RI S4 Sampling_Add Prac_Soln

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ______________________________________________ Additional Practice Questions for Chapter S4: Sampling Page 1 of 13 Additional Practice Questions for Chapter S4: Sampling (Solutions) 1 SAJC Prelim 9233/2005/02/Q28 A manufacturer of candles claimed that it produced birthday candles with a mean burning time of 6 minutes. A random sample of 150 birthday candles was tested and the burning times, X minutes, were summarized by 120)5(  x and 2( 5) 638.x  Calculate the unbiased estimates for the mean  and variance 2. [3] Solution Let 5.y x  Then 120y , 2 638y  . 1 120 0.8150y yn   Now 5y x  x  5 5.8y    2 2 2 2 2 2 1 1 1 120638149 150 3.64 (3 s.f.) Unbiased estimate for and are 5.8 and 3.64 (3 s.f.) x y yyn ns s                  

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ______________________________________________ Additional Practice Questions for Chapter S4: Sampling Page 2 of 13 2 ACJC Prelim 9233/2005/02/Q27 Two firms A and B manufacture similar components with mean breaking strengths of 6 units and 5.5 units, and standard deviations 0.4 units and 0.25 units respectively. It is given that both distributions are normal. If random samples of 100 components from firm A and 50 from firm B are tested, find the probability that the mean breaking strength of the sample from firm A minus that from firm B will be between 0.45 and 0.55 units. [3] Solution Let X and Y be the breaking strengths of a random component manufactured by firms A and B respectively. Then  2~ N 6, 0.4X and 2~ N 5.5, 0.25Y . i.e. 20.4~ N 6,100X     and 20.25~ N 5.5,50Y     . 2 20.4 0.25~ N 0.5,100 50X Y       i.e.  ~ N 0.5, 0.00285X Y  P 0.45 0.55 0.651X Y    (3 s.f.) 3 HCI Prelim 9233/2005/02/Q24 The life, in hours, of a randomly chosen light bulb produced by a manufacturer is normally distributed with mean 1100 and standard deviation 70. How large a sample is required such that the probability that the mean life in the sample shall exceed 1120 is not more than 5%? [5] Do you need to use Central Limit Theorem in your working? Explain. [1] Solution Let X be the life, in hours, of a randomly chosen light bulb.  2~ N 1100, 70X . Let the sample size be n. Then 270~ N 1100,X n     P 1120 0.05X   1120 1100P 0.05 70Z n       2P 0.05 7 nZ      From GC,  P 1.6449 0.05Z   2 1.64497 n  33.14n   Least 34n It is not necessary to use the Central Limit Theorem since X is normally distributed. 1.6449 0 0.05

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________