RI revision programme stats (hard)

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision (Zeta) Topic(s): (C) Statistics (Solutions) ___________________________________________ Y6 H2 Math Term 3 Revision Lecture: (C) Statistics Page 1 of 14 Source of Question: NJC Prelim 9758/2023/02/Q11 1 (a) In a one-period binomial model, the stock price initially at 0S , either rises to 0uS with a probability p or drops to 0 1 Su with a probability ( )1 p− at the end of one period, where 1u> . Suppose that 0 10S = , 1.2u= and 0.55p= . Show that the expected value of the stock price S at the end of the period is 10.35 and find the variance of the stock price. [3] (b) The above model can be extended to a n-period binomial model. For each period, the stock price S either rises to uS with a probability p or goes drops to 1 Su with a probability 1 p− at the end of this period. (i) In the first quarter, the initial price of the stock at the start of the first period is 0S . Suppose the stock price rises 4 times and drops twice over 6 periods. Find the stock price in terms of 0S and u at the end of the 6 periods. [1] (ii) Show that the probability for part (b)(i) to happen is ( ) 2415 1pp − . [1] Initial price Price at the end of one period … price at the end of 2nd period initial price price at the end of 1st period … … 1st period 2nd period …

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ___________________________________________ Y6 H2 Math Term 3 Revision Lecture: (C) Statistics Page 2 of 14 (iii) In the second quarter, the initial price of the stock at the start of the first period is 1S . Find the probability that the stock price is higher than 1S at the end of 6 periods in terms of p. [2] (c) A contract that gives investors the right to buy a stock at a specified price at the end of a specified period is termed “call option”. The specified price is called “strike price” and the amount needed to buy this contract is called “premium”. Each unit of a stock’s call option has a return value V at the end of the period. Its value is given by strike strike strike if , 0 if , SS S SV SS −>= ≤ where strikeS is the strike price and S is the stock price at the end of the period. Stock K, currently priced at $10, is modelled by the n- period binomial model in part (b) with 6n= , 1.2u= and 0.55p= . A call option of stock K has a strike price of $10. (i) Fill in the missing numbers in the following probability distribution table of V for this call option, giving the values to a suitable degree of accuracy. [2] The call option currently has a premium of $2.50 per unit. Peter plans to invest $10 000 on purchasing 4000 units of the call option and sell them at the end of the 6 periods. (ii) Find the expected return value of the investment, giving your answer to the nearest dollar. [2] (iii) Comment on whether Peter should proceed with this investment plan. [1] Solution: (a) ( ) ( ) 1E 0.55 1.2 10 0.45 10 10.35 1.2S = ×+ ×=  ( ) ( ) 2 22 1E 0.55 1.2 10 0.45 10 110.45 1.2S = ×+ ×=  ( ) ( ) ( ) 222Var E E 110.45 10.35 3.3275SS S=− =−=  Number of times for the stock price to rise 3 or below 4 5 6 v 0 10.736 19.85984 ( )P Vv= 0.13589 0.02768

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ___________________________________________ Y6 H2 Math Term 3 Revision Lecture: (C) Statistics Page 3 of 14 (b)(i) ( ) 2 42 00 1S u uS u  = (b)(ii) ( ) ( ) 22446The required probability 1 15 14 pp pp= −= − (b)(iii) If the stock price goes up 3 times and goes down 3 times, the end price is exactly 0S , so there must be at least 4 “up”s. Let X denote the number of “Up”s in the 6 periods. ( )~ B 6,Xp ( ) ( ) ( ) ( ) ( ) 214 56 24 56 6 6641 1 4 56 15 1 6 1 PX p p p p p p p p pp   ≥= − + − +     = − + −+ (c)(i) 210 1.2 10 4.4× −= ( )~ B 6,0.55X ( )P 3 0.55848X ≤= and ( )P 4 0.27795X = = 3 or below 4 5 6 0 4.4 10.736 19.85984 0.55848 0.27795 0.13589 0.02768 (c)(ii) ( )E 4.4 0.27795 10.736 0.13589 19.85984 0.02768 3.2316154 V = × +× + × = Alternatively, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 24 6 3 2 1 E 10 1.2 10 P 4 10 1.2 10 P 5 10 1.2 10 P 6 1.2 1 P 3 3.23159461r r VX X X Xr = =×− = + ×− = +× − = = − = +=∑ The expected return is ( )4000 3.2316154 12926.46 12926 (nearest dol lar)= = . (iii) Acceptable answer 1: Good investment as the expectation is higher than 10000. Acceptable answer 2:

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ___________________________________________ Y6 H2 Math Term 3 Revision Lecture: (C) Statistics Page 4 of 14 By GC, 4.6387546Vσ = The return 4000V has a standard deviation 4.6387546 4000 18555 (nearest dollar)×= Risky investment as the standard deviation (or variance) is large (in comparison / relative to the expectation). Acceptable answer 3: Calculate the expectation of using $10000 to buy the stock directly. ( ) ( ) 6 26 0 10000 1.2 1 P 12293 (nearest dollar)r r Xr− = −= =∑ Good investment as the expectation is higher than buying the stock directly. Source of Question: NJC Prelim 9758/2022/02/Q7 2 On average, 7.8% of markers produced by a factory are faulty. Each day, the factory’s quality manager picks a random sample of n markers and inspects the number of faulty markers found in the sample. The number of faulty markers found in the sample is the random variable X and whether the markers are faulty are independent of one another. (i) Explain what is meant by a random sample in this context. [1] (ii) Given that the probability that more than ( 4)n− markers in the random sample are found to be non-faulty is less than 0.3, find the least possible value of n. [2] Assume for the remainder of this question that n = 10. The samples in 50 randomly chosen days are collected and the number of faulty markers in each sample is recorded. (iii) Find the probability that exactly 35 of the samples each contains at least one faulty marker, given that none of the 50 samples contain more than 2 faulty markers each. [4] Solution: 2 (i) A random sample is one such that every marker produced has equal probability of being selected in this sample and the markers are selected independently of one another. (ii) Let X be the number of markers, out of n, that are faulty. ( )~ B ,0.078Xn If n – X markers are not faulty, X markers are faulty. Thus, ( ) ( ) ( ) P 4 0.3 P 4 0.3 P 3 0.3 nX n X X − >− < << ≤<

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ___________________________________________ Y6 H2 Math Term 3 Revision Lecture: (C) Statistics Page 5 of 14 Using GC, ( ) ( ) 60, P 3 0.302 0.3 61, P 3 0.290 0.3 nX nX = ≤= > = ≤= < Least value of n is 61. Alternatively, we can define W to be the number of markers, out of n, that are NOT faulty. Then ( ) ( )~ B ,1 0.078 B ,0.922Wn n −= . ( ) ( ) ( ) P 4 0.3 1 4 0.3 P 4 0.7 Wn PW n Wn >− < − ≤− < ≤− > Using GC, When n = 60, ( )P 4 0.698 0.7Wn≤− = < When n = 61, ( )P 4 0.7103 0.7Wn≤− = > Hence least n is 61. (iii) Let the number of faulty markers found in the sample of 10 be Y. ( )~ B 10,0.078Y ( )P 0 0.4439246Y = = ( )P 2 0.962450Y ≤= ( ) ( ) ( )P 1 2 P 2 P 0 0.518526Y YY≤≤ = ≤− == Thus, the required probability is P(all 50 samples each have less than or equal to 2 faulty markers and