2023 S6 Sampling and Estimation APQ Solutions (RVHS)

Statistics 6 Tutorial: Sampling Additional Practice Questions 1. [N05/P2/26] In a country, 75% of the population have height exceeding 1.50m and 10% have height exceeding 1.90m. Assuming a normal distribution of heights, show that the height exceeded by 20% of the population is 1.81m, correct to 3 significant figures. A random sample of 80 people is taken from the population. Find the probability that the sample mean exceeds 1.69m. Solution Let X be the r.v. that denotes the height of a randomly chosen person from the population. ( ) 2~,XN  Want to find a when ( ) 0.20P X a= ( ) 0.80 1.809811542 1.81 P X a a = = (note: answer should be 0.0111) 2. [UCLES] The random variable X has a normal distribution with mean 2 and variance 3. The random variable S is the sum of 200 independent observations of X, and the random variable T is the sum of a further 300 independent observations of X. Find (i) , (ii) , (iii) The random variable N is the sum of n independent observations of X. Find the approximate values of and as n becomes very large. ( )1.50 0.75 & ( 1.9) 0.1 1.5 1.9( ) 0.25 & ( ) 0.9 1.5 1.9 0.6745 & 1.282 1.6377 & 0.2045 P X P X P Z P Z       =  = −− =  = −− =− = == 20.2045~ 1.6377, 80XN   ( )1.69 0.111PX = ( )P 405S  ( )P 3 40 2ST+ ( )P 2.5 ,Nn ( )P 1.7Nn ( )P 1.7 2.5n N n

Solution (i) ( )1 200... ~ 400,600S X X N= + + ( 405) 0.419PS = (ii) 3 ~ (1200,5400) 2 ~ (1200,3600) 3 2 ~ (0,9000) S N T N S T N− (3 2 40) 0.663P S T−  = (iii) ~ (2 ,3 ) ( 2.5 ) 0.5( ) 0 as 3 N N n n P N n nP Z n  =  → → ( 1.7 ) 0.3( ) 0 as 3 P N n nP Z n  −=  → → (1.7 2.5 ) 0.3 0.5( ) 1 as 33 P n N n nnP Z n  −=   → → 3. [2011 RI/II/12(modified)] The weight, x kg, of each student in a random sample of 120 students from a secondary school is measured, and the results are summarized by , (i) Find unbiased estimates of the population mean and variance. (ii) Another random sample of n students (n ≥ 50) is taken from the school. Given that the probability of the sample mean weight exceeding 49.9 kg is at most 0.01, find the least value of n. Solution (i) Unbiased estimate of population mean = Unbiased estimate of population variance ( 50) 100x− =− 2( 50) 1158x−= ( 50) 10050 50120 149 or 49.167 (5 s.f ) 49.2 (3 s.f )6 x n − −+ = + ==  ( ) 2 2 2 ( 50)1 ( 50)1 1 ( 100) (1158 ) 9.0308 9.03 (3 s.f )119 120 x xnn  −= − −−  −= −  = 

(ii) Since n is large, by Central Limit Theorem, Method 2 : Using table of values Key in Y1 = normalcdf(49.9, E99, 49.167, )  Least value of n = 91 9.0308(49.167, ) approximately.XN n ( 49.9) 0.01 ( 49.9) 0.99 49.9 49.167( ) 0.99 9.0308 49.9 49.167From , 2.3263 9.0308 90.960 Least value of 91 Method 1 : Using Standardization PX PX PZ n GC n n n   − −   = 9.0308 /X n 90 0.01033 > 0.01 91 0.00999 < 0.01 92 0.00965 < 0.01 ( 49.9)PX 

4. A population has variance 4. Given that the probability that X differs from the population mean  by less than 0.1 is at least 0.95, find the least value of the sample size required. Solution Let n denote the sample size. X ~    n 4 ,N  , by CLT, assuming n is large. ( ) 95.01.0P −X ( ) 95.01.01.0P −− X ( ) 95.01.01.0P ++−  X 95.0 4 1.0 4 1.0P            − n Z n ( ) 025.005.0P − nZ 959963986.105.0 −− n 19927972.39n 583531.1536n So, least value of n = 1537. 5. In the study of the sodium content, x, of a cereal, a random sample of 50 similar servings of Yummy Bites were taken. The following are the measurements of x (in milligrams): and . (i) Find the unbiased estimates of the mean and variance of X. (ii) State the distribution of , the mean mass of 50 servings and any assumptions made. Hence find the least value of a such that . Solution (i) ( ) 1600190 =− x ( ) 65000190 2 =− x X ( )P 0.9Xa 190Let y x=− 190 190 1600 190 22250 yx xy x =− =+ = + = 2 2 1 (1600) 13800[65000 ] 281.6349 50 49s = − = =

(ii) approx by CLT since n=50 is large ( ) ( ) P 0.9 P 0.1 222P 0.1 281.63 50 222 1.28156 281.63 50 218.96 219 Xa Xa aZ a a Least a     −    − −  = 6. [N03/P2/27] The random variable X has the distribution N(1, 20). (i) Given that , find a. (ii) A random sample of n observations of X is taken. Given that the probability that the sample mean exceeds 1.5 is at most 0.01, find the set of possible values of n. Solution (i) 3P( ) 2 2P( ) 3 2.93 Xa Xa a = = = (ii) 20~ N 1, P( 1.5) 0.01 P( 1.5) 0.99 X n X X     20 20 1.5 1P 0.99 1.5 1 2.326 n n Z  − −  0.5 20 2.326n n ≥ 433 The set of values of n is {n: n is an integer, n ≥ 433} 281.63~ N 222, 50X   ( ) ( )P 2PX a X a =  ( ) 2 ( )P X a P X a = 

7. The random variable X has mean  and variance 9. A random sample of size n (n > 50) is taken from the population and the sample mean is denoted by . Find the least value of n such that P(| | < 0.5) > 0.96. Solution Assumptions: n is large, n > 50 or X follows a normal distribution By CLT, ~ N(, ) approximately P(| − | < 0.5) > 0.95  P(−0.5 < −  < 0.5) > 0.95  P( < Z < ) > 0.95  > 1.96  n > 138.3  Least value of n = 139 8. 2021 Prelim/ACJC/H2/P2/Q9 [Modified] The Particle Filtration Efficiency (PFE) of a mask is a measure of how well a mask filters airborne particles such as pollen or dust. A mask with higher PFE is deemed to be of better efficiency as it filters more particles. A mask with a PFE of 95% would h ave met the requirement for surgical masks. A factory manufactures Brand BEY surgical masks that is known to have expected PFE of 95.8%. During a routine check of the manufacturing process, the quality control manager suspects that the efficiency of the Brand BEY surgical masks produced is compromised such that the mean PFE is reduced. The PFE, x%, of a random sample of 50 masks is taken and the summarised results are as follows. ( 90) 289x−= 2( 90) 1670.56x−= (i) State what it means for a sample to be random in this context. [1] The manager carries out a test on the PFE of the Brand BEY masks, which requires the distribution of the sample mean. (ii) Calculate the unbiased estimates of the population mean and variance for the PFE of Brand BEY masks. [2] (iii) State the distribution of the sample mean and explain whether there is a need for the manager to make any assumption about the population distribution of the PFE of the masks. [1] X X − X n 9 X X 3 5.0 n− 3 5.0 n 3 5.0 n

Solution: 8(i) A sample is random if every Brand BEY surgical mask manufactured has an equal chance of being selected to be in the sample of the 50 taken and that the selections are independent of each other. Or A sample is random if every subset of n of Brand BEY surgical masks has an equal chance of being in the sample of the 50 taken. (ii) Let X be random variable for the PFE of Brand BEY masks. Unbiased estimate of population mean, 289 90 95.7850x= + = Unbiased estimate of population variance, ( ) 2 2 2891 1670.56 0.002857149 50s  = − =  . (iii) As 50n= is large, by Central Limit Theorem, 0.0028571~ (95.78, ) approx.50 i.e. ~ (95.78,0.000051742) XN XN There is no need for any assumptions to be made about the population distribution of PFE of masks since n = 50 is large, by central limit theorem, the sample mean PFE will follow a normal distribution approximately.

9. 2020 Prelim/NYJC/H2/P2/Q8(Part of) A manufacturer produces teacups, saucers and plates. Past records indicate that on average 5% of the teacups, 2% of the saucers and 1% of the plates produced are flawed. The quality of the teacups, saucers and pl