RI S4 Sampling Tutorial Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ___________________ Tutorial S4: Sampling Page 1 of 10 Tutorial S4: Sampling 1 VJC Prelim 9740/2014/02/5 1st part There were 500 spectators seated at the grandstand of a stadium. A surveyor took the seat plan of the grandstand and threw a die on the seat plan 20 times and interviewed the spectators at the seats on which the die landed. Give a reason why this might not result in a random sample of spectators. [1] Solution The surveyor is likely to avoid throwing at the edge of the map. Hence not all the 500 seats have an equal chance of being selected. Alternative Solution After a seat is selected, the surveyor is likely to avoid the nearby seats. Hence the selections are not independent of one another. 2 9758/2019/02/Q6 In a certain country there are 100 professional football clubs, arranged in 4 divisions. There are 22 clubs in Division One, 24 in Division Two, 26 in Division Three and 28 in Division Four. (i) Alice wishes to find out about approaches to training by clubs in Division One, so she sends a questionnaire to the 22 clubs in Division One. Explain whether these 22 clubs form a sample or a population. [1] (ii) Dilip wishes to investigate the facilities for supporters at the football clubs. but does not want to obtain the detailed information necessary from all 100 clubs. Explain how he should carry out his investigation, and why he should do the investigation in this way. [2] (iii) Find the number of different possible samples of 20 clubs, with 5 clubs chosen from each division. [3] Solution (i) These 22 clubs form the population as they are ALL the clubs in Division One whose approaches to training she is interested to find out. (ii) How Assuming no special treatment with regard to facilities for supporters of different divisions, he could randomly pick a certain number of clubs out of the 100, say 10, to do a thorough investigation. Why This is to avoid bias and it would also be more cost effective and manageable. (iii) Number of different possible samples = 22 24 26 28 18 5 5 5 5 7.24 10C C C C    

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________ Tutorial S4: Sampling Page 2 of 10 3 In each of the following cases, find unbiased estimates of the population mean and population variance of X. (a) sample size = 100, 160x , 2 265x  , (b) sample size = 150, 154 150x  ,   2154 11000,x  (c) sample size = 50, ( 10) 328x  , 2 16062x  , (d) sample size = 20, 1100x , sample variance = 10. [(a) 8 5 , 1 11 (b) 155, 10850 149 (c) 414 25, 8394 175 (d) 55, 200 19 ] Solution An unbiased estimate of population mean An unbiased estimate of population variance (a) 8 1.6100 5 xx   2 2 1 (160) 126599 100 11s       (b) Let 154y x  Then 150 1150 150 yy   154 154 155y x x y      An unbiased estimate of population mean is 155  2 2 2 2 1 1 1 (150)11000149 150 10850 72.8 (3 sf).149 y ys y n n                 2 2 y xs s So an unbiased estimate of population variance is 10850 or 72.8 (3 sf).149 Recall: If y = ax+b, then y ax b  and 2 2 2 y xs a s (c) Let 10y x  Then 328 164 50 50 25 yy   41410 10 25y x x y      An unbiased estimate of population mean is 414 or 16.5625    2 2 2 2 2 2 1 1 1 1 1 16062 50(16.56)49 8394 175 x xs xn n x nxn                    

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________ Tutorial S4: Sampling Page 3 of 10 (d) 5520 xx  2 (sample variance)1 ns n  20(10)19 200 19   Note : Sample variance =   221 1 x xn n       and   22 2 1 1 1s x xn n         so an unbiased estimate of population variance is (Sample variance)1 n n 4 The amount, x mg, of Vitamin B2 in a packet of cereals was measured. 10 packets of cereals were taken and the following data were obtained: 272, 285, 278, 293, 298, 283, 279, 281, 295, 271 Find unbiased estimates of the population mean and population variance of Vitamin B2 in packets of the cereal. [283.5, 86.7] Solution An unbiased estimate of the population mean of Vitamin B2 in a packet of cereal is 283.5mg. A unbiased estimate of the population variance of Vitamin B2 in a packet of cereal is s2 (9.3125)2mg2 = 86.7 mg2 (3 s.f.)

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________ Tutorial S4: Sampling Page 4 of 10 5 A sample of 30 salesmen was surveyed and the ages of each of the salesman are given as follows: Age 21 22 23 24 25 26 27 28 Frequency 3 3 4 6 7 4 2 1 Calculate unbiased estimates for the population mean  and population variance 2. [24.2, 3.41] Solution Unbiased estimates of mean  and variance 2 are 24.2 and (1.8458)2  3.41 6 Given that the random variables X and Y have distributions N(30,18) and N(20,16) respectively, and that X and Y denote the means of 15 independent observations of X and 8 independent observations of Y respectively. State the sampling distributions of (i) X Y , (ii) 5 3X Y , (iii) 4 2X Y , (iv) 1 2 15 1 2 8 23 X X X Y Y Y        . Solution Given ~ N(30,18)X and ~ N(20,16)Y with 15xn  and 8yn  Then, 18~ N 30, 15X     and 16~ N 20, 8Y     i.e., ~ N 30, 1.2X and ~ N 20, 2Y (i) E E E 30 20 10X Y X Y      Var Var +Var 1.2 2 3.2X Y X Y     ~ N 10, 3.2X Y (ii) E 5 3 5E 3E 5(30) 3(20) 210X Y X Y      2 2 2 2Var 5 3 5 Var +3 Var 5 (1.2) 3 (2) 48X Y X Y     5 3 ~ N 210, 48X Y

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________ Tutorial S4: Sampling Page 5 of 10 (iii) E 4 2 4E 2E 4(30) 2(20) 80X Y X Y      2 2 2 2Var 4 2 4 Var +2 Var 2 (1.2) 2 (2) 27.2X Y X Y     4 2 ~ N 80, 27.2X Y (iv) 1 2 15 1 2 8Let = 23 X X X Y Y YT            2 1 610E( ) 15E( ) 8E( ) = 23 23 1 398Var( ) 15Var( ) 8Var( )23 529 610 398~ N ,23 529 T X Y T X Y T             7 A large number of samples of size n are taken from a normal population with mean 74 and standard deviation 6. It was found that 28.2% of the samples have sample mean that exceeds 75. Estimate the value of n. [12] Solution Let X be a random variable taken from the normal population with mean 74 and standard deviation 6. Then X ~ N(74, 62) 36~ N(74, )X n P( 75) 0.282 75 74P 0.2826 P 0. From GC, 2826 P( 0.57691) 0.282 0.576916 11.982 12 X Z n nZ Z n n n                          

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ ________________