1 + λ) = λ(1 + λ) Var(X) = E(X2 ) – [E(X)]2 = λ(1 + λ) – λ2 = λ Example 26 The mean number of bacteria per millilitre of liquid is 2. By assuming that the number of bacteria follows a Poisson distribution, find the probability that in 1 ml of liquid, there are (a) 0 bacteria, (b) 3 bacteria, (c) less than 2 bacteria. Solution: Let the discrete random variable X represent the number of bacteria in 1 ml of liquid. Then, X ~ P0 (2) P(X = x) = e–22x x! , x = 0, 1, 2, … (a) P(X = 0) = e–220 0! = 0.1353 (b) P(X = 3) = e–223 3! = 0.1804 (c) P(X 2) = P(X = 0) + P(X = 1) = 0.1353 + e–221 1! = 0.1353 + 0.2707 = 0.4060 Example 27 X is a discrete random variable with X ~ P0 (2.4). Find (a) P(X 3), (b) P(1 X 2), (c) E(X), (d) the standard deviation of X.
159 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Solution: P(X = x) = e–2.42.4x x! , x = 0, 1, 2, … (a) P(X 3) = 1 – P(X 2) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)] = 1 – e–2.43 2.40 0! + 2.41 1! + 2.42 2! 4 = 1 – e–2.4(1 + 2.4 + 2.88) = 1 – 0.5697 = 0.4303 (b) P(1 X 2) = P(X = 1) + P(X = 2) = e–2.4 12.4 + 2.42 2 2 = e–2.4(2.4 + 2.88) = 0.4790 (c) E(X) = 2.4 (d) Var(X) = 2.4 Standard deviation of X = 2.4 = 1.5492 Example 28 The average number of cars which stop at a petrol station is 36 per hour. By assuming that the number of cars which stop at the petrol station follows a Poisson distribution, find the probability that (a) no cars stop at the petrol station in an interval of 10 minutes, (b) more than 3 cars stop at the petrol station in an interval of 10 minutes, (c) more than 5 cars stop at the petrol station in an interval of 20 minutes, (d) less than 2 cars stop at the petrol station in an interval of 5 minutes. Solution: (a) Mean number of cars which stop at the petrol station = 36 per hour = 6 per 10 minutes Suppose X is a discrete random variable which represents the number of cars stopping at the petrol station in a time interval of 10 minutes. Then X ~ P0 (6) P(X = x) = e–66x x! for x = 0, 1, 2, 3, … P(X = 0) = e–6 = 0.00248 (b) P(X 3) = 1 – P(X 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)] = 1 – e–6 1 60 0! + 61 1! + 62 2! + 63 3! 2 = 1 – e–6(1 + 6 + 18 + 36) = 1 – 61e–6 = 1 – 0.1512 = 0.8488
160 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (c) Mean number of cars stopping at the petrol station = 36 cars per hour = 12 cars every 20 minutes Suppose the random variable Y represents the number of cars stopping at the petrol station in a time interval of 20 minutes. Then Y ~ P0 (12) P(Y = y) = e–1212y y! for y = 0, 1, 2, 3, … P(Y 5) = 1 – P(Y 5) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) +P(X = 3) + P(X = 4) + P(X = 5)] = 1 – e–12 1 120 0! + 121 1! + 122 2! + 123 3! + 124 4! + 125 5! 2 = 1 – e–12(1 + 12 + 72 + 288 + 864 + 2 073.6) = 1 – e–12(3 310.6) = 1 – 0.02034 = 0.9797 (d) Mean number of cars stopping at the petrol station = 36 cars per hour = 3 cars every 5 minutes Suppose Z is a random variable which represents the number of cars stopping at the petrol station in a time interval of 5 minutes. Then, Z ~ P0 (3) P(Z = z) = e–33z z! , for z = 0, 1, 2, 3, ... P(Z 2) = P(Z = 0) + P(Z = 1) = e–31 30 0! + 31 1! 2 = e–3(4) = 0.1991 Exercise 3.11 1. If X ~ P0 (1.5), find (a) P(X = 0), (b) P(X 2), (c) P(3 x 6), (d) P(X 3), (e) E(X) and standard deviation of X.
161 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 2. If X ~ P0 (λ) and P(X = 0) = 0.3012, find (a) λ, (b) variance of X, (c) P(0 X 2), (d) P(X 4). 3. The number of telephone calls received at a telegraph office in a time interval of 5 minutes follows a Poisson distribution with mean 0.4. Find the probability that (a) no telephone calls are received between 1145 hours and 1150 hours, (b) more than 3 telephone calls are received in the time interval of 25 minutes. 4. The number of accidents occuring in a factory each week follows a Poisson distribution with variance 2.8. Find the probability that (a) no accidents occur in a particular week, (b) less than 3 accidents occur in a particular week, (c) more than 4 accidents occur in two weeks, (d) exactly 6 accidents occur in two weeks. 5. If the number of bacteria in 1 millilitre of water follows a Poisson distribution with mean 2.4, find the probability that (a) there are more than 4 bacteria in 1 millilitre of water, (b) there are less than 4 bacteria in 0.5 millilitre of water. 6. A book has 1000 pages with 250 printing errors. By assuming that the printing errors occur at random, find the probability that a certain page contains (a) exactly 3 printing errors, (b) more than 2 printing errors, (c) between 1 and 4 printing errors inclusive. 7. The mean number of people who borrow books from the public library in a span of 30 minutes is 4. Find the probability that (a) not a single person borrows any book in a span of 3 minutes, (b) at least two persons borrow books in a span of 1.5 minutes, (c) at most two persons borrow books in a span of 6 minutes. 8. The number of customers who patronise a coffee shop follows a Poisson distribution with a mean of 20 per hour. Find the probability that (a) no customer patronises the coffee shop in a time interval of 6 minutes, (b) at least one customer patronises the coffee shop in a time interval of 12 minutes. 9. The number of goals scored by the Penang football team follows a Poisson distribution with mean λ. If the probability that Penang’s football team does not score any goal is 0.2231, find (a) the value of λ, (b) the probability that the Penang football team scores less than 4 goals in a game, (c) the probability that the Penang football team scores less than 4 goals in two games. 10. The number of telephone calls received by an operator from 8.00 a.m. to 8.05 a.m. follows a Poisson distribution with mean λ = 4. Find the probability that (a) the operator will not receive any call in the same time interval the following day, (b) in the following two days, the operator will receive a total of 3 calls during the same time interval. 11. A factory manufactures plastic bags and finds that 1% of the output is defective or unsatisfactory. 200 plastic bags are delivered to a trader. Estimate the probability that at least 5 of the plastic bags delivered to the trader are defective or unsatisfactory. (Use a Poisson distribution as a model.)
162 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Determining the most likely value of a Poisson random variable Just like the binomial distribution, the most likely value of X or mode of X can be determined in two ways. The following example shows the two methods clearly. Example 29 A random variable X has a Poisson distribution with mean 3.5. Find the value of X most likely to occur. Solution: Method I We calculate P(X = x). This value will increase and then decrease. P(X = x) = e–λ λx x! = e–3.53.5x x! P(X = 0) = e–3.53.50 0! = 0.0302 P(X = 1) = e–3.53.5 1! = 0.1057 P(X = 2) = e–3.53.52 2! = 0.1850 P(X = 3) = e–3.53.53 3! = 0.2158 Highest value P(X = 4) = e–3.53.54 4! = 0.1888 The highest probability occurs when X = 3. Hence, the value of X most likely to occur is 3. Method II P(X = x) = e–λ λx x! .............................. a and P(X = x + 1) = e–λ λx + 1 (x + 1)! ........................ b b a : P(X = x + 1) P(X = x) = λx + 1 (x + 1)! 1 x! λx 2 = λ x + 1 If P(X = x + 1) P(X = x), λ x + 1 3.5 x + 1 x 2.5 Hence, P(X = 3) P(X = 2) P(X = 1) P(X = 0) If P(X = x + 1) P(X = x), x 2.5 Hence, P(X = 3) P(X = 4) P(X = 5)… P(X = 3) has the highest probability and the value of X most likely to occur is 3.
163 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 30 The number of white blood corpuscles on a slide has a Poisson distribution with mean 3.1. (a) Find the mode number of white corpuscles on a slide. Calculate the probability of obtaining this mode. (b) If three such slides are prepared, calculate the probability of obtaining a total of at least two white corpuscles. Solution: (a) If X represents the ‘number of white corpuscles on a slide’, then X ~ P0 (3.1). Using the Poisson distribution, P(X = x + 1) P(X = x) = λ x + 1 . If P(X = x + 1) . P(X = x), then λ . x + 1 3.1 . x + 1 2.1 . x x = 2, 1, 0 Hence, P(X = 3) . P(X = 2) > P(X = 1) . P(X = 0). If P(X = x + 1) , P(X = x), then λ , x + 1 3.1 , x + 1 2.1 , x x = 3, 4, 5, ……. Hence, P(X = 3) . P(X = 4) . P(X = 5) . P(X = 6) . ……… Since P(X = 3) has the highest probability, the mode = 3 corpuscles. P(X = 3) = e–3.13.13 3! = 0.2237 (b) Let X1 , X2 , X3 be random variables representing the number of white corpuscles on the first, second and third slides respectively. Then T = X1 + X2 + X3 , represents the total number of white corpuscles in all the three slides. Mean of T = 3.1 + 3.1 + 3.1 = 9.3 and T ~ P0 (9.3). P(T > 2) = 1 – P(T , 2) = 1 – [P(T = 0) + P(T = 1)] = 1 – [e–9.3 + e–9.3(9.3)] = 1 – 10.3(e–9.3) = 0.999 Exercise 3.12 1. For the Poisson distribution given below, determine the value of X most likely to occur. (a) X ~ P0 (1.3) (b) X ~ P0 (2.2) (c) X ~ P0 (4.5) (d) X ~ P0 (8.2)
164 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 2. The number of disruptions in the water supply to a town follows a Poisson distribution with mean of 1 every 5 weeks. Calculate the probability that (a) there are no disruptions in a week, (b) there are more than 2 disruptions in a week, (c) there is more than 1 disruption in two weeks. Determine the number of disruptions most likely to occur in a year. (Assume that there are 52 weeks in a year.) 3. A random variable X has a Poisson distribution with mean 3.7. Find the mode of X. 4. The variance of a Poisson random variable Y is 1.5. The value of Y with the highest probability is k. Determine the value of k and P(Y = k). 5. Along the North-South highway from Penang to Kuala Lumpur, breakdowns requiring the services of car foreman occurs averagely, with a frequency of 2.2 per day. Assuming that the breakdowns occur randomly following the Poisson distribution, find (a) the probability that exactly 1 breakdown occurs on a given day, (b) the most likely number of breakdowns in a day. 6. In a city, one person in 100 on the average has blood of type X. If 400 blood donors are taken at random, find (a) the mode number of people with type X blood. (b) the approximate probability that there are at least six persons having blood of type X. How many random donors must be taken in order that the probability of including at least one donor of type X shall be at least 0.94? 3.5 Normal Distribution The normal distribution is the most important probability distribution of a continuous random variable in statistics. Most measurements in our everyday life can be represented by a normal distribution. Examples of measurements which have the normal distribution are (a) times taken to walk 1 km by 100 students, (b) heights of school children aged between 7 years and 8 years old, (c) weights of babies born in the year 2000 in Penang, (d) lengths of steel rods which are produced by a factory. The graph of a normal distribution, named the normal curve, is a bell-shaped curve such that the total area under the curve is one unit, the curve is symmetrical about the mean and the two tails of the curve extend indefinitely. A continuous random variable X with probability density function f(x) = 1 σ2π e– (x – µ) — 2 2σ2 for –∞ x ∞ is said to have a normal distribution with parameters µ and σ2 . The normal distribution is also known as Gaussian distribution in honour of the contribution from Karl Friedrich Gauss.
165 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 The graphs of some normal distributions are shown in the figures below. σ2 σ1 µ1 µ2 x f(x) σ2 σ1 µ1 = µ2 x f(x) Normal curves with Normal curves with µ1 ≠ µ2 but σ1 = σ2 µ1 = µ2 and σ1 σ2 Figure 3.8(a) Figure 3.8(b) σ2 σ1 µ1 µ2 x f(x) As s increases in value, the graph of the normal curve is more widely distributed. Normal curves with µ1 ≠ µ2 and σ1 σ2 Figure 3.8(c) Mean and variance If the continuous random variable X is normally distributed with probability density function f(x) = 1 σ2π e– (x – µ) — 2 2σ2 for –∞ x ∞, then the mean of X is µ and the variance of X is σ2 . The proofs for E(X) = µ and Var(X) = σ2 are as shown below. E(X) = ∫ ∞ –∞ xf(x) dx = ∫ ∞ –∞ 1 σ2π xe – (x – µ) — 2 2σ2 dx Let t = x – µ σ ⇒ x = µ + tσ ⇒ dx = σ dt. When x → ∞, t → ∞ x → – ∞, t → – ∞
166 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Hence, E(X) = ∫ ∞ –∞ 1 σ2π (µ + tσ)e–—1 2 t 2 σ dt = µ 1 2π ∫ ∞ –∞ e –—1 2 t 2 dt + σ 2π ∫ ∞ –∞ te –—1 2 t 2 dt, 1 2π ∫ ∞ –∞ e –—1 2 t 2 dt = 1 = µ + σ 2π 3–e–—1 2 t 2 4 ∞ –∞ e –—1 2 t 2 → 0 when t → ±∞ = µ that is, mean X = µ E(X 2 ) = ∫ ∞ –∞ x2 f(x) dx = ∫ ∞ –∞ 1 σ2π x2 e – (x – µ) — 2 2σ2 dx = ∫ ∞ –∞ 1 σ2π (µ + tσ) 2 e –—1 2 t 2 σ dt = µ2 2π ∫ ∞ –∞ e –—1 2 t 2 dt + 2µσ 2π ∫ ∞ –∞ te –—1 2 t 2 dt + σ2 2π ∫ ∞ –∞ t 2 e –—1 2 t 2 dt = µ2 + 2µσ 2π 3–e–—1 2 t 2 4 ∞ –∞ + σ2 2π 3t(–e–—1 2 t 2 ) 4 ∞ –∞ – σ2 2π ∫ ∞ –∞ –e–—1 2 t 2 dt Integration by parts = µ 2 + σ2 2π ∫ ∞ –∞ e –—1 2 t 2 dt = µ2 + σ2 Var(X) = E(X2 ) – [E(X)]2 = µ2 + σ2 – µ2 = σ2 that is variance X = σ2 . If a continuous random variable is normally distributed with mean = µ and variance = σ2 , we write X ~ N(µ, σ2 ). Standardisation of a normal random variable To find P(a X b) for X ~ N(µ, σ2 ), we need to find the area of the shaded region in Figure 3.9 P(a X b) = ∫ b a f(x) dx = ∫ b a 1 σ2π e– (x – µ) — 2 2σ2 dx This integration is tedious and difficult. To evaluate this probability, a table for the standard normal distribution with mean µ = 0 and σ2 = 1 has been prepared. Since every normal curve has different values of µ and σ, we must therefore transform a normal random variable X by a process known as standardising in order to use the same set of tables. Figure 3.9 µ ba x f(x)
167 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 A normal random variable X can be standardised to a standard normal variable Z by using the formula Z = X – µ σ . If X ~ N(µ, σ2 ), then Z ~ N(0, 1). The continuous random variable Z is called the standard normal random variable and its distribution is called the standard normal distribution. We now show that E(Z) = 0 and Var(Z) = 1. E(Z) = E1 X – µ σ 2 Var(Z) = Var 1 x – µ σ 2 = 1 σ [E(X) – µ] = 1 σ2 [Var(X) + 0] = 1 σ [µ – µ] = 1 σ2 (σ2 ) = 0 = 1 Figure 3.10(a) and 3.10(b) illustrate the difference between the graph of X ~ N(µ, σ2 ) and the graph Z ~ N(0, 1). µ + 3σ µ + 2σ µ + µ σ µ – σ µ – 2σ µ – 3σ x f(x) –3 –2 –1 0 1 2 3 z φ (z) Figure 3.10(a) Figure 3.10(b) For a standard normal random variable Z, its probability density function is φ(Z) = 1 2π e – Z2 —2 , – ∞ , Z , ∞ The corresponding distribution function is denoted by Φ such that Φ(a) = P(Z < a) = ∫ a –∞ φ(z) dz We use the standard normal table which gives the value of Φ(z), where Φ(z) = P(Z < z). The shaded area in Figure 3.11 shows the area tabulated. 0 φ(z) Φ(z) z Figure 3.11
168 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 For negative values of z, use Φ(–z) = 1 – Φ(z) because the standard normal curve is symmetrical about z = 0. Examine the following examples. (a) P(Z –a) = P(Z a) = 1 – P(Z < a) = 1 – Φ(a) (b) P(Z – a) = P(Z a) = Φ(a) (c) P(–b , Z , – a) = P(a Z b) = P(Z , b) – P(Z , a) = Φ(b) – Φ(a) (d) P(– a Z 0) = P(0 Z a) = P(Z , a) – P(Z , 0) = Φ(a) – Φ(0) Example 31 If the continuous random variable Z has a standard normal distribution, use tables to find (a) P(Z , 1.26), (b) P(Z . 2), (c) P(0.2 Z 0.5), (d) P(Z –1.851), (e) P(Z –1.378), (f) P(–0.71 Z 2.53), (g) P(| Z | 2.644). Solution: (a) P(Z , 1.26) = Φ(1.26) = 0.8962 (b) P(Z . 2) = 1 – P(Z < 2) = 1 – Φ(2) = 1 – 0.9772 = 0.00228 –a 0 ⇒ 0 a Figure 3.12 –a 0 ⇒ 0 a Figure 3.13 –b –a 0 ⇒ Φ(a) 0 a b Φ(b) Figure 3.14 –a 0 ⇒ 0 a Figure 3.15 0 2 0 1.26
169 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (c) P(0.2 Z 0.5) = P(Z , 0.5) – P(Z , 0.2) = Φ(0.5) – Φ(0.2) = 0.6915 – 0.5793 = 0.1122 0 0.2 0.5 ⇒ 0 0.5 – 0 0.2 (d) P(Z –1.851) = P(Z 1.851) = 1 – P(Z < 1.851) = 1 – Φ(1.851) = 1 – 0.9679 = 0.0321 –1.851 0 ⇒ 0 1.851 (e) P(Z –1.378) = P(Z 1.378) = Φ(1.378) = 0.9160 –1.378 0 ⇒ 0 1.378 (f) P(–0.71 Z 2.53) = P(Z , 2.53) – P(Z , –0.71) = P(Z 2.53) – P(Z . 0.71) = Φ(2.53) – [1 – Φ(0.71)] = 0.9943 – 1 + 0.7611 = 0.7554 (g) P(| Z | 2.644) = P(Z –2.644 or Z 2.644) = P(Z –2.644) + P(Z 2.644) = 2P(Z . 2.644) = 2[1 – P(Z < 2.644)] = 2[1 – Φ(2.644)] = 2(1 – 0.9959) = 0.0082 –0.71 0 2.53 –2.644 0 2.644
170 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 32 If Z ~ N(0, 1), find the value k if (a) P(Z k) = 0.4153, (b) P(Z k) = 0.9738, (c) P(Z k) = 0.0187. Solution: (a) P(Z k) = 0.4153 1 – P(Z < k) = 0.4153 1 – Φ(k) = 0.4153 Φ(k) = 0.5847 From tables, k = 0.214 (b) P(Z k) = 0.9738 Because the probability exceeds 0.5, k must be negative. 1 – P(Z < k) = 0.9738 1 – Φ(k) = 0.9738 Φ(k) = 0.0262 1 – Φ(–k) = 0.0262 Φ(–k) = 0.9738 From tables, – k = 1.94 k = –1.94 (c) P(Z k) = 0.0187 Take note that in the figure below, k must be negative. Φ(k) = 0.0187 1 – Φ(–k) = 0.0187 Φ(–k) = 0.9813 From tables, – k = 2.082 k = –2.082 k 0 ⇒ 0 –k We can also use the standard normal table for the random variable X with X ~ N(µ, σ2 ). See the examples below. Example 33 The random variable X ~ N(100, 16). Find (a) P(X 104), (b) P(X 92), Solution: (a) P(X 104) = P1 X – 100 4 104 – 100 4 2 = P(Z 1) = 1 – Φ(1) = 1 – 0.8413 = 0.1587 0 k k 0 0 –k 0 1 Standardise X
171 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (b) P(X 92) = P1 X – 100 4 , 92 – 100 4 2 = P(Z –2) = P(Z . 2) = 1 – Φ(2) = 1 – 0.9772 = 0.0228 Example 34 If X ~ N(50, 9), find (a) P(47 X 53), (b) the value of k if P(X k) = 0.2543. Solution: (a) P(47 X 53) = P1 47 – 50 3 , X – 50 3 , 53 – 50 3 2 = P(–1 Z 1) = 2Φ(1) – 1 = 0.6826 (b) P(X k) = 0.2543 P1 X – 50 3 k – 50 3 2 = 0.2543 P1Z k – 50 3 2 = 0.2543 1 – Φ1 k – 50 3 2 = 0.2543 Φ1 k – 50 3 2 = 0.7457 From tables, k – 50 3 = 0.661 k = 50 + 3(0.661) = 51.98 (correct to 2 decimal places) Exercise 3.13 1. The continuous random variable Z follows a standard normal distribution. Using tables, find (a) P(Z 1.11), (b) P(Z 2.46), (c) P(1.0 Z 2.0), (d) P(Z –1.11), (e) P(Z – 2.46), (f) P(–2 Z –1), (g) P(– 0.77 Z 1.25), (h) P(– 0.705 Z – 0.452), (i) P(| Z | 0.56), (j) P(| Z | 0.687). 2. If Z ~ N(0, 1), find (a) P(0 Z 0.851), (b) P(– 0.264 Z 0), (c) P(–1.872 Z – 0.115), (d) P(| Z | 1.645), (e) P(| Z | 1.981). 0–2 Standardise X –1 0 1
172 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 3. If Z ~ N(0, 1), find the value of k if (a) P(Z k) = 0.3806, (b) P(Z k) = 0.7814, (c) P(Z k ) = 0.0783, (d) P(Z k) = 0.9893, (e) P(| Z | k ) = 0.80, (f) P(| Z | k) = 0.02, (g) P(| Z | k) = 0.6872, (h) P(| Z | k) = 0.0484. 4. If X is a continuous random variable which has a normal distribution with mean 10 and variance 16, find (a) P(X 13), (b) P(8 X 12), (c) P(6.5 X 11), (d) P(X 12.5). 5. If Y ~ N(50, 100), find (a) P(Y 65), (b) P(Y 44), (c) P(46 Y 47), (d) P(40 Y 52), (e) P(48.5 Y 49.5), (f) P(| Y – 50 | 1.4). 6. The random variable X has a normal distribution with mean 40 and variance 4. Find the value k if (a) P(X k) = 0.25, (b) P(X k) = 0.6331, (c) P(X k) = 0.0018, (d) P(X k) = 0.050. 7. If X ~ N(240, 64), find (a) m such that P(| X – 240 | m) = 0.80, (b) q such that P(| X – 240 | q) = 0.96, (c) r such that P(| X – 240 | r) = 0.08, (d) s such that P(| X – 240 | s) = 0.975. Solving problems using the normal distribution The following examples show some problems that can be solved using the normal distribution. Example 35 The distribution for the lengths of a particular type of fish in Perdana Lake can be regarded as normal with mean 12.5 cm and standard deviation 3.6 cm. Determine the probability that a fish caught in this lake has a length which is (a) more than 15 cm, (b) less than 11.4 cm, (c) between 10 cm and 13 cm. Solution: Assume that X is the continuous random variable which represents the length of the fish caught in Perdana Lake. Then, X ~ N(12.5, 3.62 ). (a) P(X 15) = P1Z 15 – 12.5 3.6 2 Standardise X = P(Z 0.694) = 1 – Φ(0.694) = 1 – 0.7562 = 0.2438 (b) P(X 11.4) = P1Z 11.4 – 12.5 3.6 2 Standardise X = P(Z – 0.306) = P(Z 0.306) = 1 – Φ(0.306) = 1 – 0.6201 = 0.3799 0–0.306 0 0.694
173 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (c) P(10 X 13) = P1 10 – 12.5 3.6 Z 13 – 12.5 3.6 2 = P(–0.694 Z 0.139) = Φ(0.139) – Φ(–0.694) = Φ(0.139) – [1 – Φ(0.694)] = 0.5553 – [1 – 0.7562] = 0.3115 Example 36 The time taken by Kadir to distribute newspapers everyday to residents staying at Burma Road follows a normal distribution with mean 32 minutes and variance 6.25 minutes2 . Find the probability that on Wednesday, Kadir takes (a) less than 28 minutes, (b) more than 40 minutes, to distribute newspapers at Burma Road. Solution: Assume that the continuous random variable X represents the time taken by Kadir to distribute newspapers at Burma Road. Then, X ~ N(32, 6.25). (a) P(X 28) = P1Z 28 – 32 6.25 2 Standardise X = P(Z –1.6) = P(Z 1.6) = 1 – Φ(1.6) = 1 – 0.9452 = 0.0548 (b) P(X 40) = P1Z 40 – 32 6.25 2 Standardise X = P(Z 3.2) = 1 – Φ(3.2) = 1 – 0.99931 = 0.00069 Exercise 3.14 1. The marks obtained by students in a Mathematics test are normally distributed with mean 68 and standard deviation 16. Calculate the probability that a randomly selected student obtained marks which are (a) less than 58, (b) more than 75, (c) between 62 and 76, (d) between 70 and 80. 2. If the lifespans of a particular type of tyre from a factory are normally distributed with mean 2.0 years and standard deviation 3 months, find the probability that a tyre randomly selected from the factory will last (a) less than 1.5 years, (b) at least 2.5 years, (c) between 1.25 years and 1.8 years, (d) between 1 year 6 months and 2 years 3 months. –0.694 0 0.139 0–1.6 0 3.2
174 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 3. The lengths of houseflies have a normal distribution with mean 2 cm and variance 0.09 cm2 . Find the probability that the length of a housefly caught at random is (a) more than 2.28 cm, (b) less than 1.45 cm. 4. The masses of loaves of bread produced by XYZ Company have a normal distribution with mean 400 g and standard deviation 18 g. Find the probability that a loaf of bread chosen at random will have a mass of (a) more than 450 g or less than 380 g, (b) between 360 g and 440 g. 5. The thickness of paper produced by a type of machine has a normal distribution. If the mean thickness of the paper is 1.05 mm and its standard deviation is 0.02 mm, determine the probability that the thickness of a sheet of paper taken at random is (a) between 1.10 mm and 1.12 mm, (b) more than 1.06 mm or less than 0.994 mm. 6. The mass of a mango taken from Rahman’s estate is known to be normally distributed with mean 820 g and standard deviation 100 g. (a) Find the probability that a mango chosen at random will have a mass of at least 700 g. (b) In a basket which contains 200 mangoes from Rahman’s estate, how many mangoes are expected to have a mass of at least 700 g? 7. In a Statistics test, the average mark is 82 and the standard deviation is 5. All the students from 88 marks up to 94 marks will receive a B grade. If the marks have a normal distribution and 8 students receive B grade, determine the total number of students who sat for the test. 8. The Intelligence Quotients (IQ) for 600 applicants to a college is known to have a normal distribution with mean 115 and standard deviation 12. If the college requires an IQ of at least 95, calculate the number of applicants who are unsuccessful in entering the college based on their IQ. Solving problems involving the determination of the value µ or σ or both Example 37 If the random variable X has a normal distribution with mean 45 and variance σ2 and P(X 51) = 0.288, find the value of σ. Solution: P(X 51) = 0.288 P1Z 51 – 45 σ 2 = 0.288 P1Z 6 σ 2 = 0.288 1 – Φ1 6 σ 2 = 0.288 Φ1 6 σ 2 = 0.712 From tables, 6 σ = 0.559 σ = 10.73 0 6 σ _ 0.288
175 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Example 38 If X~ N(µ, 12) and P(X 32) = 0.8438, find the value µ. Solution: P(X 32) = 0.8438 P1Z 32 – µ 12 2 = 0.8438 Φ1 µ – 32 12 2 = 0.8438 From tables, µ – 32 12 = 1.01 µ = 32 + 1.0112 = 35.50 Example 39 The masses of textbooks in a school bag of a student have a normal distribution with mean µ and standard deviation σ. 10% of the books have masses exceeding 900 g and 5% have masses less than 750 g. Find the value of µ and σ. Solution: Assume that X is a random variable which represents ‘the mass of a textbook’. Hence X ~ N(µ, σ2 ). P(X 900) = 0.10 10% P1Z 900 – µ σ 2 = 0.10 1 – Φ1 900 – µ σ 2 = 0.10 Φ1 900 – µ σ 2 = 0.90 From tables, 900 – µ σ = 1.282 900 – µ = 1.282σ …………………… a P(X 750) = 0.05 5% P1Z 750 – µ σ 2 = 0.05 P1Z . µ – 750 σ 2 = 0.05 1 – P1Z < µ – 750 σ 2 = 0.05 P1Z < µ – 750 σ 2 = 0.95 From tables, µ – 750 σ = 1.645 µ – 750 = 1.645σ ………………… b a + b, 150 = 2.927σ σ = 150 2.927 = 51.2 (3 significant figures) 32 – μ 0 0.8438 12 _____ ⇒ 0 μ – 32 0.8438 12 _____ 0 1.01 0.8438
176 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Substituting σ = 51.2 in a: 900 – µ = 1.282(51.2) 900 – µ = 65.64 µ = 834 (3 significant figures) Exercise 3.15 1. If the random variable X has a normal distribution with mean 21 and variance σ2 and P(X 27) = 0.9332, find the value of σ. 2. If the random variable X has a normal distribution with mean 45 and variance σ2 and P(X 51) = 0.751, find the value of σ. 3. If T ~ N(µ, 25) and P(T 27.5) = 0.6915, find the value of µ. 4. If X ~ N(µ, σ2 ), P(X 80) = 0.0113 and P(X 30) = 0.0287, find the values of µ and σ. 5. If Y ~ N(µ, σ2 ), P(Y 102) = 0.58 and P(Y 97) = 0.75, find the values of µ and σ. 6. The random variable X has a normal distribution with mean µ and variance σ2 . Given that P(X 60) = 0.02 and P(X 42) = 0.01, find the values of µ and σ. 7. The continuous random variable, X has a normal distribution with mean 80 and standard deviation σ. Given that P(60 X 100) = 0.6, find the value of σ. 8. The marks in an examination for a Physics paper have a normal distribution with mean µ and variance σ2 . 10% of the students obtain more than 75 marks and 20% obtained less than 40 marks. Find the values of µ and σ. The normal distribution as an approximation to the binomial distribution A binomial distribution with parameters n and p can be approximated by a normal distribution with µ = np and σ2 = npq (q = 1 – p), if n is large (n 30) and p is approximately 0.5. If X ~ B(n, p), for large n and p approximately 0.5, then as an approximation X ~ N(np, npq). As a guide, if np 5 and nq 5, the normal approximation is appropriate. Example 40 Find the probability of obtaining between 4 to 6 ‘heads’ inclusive when a fair coin is tossed 16 times using the (a) binomial distribution, (b) the normal approximation to the binomial distribution. Solution: Let the random variable X represent the number of ‘heads’ obtained from 16 tosses of a fair coin. Assume that the event of obtaining a ‘head’ is a success. Then, X ~ B(n, p) with n = 16 and p = 1 2 .
177 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (a) P(X = x) = n Cx px qn – x, x = 0, 1, 2, …, n P(X = x) = 16Cx1 1 2 2 x 1 1 2 2 16 – x , x = 0, 1, 2, …, 16. P(4 X 6) = P(X = 4) + P(X = 5) + P(X = 6) = 16C41 1 2 2 4 1 1 2 2 12 + 16C51 1 2 2 5 1 1 2 2 11 + 16C61 1 2 2 6 1 1 2 2 10 = 0.02777 + 0.06665 + 0.12219 = 0.2166 Hence, the probability of obtaining between 4 to 6 ‘heads’ inclusive is 0.2166. The probability distribution for the number of ‘heads’ in 16 tosses of a fair coin has been calculated and is shown in the figure below. The required probability is represented by the area of the shaded rectangle. This can be approximated by the area under the appropriate normal curve. 5.0 × 10–2 10.0 × 10–2 15.0 × 10–2 20.0 × 10–2 3 65 7 98 10 11 12 13 1514 16 3.5 6.5 0 1 2 4 P(X = x) Number of 'heads' (b) To use the normal approximation, we calculate µ = np = 161 1 2 2 = 8 and σ2 = npq = 161 1 2 21 1 2 2 = 4 Thus, X ~ N(8, 4). Before using the normal approximation, a correction for continuity must be carried out because a continuous distribution is used as an approximation for a discrete distribution. In this example, P(4 X 6) becomes P(3.5 X 6.5) P(3.5 X 6.5) = P1 3.5 – 8 2 Z 6.5 – 8 2 2 = P(–2.25 Z – 0.75) = P(0.75 Z 2.25) = P(Z 2.25) – P(Z 0.75) = 0.2266 – 0.0122 = 0.2144 Hence, the probability of obtaining between 4 to 6 ‘heads’ inclusive is 0.2144.
178 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Below are a few examples of continuity corrections. P(X 4) means the area of the rectangle is included. P(X 4) → P(X 3.5) P(X 4) means the area of the rectangle is not included. P(X 4) → P(X 4.5) P(X = 4) means only the area of the rectangle is considered. P(X = 4) → P(3.5 X 4.5) P(X 5) → P(X 5.5) P(X 5) → P(X 4.5) P(4 X 8) → P(3.5 X 8.5) P(4 X 8) → P(4.5 X 7.5) P(X 0) → P(X – 0.5) P(X 0) → P(X 0.5) P(X = 0) → P(–0.5 X 0.5) In general, if a and b are two integers, (a) P(X a) = P(X a – 0.5) (e) P(a X b) = P (a – 0.5 X b + 0.5) (b) P(X a) = P(X a + 0.5) (f) P(a X b) = P(a + 0.5 X b – 0.5) (c) P(X a ) = P(X a + 0.5) (g) P(X = a) = P(a – 0.5 X a + 0.5) (d) P(X a) = P(X a – 0.5) Example 41 A regular tetrahedral shaped dice with its faces labelled 1, 2, 3 and 4 is tossed 200 times. Find the probability of obtaining (a) more than 60 times the digit 4, (b) at most 50 times the digit 4, (c) exactly 50 times the digit 4. Solution: Let X be the random variable which represents the number of times the digit 4 appears and the event of obtaining the digit 4 as ‘success’. Thus, X ~ B1200, 1 4 2. Since n is large, we use the normal approximation. µ = np = 2001 1 4 2 = 50 σ2 = npq = 2001 1 4 21 3 4 2 = 37.5 Therefore, X ~ N(50, 37.5) 0 3.5 4 4.5 0 3.5 4 4.5 0 3.5 4 4.5
179 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (a) P(X 60) → P(X 60.5) continuity correction = P1Z 60.5 – 50 37.5 2 = P(Z 1.715) = 1 – P(Z < 1.715) = 1 – 0.9568 = 0.0432 The probability of obtaining more than 60 times the digit 4 is 0.0432. (b) P(X 50) → P(X 50.5) continuity correction = P(Z 50.5 – 50 37.5 2 = P(Z 0.0816) = Φ(0.0816) = 0.5327 The probability of obtaining at most 50 times the digit 4 is 0.5327. (c) P(X = 50) → P(49.5 X 50.5) = P1 49.5 – 50 37.5 Z 50.5 – 50 37.5 2 = P(– 0.0816 Z 0.0816) = 2P(Z , 0.0816) – 1 = 2(0.5327) – 1 = 0.0654 The probability of obtaining exactly 50 times the digit 4 is 0.0654. Exercise 3.16 1. If X ~ B(200, 0.3), use the normal approximation to find (a) P(X 50), (b) P(X 60), (c) P(36 X 54), (d) P(52 X 59), (e) P(66 X 76), (f) P(X = 70). 2. A fair coin is tossed 400 times. Find the probability that (a) less than 230 ‘heads’ are obtained, (b) exactly 205 ‘heads’ are obtained, (c) between 180 and 190 ‘heads’ are obtained. 3. A biased dice is thrown 240 times. If the probability of obtaining the digit 1 is 0.35, find the probability of obtaining the digit 1 on the dice, (a) exactly 75 times, (b) less than 80 times, (c) between 65 and 70 times inclusive. 0 1.715 Continuity correction 0 0.0816 0 0.0816 –0.0816
180 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 4. In 2002, it is estimated that 5 8 of the population of Malaysia watched the televised Thomas Cup finals between Malaysia and Indonesia. If random samples of 400 people are taken, calculate the mean and variance of the number of people from these samples, who watched the televised Thomas Cup finals. By using the normal approximation, find, correct to three significant figures, the probability that (a) more than 280 people, (b) between 240 and 280 people, watched the televised Thomas Cup finals, from a random sample of 400 people. 5. 10% of the biscuits produced in a factory break even before it is packed. From a sample of 500 biscuits, find the probability that the number of biscuits broken before being packed is (a) less than 40 pieces, (b) at least 40 pieces, (c) between 50 to 56 pieces inclusive, (d) at most 30 pieces. 6. The probability that a dog catches a rabbit before the rabbit disappears in the bushes is 2 5 at each attempt. If the dog attempts 360 times in a year to catch a rabbit, find the probability that the dog is successful in catching a total of (a) 160 rabbits, (b) more than 170 rabbits, (c) between 130 and 150 rabbits. 7. A type of flowering seed is sold in small packages of about 800 seeds. It is known that 45% will produce red flowers and 55% yellow flowers. If 100 of these seeds are planted, use the normal approximation to estimate the probability that (a) at least 32 seeds will produce red flowers, (b) more than 60 seeds will produce yellow flowers, (c) between 40 to 50 seeds will produce red flowers. 8. Two fair dice are tossed. Find the probability of obtaining a total score of 7. Find also the probability of obtaining 22 times the total score of 7 if these two fair dice are tossed a total of 120 times. Determine the number of tosses needed if the probability of obtaining the score of 7 at least once is 0.85 or more. 9. The probability of a guided missile hitting its targets is 0.64. Find the probability that out of 500 shots of these guided missiles, the target will be hit (a) at least 350 times. (b) not more than 340 times. Determine the number of shots needed such that the probability of at least one missile hitting its target is more than 0.95. 10. A lorry load of tapiocas contains an average of 3 spoiled tapiocas out of every 20. A wholesaler tests a sample containing 100 tapiocas and if it is found that there are more than 17 spoiled tapiocas, then he will not buy tapiocas from this lorry at all. Find the probability that he will accept the whole lorry load of tapiocas.
181 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 Summary 1. A random variable is a variable with values that are determined by the outcomes of a random experiment. 2. A discrete random variable is a random variable with values which are exact. 3. A discrete probability distribution is a table or formula which lists all the possible values of a random variable and their corresponding probabilities. Basic properties of probability distribution functions are (a) 0 P(X = x) 1 for all values of x, (b) ∑P(X = x) = 1. all x 4. For a discrete random variable X, (a) E(X) = ∑x P(X = x), (b) E[g(X)] = ∑g(x)P(X = x), (c) (i) E(a) = a, (ii) E(aX + b) = aE(X) + b, where a and b are constants. 5. For a discrete random variable X, (a) Var(X) = E[(X – µ)]2 = E(X2 ) – µ2 , (b) (i) Var(a) = 0, (ii) Var(aX + b) = a2 Var(X), where a and b are constants. 6. A continuous random variable cannot take exact values but take values in an interval. 7. If X is a continuous random variable, the probability density function, f, satisfies the following properties. (a) f(x) 0 for all values of x valid in its interval, (b) ∫ ∞ –∞ f(x) dx = 1. 8. (a) P(a X b) = ∫ b a f(x) dx (b) P(X = a) = ∫ a a f(x) dx = 0 (c) P(a X b) = P(a X b) = P(a X b) = P(a X b) 9. The cumulative distribution function, F, for a continuous random variable X is the probability that X is less than or equal to a certain value, x, and is given by F(x) = P(X x) = ∫ x –∞ f(x) dx The basic characteristics of a cumulative distribution function are as follows. (a) 0 F(x) 1 (b) P(a X b) = F(b) – F(a) (c) d dx F(x) = F(x) = f(x)
182 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 10. For a continuous random variable X, (a) E(X) = ∫ ∞ –∞ xf(x) dx, (b) E[g(X)] = ∫ ∞ –∞ g(x)f(x) dx, (c) (i) E(aX) = aE(X), where a is a constant, (ii) E(aX + b) = aE(X) + b, where a and b are constants. 11. For a continuous random variable X, (a) variance X = Var(X) = E[(X – µ) 2 ] = E(X2 ) – µ2 , (b) (i) Var(aX) = a2 Var(X), where a is a constant, (ii) Var(aX + b) = a2 Var(X), where a and b are constants. 12. The median, m, of a continuous random variable X with probability density function f is given by ∫ m –∞ f(x) dx = ∫ ∞ m f(x) dx = 1 2 If F is the cumulative distribution function of X, then F(m) = 1 2 . 13. The first quartile, q1 and third quartile, q3 are respectively given as P(X q1 ) = F(q1 ) = 1 4 P(X q3 ) = F(q3 ) = 3 4 . 14. The probability distribution function of a random variable X which has a binomial distribution with parameters n and p is P(X = x) = n Cx px qn – x for x = 0, 1, 2, …, n and q = 1 – p (a) E(X) = np, (b) Var(X) = npq, (c) P(X = x + 1) P(X = x) = 1 n – x x + 1 21 p q 2. 15. The probability distribution function of a random variable X which has a Poisson distribution with parameter λ is P(X = x) = e–λ λx x! for x = 0, 1, 2, … (a) E(X) = λ, (b) Var(X) = λ, (c) P(X = x + 1) P(X = x) = λ x + 1 . 16. If the random variable X has a normal distribution with mean µ and variance σ2 , we write X ~ N(µ, σ2 ). If X ~ N(µ, σ2 ) and Z = X – µ σ , then Z ~ N(0, 1). 17. A binomial distribution with parameters n and p can be approximated by a normal distribution with µ = np and σ2 = np(1 – p) if n is large and p approaches 0.5. As a guide, if np 5 and nq 5, then the normal approximation is suitable.
183 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 STPM PRACTICE 3 1. Two fair dice are thrown simultaneously. The number A and the number B on the top most face of the dice are recorded. Score X for each throw is defined as X = A + B if A = B, |A – B| if A ≠ B. (a) Construct a probability distribution table for X. (b) Find E(X) and Var(X). 2. The probability distribution for a random variable X is as follows. x –1 0 1 P(X = x) p 1 4 2p (a) Find the value of p. (b) Determine the mean and standard deviation of X. 3. A random variable X takes values 1, 2, …, n with equal probabilities. Determine the expectation, µ, of X and show that the variance, σ2 , is given by 12σ2 = n2 –1. Hence, find P(|X – μ| σ) in the case n = 100. 4. The number of matchsticks not used in a box branded ‘Super’ is represented by X. The probability distribution of X is as follows. x 0 1 2 3 P(X = x) 5k 5k k k Determine the value of the constant k, and find the expectation and variance of X. 5. A random variable X takes values –2, 0 and 2 respectively with corresponding probabilities 1 4 , 1 2 and 1 4 . Find Var(X) and E(|X|). 6. A bag contains 6 balls. There are two white balls and four red balls. Three balls are drawn one by one without replacement. If X represents the number of red balls drawn, find the probability distribution of X. Find also E(X) and Var(X). If Y represents the number of white balls drawn, without finding the distribution of Y, determine E(Y) and Var(Y). 7. The faces on two fair dice which are tetrahedral in shape is marked with the numbers 1, 2, 3 and 4. When the dice are cast on a table, the number on the faces of the dice in contact with the table is recorded. The random variable X is defined as the product of the numbers on the faces in contact with the table. Obtain the probability distribution of X. Hence, find (a) P(X 8), (b) E(X), (c) Var(X).
184 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 8. A test consists of five objective questions and each question has three choices, and only one out of the three choices is correct. For each correct answer, a candidate is given 2 marks but 1 mark is subtracted if the answer is wrong. A candidate answers all questions by guessing randomly. If X is the marks obtained by the candidate, tabulate all the possible values of X with their corresponding probabilities. Show that the expected mark is 0 and find the variance for this distribution. 9. A random variable X can take values 0, 1, 2, 3, 4 and 5. The probability distribution for X is given as P(X = 0) = P(X = 1) = P(X = 2) = a, P(X = 3) = P(X = 4) = P(X = 5) = b, P(X 2) = 3P(X 2), where a and b are constants. (a) Determine the values of a and b. (b) Show that the expectation of X is 23 8 and determine the variance of X. (c) Determine the probability that the sum of two observations from this distribution is more than 7. 10. A sample of 4 mice is chosen from a group of 10 mice (6 males and 4 females). Let X represent the number of male mice in the sample. Calculate the values of P(X = r) for r = 0, 1, 2, 3, 4. Hence, find the mean and standard deviation of X. Deduce the mean and standard deviation for the number of female mice in the sample. 11. 1 3 2 1 32 A circular card is divided into three sectors with score 1, 2 and 3 and their respective angle sizes are 135°, 90° and 135°. On another circular card, the scores for three sectors are also 1, 2 and 3 but their respective angle sizes are 180°, 90° and 90°. Each card has a pointer hinged to the centre of the card. When the pointers are rotated, they will stop at random, independently. Find the probability that (a) the score on each card is 1, (b) the score on at least one of the cards is 3. The random variable X is the bigger score between the two scores if they are different. If the two scores are the same, then X is the common value score. Show that P(X = 2) = 9 32 . Show also E(X) = 75 32 and find Var(X). 12. A stall which sells chicken rice has 6 helpers. At any instant, the probability that one of the helpers is busy is 1 5 . Calculate the probability that at any instant, (a) all the helpers are busy, (b) less than 5 helpers are busy, (c) at least two helpers are busy. Determine the most probable number of helpers who are busy at any instant and find the probability that this number of helpers is busy at any instant. 13. The probability that a shooter strikes a target is p. (a) Find the probability that he will strike the target at least 5 times out of 6 shots. (b) Find the probability that he will strike the target at least twice out of n shots (n 2). (c) By using a suitable approximation, find the probability that he will strike the target more than 220 times out of 500 shots when p = 0.4.
185 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 14. A population consists of three types of individuals, X, Y and Z in the ratio 1 : 3 : 6. (a) A sample of 5 individuals is chosen from the population. Determine the probability that (i) exactly two individuals of type X have been chosen, (ii) more than two individuals of type X have been chosen, (iii) less than four individuals of type Z have been chosen. (b) A sample of 150 individuals is taken at random from the population. Determine the approximate value for the probability that exactly 50 individuals are of type Y. 15. A company has five telephone lines. The probability that a line is used at any time is 1 3 . Find the probability that (a) at least one line is used at any time, (b) more than three lines are used at any time, (c) all lines are used at any time. Find the most probable number of lines used at any time. What is the probability that this number of lines is being used? 16. A gambling machine shows one out of five colours: red, orange, yellow, green and blue. When the machine is working properly, every colour has an equal chance of appearing and the colour shown at one instant is independent of the colour shown earlier. Calculate the probability that (a) the machine shows different colours for 5 consecutive times, (b) the colours orange and red do not appear for 5 consecutive times, (c) the green colour appears exactly 3 times out of 5 consecutive times. 17. A crossword puzzle appears in the newspapers everyday except Sunday. On average, a student can complete 8 out of 10 crossword puzzles. (a) Find the expectation and standard deviation of the number of crossword puzzles completed by the student in a week. (b) Show that the probability the student can complete at least 5 crossword puzzles in a week is 0.655 (correct to 3 significant figures). (c) Given that the student completed Monday’s crossword puzzle, find the probability that he can complete at least 4 crossword puzzles for the other days in the week (correct to 3 significant figures). (d) Find the probability that, in 4 weeks, he can complete 4 or less crossword puzzles, in only one of the 4 weeks. 18. A plane has 116 seats. The airline knows that on average, 2.5% of the passengers with tickets for a particular flight do not arrive for the flight. If the airline sells 120 tickets for a certain flight, determine, by using a suitable approximation, the probability that more than 116 passengers arrive for that flight. Determine also the probability that there are empty seats on the flight. 19. Gnat larvae are distributed at random in a pond. Suppose the number of larvae found in a random sample of 10 cm3 of pond water is a random variable which has a Poisson distribution with mean 0.2. Ten independent random samples, each of 10 cm3 of pond water, are taken by a scientist. Determine (correct to 3 significant figures), (a) the probability that not a single sample contains larvae, (b) the probability that one sample contains larvae and the others do not contain any larvae, (c) the probability that one sample contains two or more larvae and the others do not contain any larvae, (d) the expected total number of larvae in the ten samples, (e) the expected number of samples that do not contain larvae.
186 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 20. The number of emergency admissions to a hospital each day has a Poisson distribution with mean 2. (a) Determine the probability that there are no emergency admissions on a certain day. (b) Determine the probability that there are more than two emergency admissions on a certain day. (c) The hospital has 4 beds for emergencies at the beginning of each day. Calculate the probability that this number is insufficient for that day. (d) Calculate the probability that there are exactly 3 emergency admissions on two consecutive days. 21. A gold prospector examines 800 gold bearing ingots each month. The contents of each ingot can be assumed to be independent and the probability of one ingot containing gold is 0.005. The prospector is considered having a lucky month if 4 or more ingots examined by him contain gold. Show that the probability a month chosen at random is a lucky month is 0.567 (correct to 3 significant figures). Determine the probability that (a) out of four months chosen at random, there are more than two lucky months, (b) in 24 months chosen at random, there are more than 12 lucky months. 22. In a certain town, the probability that a letter is sent to the wrong address is 0.003. (a) In a day, 1000 letters are sent independently. If X represents the number of wrong address deliveries, state the distribution of X. (b) Determine the minimum number of letters which must be delivered so that the probability that no letter is wrongly delivered is less than 0.2. 23. A building is fixed with five thousand new 25 watt fluorescent tubes. The probability that each tube is still in good condition after being in use for 10 hours is 0.9992. By using the Poisson distribution, find correct to 4 decimal places, (a) the probability that more than 3 fluorescent tubes are spoilt after operating for 10 hours, (b) the probability that between 2 and 4 fluorescent tubes inclusive are spoilt after operating for 10 hours. 24. It is known that 4% of the mangosteens from a fruit farm are bad. By using Poisson distribution, calculate the probability that, out of 60 mangosteens chosen at random from the fruit farm, less than 4 mangosteens are bad. 25. A family consisting of three persons whose names are David, Ivy and Jack live in an apartment which has one telephone line. When the phone rings, the probabilities that the calls are for David, Ivy and Jack are 0.25, 0.35 and 0.40 respectively. The probabilities that David, Ivy and Jack are in the apartment when the phone rings are 0.6, 0.4 and 0.8 respectively. (a) By assuming that all the probabilities are independent, find the probability that, when the phone rings, (i) only Jack is in the apartment, (ii) the person wanted by the call is not in the apartment, (iii) the call is for Ivy and Ivy is the only one in the apartment. (b) If the number of telephone calls received on a particular day follows a Poisson distribution with mean 1.2, calculate, correct to four decimal places, the probability that (i) there are phone calls on a particular day, (ii) there are less than four phone calls on a particular day, (iii) there are no phone calls for three consecutive days.
187 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 26. For each function f, g and h defined below, determine whether or not the function is a probability density function. (a) f(x) = x 64 (16 – x2 ) for 0 x 4 (b) g(x) = 2 3 (x – 2) for 1 x 4 (c) h(x) = 2x + 3 for –1 x 2 (The value of each function is zero outside the stated interval.) 27. The continuous random variable X has a probability density function f(x) = 1 4 , 0 x 1, 3 8 (3 – x), 1 x 3. (a) Sketch the graph y = f(x) and verify that the graph y = f(x) satisfies the conditions for a probability density function. (b) Find (i) P(X 2.5), (ii) P(| X – 2 | 0.5). 28. The continuous random variable X is defined on the interval 4 X 8 with the probability density function cx + k, 4 x 6, f(x) = 0.3, 6 x 8. If f is continuous at x = 6, find the values of c and k. 29. The continuous random variable X has probability density function k, 0 x 1, f(x) = 1 2 k(3 – x), 1 x 3, 0, otherwise where k is a constant. (a) Find the value k. (b) Prove that E(X) = 13 12 and find the variance for X. (c) Find P(X 13 12 ). 30. The continuous random variable X has probability density function 0, x 0, f(x) = kx, 0 x 2, 16k x3 , x 2. Calculate the value of the constant k. Find the value of the median and the expected value of X. Find also the value a such that P(X a) = 0.005.
188 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 31. The continuous random variable X has probability density function ax2 + bx, 0 x 2, f(x) = 0, otherwise. Given also E(X) = 1.25. (a) Show that b = 3 4 and find the value of a. (b) Find the variance of X. (c) Verify that the median of X is approximately 1.3. (d) Find the mode. 32. The continuous random variable X has probability density function k(3 – x) 2 , 0 x 3, f(x) = 0, otherwise. (a) Show that k = 1 9 and sketch the graph f(x). (b) Calculate E(X) and show that Var(X) = 27 80 . (c) Obtain the distribution function, F(x), with F(x) = P(X x). 33. The continuous random variable X has probability density function k(x —1 2 – x), 0 x 1, f(x) = 0, otherwise where k is a constant. (a) Prove that k = 6. (b) Calculate E(X) and Var(X). (c) Find the distribution function. (d) Show that the median, m, for X satisfies the equation 6m2 – 8m—3 2 + 1 = 0. 34. The continuous random variable X has probability density function k(4 – x2 ), 0 x 2, f(x) = 0, otherwise where k is a constant. Show that k = 3 16 and find E(X) and Var(X). Find the cumulative distribution function for X and verify that the median X lies between 0.69 and 0.70. Find also P(0.69 X 0.70). Give your answer correct to one significant figure. 35. On any day, the time spent by a housewife to watch television is a continuous random variable T with cumulative distribution function. 0, t 0, F(t) = 1 – k(15 – t) 2 , 0 t 15, 1, t 15 where k is a constant.
189 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 (a) Show that k = 1 225 and find P(5 T 10). (b) Show that, for 0 t 15, the probability density function is given by f(t) = 2 15 – 2 225 t. (c) Find the median of T. (d) Find Var(T). 36. A continuous random variable X has probability density function f with its graph as shown below. f(x) A x 0 1 2 3 (a) Find the value of A. (b) Find E(X) and Var(X). (c) Find the median of X. A sample X1 , X2 and X3 is obtained. What is the probability that at least one of these observations is more than the median? 37. A continuous random variable X takes values in the interval 0 x 4. The probability that X exceeds x is αx2 + β, 0 x 4. (a) Determine the value of the constants α and β. (b) Find the probability density function of X. (c) Show that the expectation, µ, for X is 8 3 . (d) Show that the standard deviation, σ, for X is 22 3 . (e) Show that P(µ – σ X µ + σ) = 4 9 2 . 38. A petrol station is supplied with petrol every Sunday morning and its weekly sales, in thousand litres, is a random variable, X, which has a probability density function, f(x), given by f(x) = 3 125 (5 – x) 2 , 0 x 5. Find, (a) the probability that the weekly sales of the petrol station is less than 3 000 litres. (b) the probability that the petrol station cannot fulfil its weekly order, if the capacity of the tanks in the petrol station is 4 000 litres. 39. A particular species of bird is known to have a mean lifespan of 20 years with standard deviation σ years. By assuming that the birds’ lifespans have a normal distribution, find the value σ if 10% of these birds live more than 24 years. Determine the probability that a bird of this species, chosen at random, will have a lifespan of less than 17 years. 40. A farmer plants 400 watermelon seeds in his farm. If the average rate of germination is 68%, find the probability that (a) more than 300 seeds will germinate, (b) less than 250 seeds will germinate.
190 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 41. A clerk drives to his workplace in a university. On average, his travelling time takes 24 minutes with a standard deviation of 3.8 minutes. By assuming that his travelling times are normally distributed, find (a) the probability that his travelling time will be at least half an hour on a particular day, (b) the percentage of times he will be late if office starts at 8.00 a.m. and he leaves his house at 7.45 a.m. every working day, (c) the probability that he will be late for coffee, if he leaves his house at 7.35 a.m. and coffee is served in the office from 7.50 a.m. to 8.00 a.m. 42. Diameters of a type of steel pipes produced in a factory are normally distributed with mean 0.95 cm and standard deviation σ cm. If at least 88% of the steel pipes produced have diameters which are less than 0.98 cm, find the largest possible value of σ. 43. The operational period of a digital camera sold by Vision Electronic Company has a normal distribution with mean 64 months and standard deviation 12 months. The warranty period given by the company is 2 years. A total of 7540 digital cameras were sold by the company. Determine the number of digital cameras that are expected to stop working during the warranty period. 44. The masses of suspended particles in a sample of water taken from a lake can be assumed to be a random variable which is normally distributed. The probability that a sample of lake water contains less than 1.5 mg of suspended particles is 1 4 and the probability that a sample of lake water contains more than 3 mg suspended particles is 1 5 . (a) Determine the mean and variance of the mass of suspended particles in each sample of lake water. (b) Show that the probability of a sample of lake water containing less than 1.8 mg of suspended particles is 0.3542. 45. Medicine M produced by a researcher can cure a type of eye disease with probability 0.9. (a) If eight people having the same eye disease are treated with medicine M, find the probability that (i) exactly 6 people are cured after the treatment, (ii) at most 7 people are cured after the treatment. (b) If 200 people having the same eye disease are treated with medicine M, find (i) the value of n so that the probability that at least n people are cured after treatment is 0.96, (ii) the probability that between 175 to 188 people inclusive are cured after treatment. 46. The score, S, obtained by a sharp shooter in one shot is a random variable with probability distribution as follows. P(S = 8) = 0.01, P(S = 9) = 0.29, P(S = 10) = 0.70. (a) Use the normal approximation to determine the probability that the sharp shooter obtains 8 points for 6 or more shots in 900 independent shots. (b) Find the expected value and variance of S. (c) Use the normal approximation to determine the probability that the sharp shooter obtains less than 96 points in 10 independent shots. 47. A basket contains 100 balls which consists of 5 yellow balls and the rest black balls. A ball is selected at random from the basket and its colour is noted. The ball is then put back into the basket. Write down an expression for the probability that out of a total of 10 selections, a yellow ball is selected only once. By using the appropriate approximation in each case, determine, correct to three significant figures, the probability that (a) out of a total of 80 such selections, a yellow ball is selected 4 times, (b) out of a total of 6400 such selections, a yellow ball is selected between 300 and 350 times inclusive.
191 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 48. A man plants 20 seeds of delphinium, 50 seeds of cactus and 100 seeds of lupin. The germination of any seed is independent of the others. The probability that a delphinium seed germinates is 0.4, a cactus seed germinates is 0.03 and lupin seed germinates is 0.9. (a) Determine the probability that exactly 9 delphinium seeds germinate. (b) By using the Poisson distribution, determine the probability that exactly 2 cactus seeds germinate. (c) By using the normal distribution, determine the probability that 88 lupin seeds germinate. The man buys 5 packets of cactus seeds with each packet containing 50 seeds. If all the seeds are planted, determine the probability that exactly one from each packet germinates. 49. There are one or two flowers on the faces of 50 cents stamps. 90% of these 50 cents stamps have two flowers while the rest of the stamps have single flower. From the stamps which have single flower, 95% of these stamps have a flower at the centre of the stamps while the rest have a flower on the left side of the stamps. (a) Determine the probability that exactly 8 stamps have two flowers out of a random sample consisting of 10 pieces of 50 cents stamps. (b) Determine by using the Poisson approximation, the probability that less than 3 stamps have only one flower on the left side of the stamp out of a random sample of 100 pieces of 50 cents stamps. (c) Determine by using the normal approximation, the probability that between 5 and 15 stamps inclusive have one flower out of a random sample of 100 pieces of 50 cents stamps. (Give your answers correct to 3 significant figures.) 50. A number of different species of fungus are distributed randomly in a field. 80% of these fungus are mushrooms while the rest are toadstools. 5% of the toadstools are poisonous. A man who cannot differentiate between a mushroom and a toadstool picked a total of 100 fungus. Determine, correct to two significant figures, by using an appropriate approximation, the probability that the man picked (a) at least 20 toadstools, (b) exactly two poisonous toadstools. 51. The probability that it is sunny in a day is 0.30 and the probability that a student wears a hat is 0.65. The probability that it is sunny or the student does not wear a hat is 0.46. If it is sunny on a particular day, find the probability that the student does not wear a hat. 52. The random variable X is normally distributed with mean 54 and standard deviation 9. Find the least integer k such that P(|X – 54| . k) , 0.34. 53. A discrete random variable X has cumulative distribution function 0, x , 0, F(x) = 0.2, 0 < x , 2, 0.7, 2 < x , 4, 1, x > 4. (a) Construct the probability distribution table for X. (b) Find the mean and variance of X.
192 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 54. The lifespan, in years, of a type of bicycle tube is a random variable X. The probability density function is given by kxe x – —2 , 0 < x < ∞, f(x) = 0, elsewhere (a) Determine the value of k. (b) Find the cumulative distribution function of X. (c) Determine the probability that a randomly chosen bicycle tube has a lifespan of more than 8. 55. The probability that a piece of cloth placed into a tub containing bleaching powder does not bleach completely is 0.02. (a) Determine the maximum number of pieces of cloth that may be placed in the tub so that the probability that all the pieces of cloth are bleached completely is more than 0.65. (b) Using a suitable approximation, find the probability that at least two out of 100 pieces of cloth do not bleach completely. 56. The lifespan of a certain brand of bicycle tyre is normally distributed with a mean of 1200 km and a standard deviation of 80 km. (a) Find the probability that a randomly chosen tyre has a lifespan of less than 1100 km. (b) Determine the value of a, to the nearest integer, if it is found that 4% of the tyres are not roadworthy after a km. 57. A random variable X takes the integer value x with probability P(x) where kx2 , x = 1, 2, 3, 4 P(x) = 0, elsewhere Find (a) the value of k, (b) the mean and standard deviation of the distribution, (c) the mean and standard deviation of 3X – 2. 58. The continuous random variable X has probability density function f(x) = kx , 0 x < 1, k x2 , x . 1. (a) Show that k = 4 3 . (b) Calculate the mean and variance of X. 59. A farmer plants 10 seeds of type A, 40 seeds of type B and 80 seeds of type C. The probability that a type A seed germinates is 0.3, a type B seed germinates is 0.04 and a type C seed germinates is 0.8. The germination of any seed is independent of the others. (a) Determine the probability that exactly 6 seeds of type A germinate. (b) By using the Poisson distribution, determine the probability that exactly 2 seeds of type B germinate. (c) By using the normal distribution, determine the probability that exactly 70 type C seeds germinate. The farmer buys 6 packets of type B seeds and each packet contains 40 seeds. If all the seeds are planted, determine the probability that exactly one from each packet germinates.
193 Mathematics Semester 3 STPM Chapter 3 Probability Distributions 3 60. The continuous random variable X has probability density function 1 8 , 0 < x , 2, f(x) = kx , 2 < x , 4 0 , elsewhere (a) Find the value of k. (b) Find the cumulative distribution function. (c) Sketch the graph of the cumulative distribution function. (d) Find P(1 < X < 3). 61. The number of students in standard six who are unable to swim is not more than four in a class. The probability that there are no students in a class who cannot swim is 0.35 and the probabilities that there are at most one, two and three non-swimmers in a class are 0.65, 0.8 and 0.9 respectively. (a) Construct a probability distribution table for the number of non-swimmers in a class. (b) Calculate the mean number of non-swimmers in a class. 62. In a given population, 20% of the population choose swimming as their leisure activity. (a) Find the probability that, of 16 people selected at random, (i) exactly 4 people choose swimming as their leisure activity, (ii) at most 13 people do not choose swimming as their leisure activity. (b) Using a suitable approximation to find the probability that, out of 50 people selected at random, 10, 11, 12 or 13 choose swimming as their leisure activity. Justify the choice of your approximation. 63. A continuous random variable X has probability density function f(x) = 2k – x 2k2 , 0 < x < 2k, where k is a positive constant 0 , otherwise. (a) Find the mean of X in terms of k. (b) Show that the standard deviation of X is equal to √2 3 k. (c) Find the P(X , µ + 1 √2 s), where µ and s are the mean and standard deviation of X respectively. 64. The table below shows the probability distribution for a random variable X. X 0 1 2 3 P(X=x) c c2 c 2 + c 2c 2 + c Calculate (a) the value of c, (b) E(X). Hence, show that 256Var(X) = 359. 65. The masses of mangoes in a farm are normally distributed with mean, µ and standard deviation, s. The mass percentages of mangoes less than 500 g is 13.57% and the mass percentages of mangoes more than 608 g is 5.48%. (a) Determine the values of µ and s. (b) Find the probability that a mango selected has a mass of at most 608 g given that it is more than 500 g. A farmer feels that the number of mangoes, Y, on a branch of mango trees in the farm may be modelled by a Poisson distribution. (c) If P(Y=0) = 0.1823, find the parameter of the Poisson distribution. (d) Give one reason why Poisson distribution cannot be an exact model for Y.
Mathematics Semester 3 STPM Chapter 4 Sampling and Estimation 4 CHAPTER SAMPLING AND 4 ESTIMATION Subtopic Learning Outcome 4.1 Sampling (a) Distinguish between a population and a sample, and between a parameter and a statistic. (b) Identify a random sample. (c) Identify the sampling distribution of a statistic. (d) Determine the mean and standard deviation of the sample mean. (e) Use the result that X – has a normal distribution if X has a normal distribution. (f) Use the central limit theorem. (g) Determine the mean and standard deviation of the sample proportion. (h) Use the approximate normality of the sample proportion for a sufficiently large sample size. 4.2 Estimation (a) Calculate unbiased estimates for the population mean and population variance. (b) Calculate an unbiased estimate for the population proportion. (c) Determine and interpret a confidence interval for the population mean based on a sample from a normally distributed population with known variance. (d) Determine and interpret a confidence interval for the population mean based on a large sample. (e) Find the sample size for the estimation of population mean. (f) Determine and interpret a confidence interval for the population proportion based on a large sample. (g) Find the sample size for the estimation of population proportion. central limit theorem – teorem had memusat confidence interval – selang keyakinan distribution – taburan estimation – penganggaran normal distribution – taburan normal sampling – persampelan parameter – parameter population – populasi proportion – perkadaran random sample – sampel rawak Bilingual Keywords
195 Mathematics Semester 3 STPM Chapter 4 Sampling and Estimation 4 4.1 Sampling Population and sample A population is a set of objects that possesses specific characteristics. For example, according to a survey, the mean income in Malaysia in 2012 is about RM48 000 per annum. The income is the specific characteristic and the population consists of all working Malaysians. In order to determine the mean income, we have to gather information about salaries of all working Malaysians in the country. This will definitely consume time and resources and perhaps not practical as they are too many people. However, to overcome these difficulities, a smaller group of individuals is chosen. The mean income of this group, which is called a sample, is determined and then the mean income of the population can be estimated from it. We can draw a conclusion about a population characteristic from the characteristic of a sample. The process of choosing subsets from a universal set is known as sampling. A sample is a subgroup chosen from a population. A random sample is one that is chosen at random, that is, each element has an equal chance of being chosen. If all samples of the same size selected from a population have the same chance of being selected, it is called simple random sampling. Such a sample is called a simple random sample. Parameter and statistic A parameter is a numerical measure, such as the mean, for a population. A statistic is a summary measure calculated for a sample data set. For example, if the mean income in Malaysia is RM48 000 per annum, then the mean income of RM48 000 per annum is a population parameter. If the mean income in Kuala Lumpur is RM100 000 per annum, then the mean income of RM100 000 per annum is a sample statistic. The symbols of some parameters and statistics are as follows: Quantity Parameter Statistic Size Mean Variance Proportion N µ s 2 p n x – s 2 p ^ Table 4.1 The study of sampling theory involves taking samples and measuring the statistic. In general, there are 4 types of methods of selecting samples. 1. Simple random sampling In simple random sampling, each member from a population has the same probability of being chosen. One way of ensuring an equal probability of being chosen is to use random numbers. A random number is a number that has the same probability of appearing. Random numbers can be generated by a computer or a scientific calculator. For example, to select 2 students from a class of 30 students, we assign numbers 1 to 30 to each student. Then, we obtain a random number from the calculator. Table 4.2 shows a set of random numbers that has been obtained from the calculator by pressing the ‘Ran#’ button.