Statistics 4 APQ Solutions (RVHS)

Statistics 4 Tutorial: Special DRV – Binomial Distribution Additional Practice Questions 1. [Modified 2010 HCI CT/11] [For this question, give your answers correct to 4 decimal places.] At a college canteen drinks stall, the probabilities of a student buying a cup of tea and a cup of coffee are 0.6 and 0.05 respectively. Sale of drinks takes place independently at random times. (i) Ten students queue up to buy a cup of drink each. State a suitable distribution for the number of students who do not order coffee. B(10, 0.95) (ii) Three students queue up to buy a cup of drink each. Find the probability that exactly one of them buys a cup of tea and exactly one of them buys a cup of coffee. Probability = (0.6)(0.05)(0.35) 3! = 0.0630 (iii) Five students queue up to buy a cup of drink each. Show that the probability that none of these 5 students bought tea is 0.0102. Let X denote no. of students buying tea. There are 80 classes in the college and 5 students from each class each buys a drink from the stall. Find the probability that there are more than 75 classes with at least 1 student from the class buying tea. Let W denote the no. of classes, out of 80, with no student buying tea. 2. [Modified 2009 JJC/I/9] In a large batch of items from a production line, it was found that the probability that an item is faulty is p, where 0 0.5p . A random sample of 10 items is taken and the random variable X is the number of faulty items. Given that P(X = 7) = 0.0000221, find the value of p, giving your answer correct to 3 decimal place. X ~B(10, p) P(X = 7) = ( ) 3710 17 pp − = 0.0000221 ~ B(5,0.6) ( 0) 0.01024 0.0102 X PX = = = ~ B(80,0.01024) ( 4) 0.9986 (4 d.p.) W PW  =

From GC, p = 0.115 Assume that 0.345p= . Find the probability that in a random sample of 60 items, there are more than 20 items that are faulty. Let Y denote the number of faulty items in the sample. Then Y~B(60, 0.345) P(Y > 20) = 0.516 (3sf) (from GC) 3. [Modified 2009 MJC/I/9) On average, 70% of teenagers in Singapore have myopia. A junior college in Singapore has 1200 students, with 20 students in each class. (i) State the expected number of students who do not have myopia in a randomly selected class. Expected value = 20 * 0.3 = 6 (ii) Find the probability that there are less than 5 students who do not have myopia in a randomly selected class. Let X denote the number of students in a class who do not have myopia. Then X~B(20, 0.3) P(X < 5) = 0.238 (3sf) (from GC) (iii) Find the least number of classes to be selected such that the probability of at least one class with less than 5 students who do not have myopia exceeds 0.8. Let Y denote the number of classes that have less than 5 students with myopia, and let y  60 denote the number of classes sampled. Then Y~B(y, 0.23751) P(Y ≥ 1) > 0.8 1 – P(Y = 0) > 0.8 P(Y = 0) < 0.2 ( ) ( ) 0 0.23751 1 0.237510 yy − <0.2 y ln(0.76249) < ln(0.2) y > ln(0.2) ln(0.76249) = 5.935 Therefore, at least 6 classes need to be sampled.

4. [2011 RJC/I/10] In a certain country, 27% of the adults are myopic . Eight adults are chosen at random. Let X denotes the number of adults who are myopic. (i) Show that P( 3)X  = 0.372 Let X denote the number of adults out of 8 who are myopic. Then X ~ B(8, 0.27) P( 3)X  = 1 – P(X  2) = 0.371826 = 0.372. Ten such samples of eight adults are taken. (ii) Find the probability that four of these samples each have less than three adults who are myopic. Let Y be the number of samples out of 10 that has less than 3 adults who wear glasses. Then Y ~ B (10, 0.628) P( Y = 4) = 0.0866 Nine adults are randomly selected and asked if they are myopic. Find the probability that (iii) the 9th adult is the 5th adult who is myopic. P(the 9th adult is the 5th adult who is myopic) = (0.27)(P(X = 4)) = 0.0285 (iv) the 9th adult is the first adult who is myopic P(the 9th adult is the first adult who is myopic) = (0.73)8(0.27) = 0.0218

5. The probability that a certain type of cactus seed will germinate is p. In a long-term study, 1500 of these seeds were planted, of which 600 germinated. (i) Write down an estimate of p. 600 2 1500 5p= Suzy plants 24 such seeds at the beginning of the growing season in three batches, each batch consisting of 8 seeds. (ii) Assuming that the seeds germinate independently, use the value of p found in (i) to find (a) the probability that at least 12 seeds germinate, Let X rep the no. of seeds that will germinate out of 24, X ~ B(24, 2 5 ) P(X  12) = 1 − P(X  11) = 0.213 (b) the most probable number of seeds that will germinate in a batch of 8, P(X = 2) = 0.209 P(X = 3) = 0.279 P(X = 4) = 0.232 Most probable no. that will germinate in a batch of eight is 3. (c) the probability that only one batch produces more than 4 germinating seeds. Let Y rep the no. of seeds that will germinate in a batch of 8, Y ~ B(8, 2 5 ) P(Y > 4) = 1− P(Y  4) = 0.1736704. P(only one batch produces more than 4 germinating seeds) = 3 P(Y > 4) [P(Y  4)] 2 = 3(0.1736704) [1 − 0.1736704] 2 = 0.356

6*. An urn contains n white balls and m black balls. Suppose k balls are drawn at random with replacement after each ball is drawn. Let the random variable X denote the number of black balls drawn. (i) State the distribution of X, together with its parameter(s). Let X be the rv of number of black balls drawn from a total of k draws, ~ B , mXk nm   + (ii) Find the probability that at least one black ball is drawn. ( 1) 1 P( 0) 1 k nP X X nm  = − = = −  + (iii) Given that the number of white balls is 9 times that of the black balls, find the least number of balls one must draw, so that the probability that at least one black ball is drawn exceeds 0.5. Since n = 9m, 1~ B , 10Xk   . ( ) ( 1) 0.5 91 0.510 0.9 0.5 6.58 Least value of k = 7. k k PX k  −   In a new sampling scheme, an urn contains a very large number (of the order of 106) of white and black balls in the proportion of 99 white balls to 1 black ball. Suppose 120 balls are randomly drawn without replacement after each ball is drawn. Let Y denote the number of black balls drawn. (iv) What would you consider to be the distribution that best fits Y? Justify your conclusion. 1~ B 120, 100Y   . Since the urn contains large number of balls ( at least 610 ), the probability of black ball will remain almost constant even if 120 balls are taken out without replacement. Thus a binomial distribution is a suitable model. (v) The above new sampling scheme is repeated 10 times. Find the probability that at least 3 black balls are drawn exactly twice.

Let W be the rv of the number of times out of 10 when at least 3 black balls are drawn, W~B(10, 0.1196365407). P(W= 2) = 0.232. 7. [Modified 2010 IJC/I/8] At a particular university, the probability that a student wears glasses is 0.7. (i) A random sample of 8 students is taken. Find the probability that as many students wear glasses as do not wear glasses. Let X be the number of students in the sample who wear glasses. Then X~B(8, 0.7) P(X = 4) = ( ) ( ) 448 0.7 0.34   = 0.136 (ii) Three random samples of 8 students each are taken. Find the probability that each of these samples has more than 2 students who wear glasses. P(X > 2) = 1 – P(X ≤ 2) = 1 – (0.00006561 + 0.00122472 + 0.01000188) =0.98870779 Required probability = (0.98870779)3 = 0.967 (3sf) (iii) At another university, p% of the students wear glasses, where p > 80 . A random sample of 12 students is taken and the number of students that wear glasses is denoted by G. Gi