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 =