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 __________________________________________________________________________________________ ______________________________________________ Additional Practice Questions for Chapter S4: Sampling Page 3 of 13 4 AJC Prelim 9233/2003/02/Q29 Or (part) The times taken by two runners A and B in a 400 metre race are independent and may be assumed to be normally distributed. The times (in seconds) taken by A has mean 46 and standard deviation 0.5 and the times taken by B has mean 46.2 and standard deviation 0.8. LetX denote the average time for A to run the 400 metre track on 10 different occasions and Y denotes the average time for B to run the same track on 16 different occasions. Find the value of t such that P 0.2 0.1X Y t    . [4] Solution Since 20.5~ N 46,10X     and 20.8~ N 46.2,16Y     ,  ~ N 0.2, 0.065X Y  . P 0.2 0.1X Y t     Since 0.2 ~ N 0,0.065 ,X Y  0.419t  (3 s.f.) GC screenshots for reference: Use center tail : OR : Use right tail 0.9 0.05 0.419 0 0.419

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ______________________________________________ Additional Practice Questions for Chapter S4: Sampling Page 4 of 13 5 TJC Prelim 9233/96/02/Q9 In a certain examination with a very large entry, the marks obtained by the male candidates were found to follow a normal distribution with a mean of 54 and a standard deviation of 16.X denotes the mean mark scored by a sample of 4 male candidates. (i) State the sampling distribution of X. (ii) Find the probability that X will exceed 70. (iii) Given that there is a probability of 0.95 that X differs from its mean mark by less than c, find the value of c. In the same examination the marks obtained by the female candidates were found to follow an independent normal distribution with a mean of 59 and a standard deviation of 20. (iv) Find the probability that the total marks obtained by a randomly chosen male candidate and a randomly chosen female candidate exceed 120. It is given that Y denotes the mean mark scored by a random sample of 5 female candidates. (v) State the sampling distribution of Y X . (vi) Find P(Y >X). Solution Let X be the mark obtained by a male candidate in a certain exam. Then 2~ N(54,16 )X (i) 216~ N 54, 4X     i.e  2~ N 54, 8X (ii) P( 70) 0.0228X  (3 s.f) (iii) P( 54 ) 0.95X c   P( 54 ) 0.95c X c     P(54 54 ) 0.95c X c     From GC, P(38.320 69.680) 0.95X   54 69.680c   (5 s.f.) 15.7c (3 s.f.) [ Use InvNorm(0.95, 54, 8, CENTRE) ] OR : P( 54 ) 0.95X c   54P 0.958 8 X c      P 0.958 P 0.958 8 cZ c cZ             From GC :  P 1.960 1.960 0.95Z    [ Use InvNorm(0.95,0,1,CENTRE) ] 38.320 54 69.680 -1.960 0 1.960

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ______________________________________________ Additional Practice Questions for Chapter S4: Sampling Page 5 of 13 1.9608 15.7 (3 s.f.) c c   Let Y be the marks obtained by a female candidate in the same exam. 2~ N(59, 20 )Y (iv) P( 120) 0.392X Y   (3 s.f.) where ~ N(113, 656)X Y (v) 220~ N 59, 5Y     i.e  ~ N 59, 80Y Then  ~ N 5, 144Y X (vi) P( )Y X P( 0) 0.662Y X    (3 s.f.) 6 CJC Prelim 9740/2012/02/Q11 A sample of n observations was taken from a population of mean and variance 10. Find the least sample size required such that the probability of the sample mean lying between 5.0 and 5.0 is more than 0.89. State an assumption that you have to make in order to proceed with this calculation. [6] Solution Let X be the random variable. Assume that n is sufficiently large for Central Limit Theorem to hold. Then, )10,(~ nNX  approximately P( 0.5 0.5) 0.89X      0.5 0.5P 0.8910 10Z n n                  P 0.5 0.5 0.8910 10 n nZ        From GC,  P 1.59819 1.59819 0.89Z    From the diagram, 0.5 1.5981910 n  19638.310n 21685.1010n 1685.102n Least n = 103 1.59819 0 1.59819

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ______________________________________________ Additional Practice Questions for Chapter S4: Sampling P