RI S2B Binomial Distributions_Add Prac_Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 _____________________________________________________ Additional Practice Questions for Chapter S2B: Binomial Distribution Page 1 of 16 Additional Practice Questions for Ch apter S2B: Binomial Distribution (Solutions) 1 Published articles in medical journals indicate that, on average, 35 out of 100 patients having a lumbar puncture will suffer SSH (‘Severe Spinal Headache’). Twelve patients are given a lumbar puncture. (i) Using a binomial model, find the expected number of patients who will suffer SSH, and find also the standard deviation. [3] (ii) Find the probability that four or more of the twelve patients will suffer SSH. [2] Solution: (i) Let X be the number of patients, out of 12, who will suffer SSH. Assumptions: (1) Whether a patient suffers SSH is independent of another. (2) The probability of a patient suffering SSH remains constant at 0.35. Then ~ B(12,0.35)X Expected number of patients who will suffer SSH = 12(0.35) = 4.2 Standard deviation = 12(0.35)(1 0.35) 1.65 (3.s.f) (ii) P( 4) 1 P( 3)XX   =0.653 (3.s.f) 2 The random variable X ~ B (16, p), where p < 0.5. If the variance of X is 3.36, find the value of p. Find also the probability that X is less than its mean. [4] Solution: Variance = (16)(p)(1p) = 3.36, which gives p = 0.3 or 0.7 (rejected since p < 0.5) Hence, p = 0.3 Mean = 16(0.3) = 4.8 P( 4.8) P( 4)XX = 0.450 (3 s.f.)

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ _____________________________________________________ Additional Practice Questions for Chapter S2B: Binomial Distribution Page 2 of 16 3 TPJC Prelim 2007/02/Q5 A factory produces chocolate which are packed into boxes of 20 and delivered to shops for sale. A chocolate will not meet the minimum criteria for packing for sale if it weighs less than 20 grams. On average, 2% of the chocolate produced did not meet the minimum criteria. (i) Find the probability that a randomly chosen box contain at least 1 chocolate that does not meet the minimum criteria. [2] (ii) Find the probability that out of 4 randomly chosen boxes of chocolate, there are exactly 2 boxes with at least 1 chocolate that does not meet the minimum criteria. [2] Solution: (i) Let X be the number of chocolates, out of 20, not meeting the minimum criteria. Assumptions: (1) Whether a chocolate does not meet the minimum criteria is independent of another. (2) The probability of a chocolate not meeting the minimum criteria remains constant at 0.02. Then ~B ( 2 0 , 0 . 0 2 )X P( 1) 1 P( 0) 0.33239 0.332 (3 s.f.)XX    (ii) Let Y be the number of boxes of chocolates, out of 4, with at least 1 c